A simple problem in physics.

[Marvin Sebourn]

We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.

Both ice and water are in separate, perfectly insulated containers with no void for air.

The ice and the water are then combined into one single, perfectly insulated container with no void for air.

Given:

The specific heat of water = 1 calorie per gram

The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.

Atmospheric pressure is normal, and constant at 14.7 psi.

When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?

Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

Thanks
Marvin Sebourn
***@aol.com

[Olrik]

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

Post by Marvin Sebourn
Thanks
Marvin Sebourn

[Marvin Sebourn]

Post by Olrik

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

"I like that answer very much, but it depresses me, as any answer would", said Marvin the Paranoid Android.

Marvin

Post by Olrik

Post by Marvin Sebourn
Thanks
Marvin Sebourn

--
Olrik
aa #1981
EAC Chief Food Inspector, Bacon Division

[Intelligence]

Post by Marvin Sebourn

Post by Olrik

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

"I like that answer very much, but it depresses me, as any answer would", said Marvin the Paranoid Android.
Marvin

Post by Olrik

Post by Marvin Sebourn
Thanks
Marvin Sebourn

--
Olrik
aa #1981
EAC Chief Food Inspector, Bacon Division

marvin was depressed, you flunked

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by Olrik

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

"I like that answer very much, but it depresses me, as any answer would", said Marvin the Paranoid Android.
Marvin

Post by Olrik

Post by Marvin Sebourn
Thanks
Marvin Sebourn

--
Olrik
aa #1981
EAC Chief Food Inspector, Bacon Division

marvin was depressed, you flunked

Marvin tPA was chronically depressed, stupid. Olrik's answer was among the best, and most clever. High marks!

Olrik's answer reminds me of a passage (p248) from "The Ultimate Hitchiker's Guide to the Galaxy", with Marvin in conversation with-guess who?

“You mean,” said Intelligence, “you mean you can see into my mind?”
“Yes,” said Marvin.
Intelligence stared in astonishment. “And…?” he said.
Marvin: “It amazes me how you can manage to live in anything that small.”

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Olrik

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

"I like that answer very much, but it depresses me, as any answer would", said Marvin the Paranoid Android.
Marvin

Post by Olrik

Post by Marvin Sebourn
Thanks
Marvin Sebourn

--
Olrik
aa #1981
EAC Chief Food Inspector, Bacon Division

you were doing marvin the paranoid

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by Olrik

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

"I like that answer very much, but it depresses me, as any answer would", said Marvin the Paranoid Android.
Marvin

Post by Olrik

Post by Marvin Sebourn
Thanks
Marvin Sebourn

--
Olrik
aa #1981
EAC Chief Food Inspector, Bacon Division

you were doing marvin the paranoid

Remarkable how you realized that within 24 hours and a dozen (more or less) postings. And how even then you couldn't write his whole name.


Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Olrik

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

"I like that answer very much, but it depresses me, as any answer would", said Marvin the Paranoid Android.
Marvin

Post by Olrik

Post by Marvin Sebourn
Thanks
Marvin Sebourn

--
Olrik
aa #1981
EAC Chief Food Inspector, Bacon Division

you were doing marvin the paranoid

Remarkable how you realized that within 24 hours and a dozen (more or less) postings. And how even then you couldn't write his whole name.

Remarkable how it took you less than a day to be caught lying, as usual

[Intelligence]

Post by Olrik

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.

That would be 42.

Post by Marvin Sebourn
Thanks
Marvin Sebourn

--
Olrik
aa #1981
EAC Chief Food Inspector, Bacon Division

You flunked

[mr_antone]

On Sun, 10 Feb 2019 19:21:38 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

32 degrees


mr_antone

[Intelligence]

Post by mr_antone
On Sun, 10 Feb 2019 19:21:38 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

32 degrees
mr_antone

you flunked

[Peter Pan]

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.

Duke has the *real* answer sitting by his computer.

[Intelligence]

0° C.

You should not have blurted it out

better question is how long it took

[Marvin Sebourn]

Post by Intelligence

0° C.

You should not have blurted it out

I asked for the numerical answer, which we received from mr_antone and Peter Pan. An unanswered question wouldn't have any point, would it?

I bet it took very little time for either of them to find and answer. It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Post by Intelligence
better question is how long it took

Better question is why you are here.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

0° C.

You should not have blurted it out

I asked for the numerical answer, which we received from mr_antone and Peter Pan. An unanswered question wouldn't have any point, would it?

Show the work involved

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

To point out that you are a fraud

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.
Marvin Sebourn

"Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
"

[Don Martin]

On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

[Intelligence]

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn
I await this self-proclaimed genius

If you are trying to refer to me, I certainly have never claimed anything like that. I have claimed I have received IQ tests, and that the results consistently in my life are above average.
I will chalk it up to further lying and deception on your part, which is fairly typical for you.

[Christopher A. Lee]

On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.

All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

[t***@gmail.com]

Post by Christopher A. Lee
You don't even need to do any math.

If the amount of ice and water were different, to where the answer is non zero temperature, you would. Now do the same math calculations, which shows you know how to do these type problems, despite the easy answer to bypass showing the math involved.

[Don Martin]

On Tue, 12 Feb 2019 14:49:22 -0600, Christopher A. Lee

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I am not going to unbate me breath until this numbskull weighs in.

[Marvin Sebourn]

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?

Thank you,

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

how long is a typical class period at school, which is how long somebody would have for a test, obviously ...

[Christopher A. Lee]

On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

[Intelligence]

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

[Marvin Sebourn]

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher may have an occasional "eye issue", but he never shows lack of courage, lack of truthfulness, or lack of sound reasoning. These laudable characteristics are highly desirable, especially when dealing with someone who lacks courage, truthfulness, or reasoning, such as...

Christopher had the courage to put forth an answer, unlike certain others here. And also the courage to admit a mistake.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher never shows lack of truthfulness

that is not even his name

[Marvin Sebourn]

Post by Intelligence

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher never shows lack of truthfulness

that is not even his name

A couple of years ago he was Christopher. Today he is Christopher. Next year he'll likely be Christopher.

BUT-what about you?
This minute you're one nym, next minute another.

And speaking of truthfulness, your use of the nym "Intelligence" is almost a criminal offense. Hardly fits. But it's laughable.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher never shows lack of truthfulness

that is not even his name

A couple of years ago he was Christopher. Today he is Christopher. Next year he'll likely be Christopher.

I do not know what jehovah's timeline is for you, but once you are laid to rest, all your sockpuppets go with it, wouldn't you say?

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher never shows lack of truthfulness

that is not even his name

A couple of years ago he was Christopher. Today he is Christopher. Next year he'll likely be Christopher.

I do not know what jehovah's timeline is for you, but once you are laid to rest, all your sockpuppets go with it, wouldn't you say?

I have no sockpuppets. How about you?

This "discussion" is boring me. I don't think that our interchanges bring you any credit. Time to bring down the curtain on your show. Responses such as your "how long is a typical class period at school, which is how long somebody would have for a test, obviously" show the lack of any argument from you. Your reply is not only way off topic, it is irrational.

Please list my sockpuppets here:

Or anywhere if you reply.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher never shows lack of truthfulness

that is not even his name

A couple of years ago he was Christopher. Today he is Christopher. Next year he'll likely be Christopher.

I do not know what jehovah's timeline is for you, but once you are laid to rest, all your sockpuppets go with it, wouldn't you say?

I have no sockpuppets. How about you?

Since you are no good at math anyway, what good would it do to tell you how many?

[Don Martin]

On Tue, 12 Feb 2019 17:54:21 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher never shows lack of truthfulness

that is not even his name

A couple of years ago he was Christopher. Today he is Christopher. Next year he'll likely be Christopher.
BUT-what about you?
This minute you're one nym, next minute another.
And speaking of truthfulness, your use of the nym "Intelligence" is almost a criminal offense. Hardly fits. But it's laughable.

Oxymorons pretty much always are.

[Christopher A. Lee]

On Tue, 12 Feb 2019 17:01:45 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Christopher A. Lee
On Tue, 12 Feb 2019 15:55:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Christopher A. Lee
On Tue, 12 Feb 2019 08:13:03 -0500, Don Martin

Post by Don Martin
On Mon, 11 Feb 2019 18:27:11 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence
better question is how long it took

Better question is why you are here.
Marvin Sebourn

Intelligence: (chortle)

Post by Intelligence
To point out that you are a fraud

Go ahead, prove it. Show your work.

I await this self-proclaimed genius' demonstration with bated breath.

You don't even need to do any math. The containers are perfect
insulators so there's nowhere for the heat to go.
All the "givens" obfuscate the problem. Which makes it a bit like the
Car Talk Puzzlers.

I'm obviously missing something in what you say, Christopher, so I'll ask for clarification. On the given "The specific heat of water = 1 calorie per gram", and "The heat of fusion of H2O = 80 calories per gram", if the ratio of the heat of fusion to the specific heat were different, then we would have a different result, so how is the original given an obfuscation? Although not listed as a given but considered by some as such, if we changed the ratio of the mass of water to the mass of ice without another change, how could the result be the same?
Thank you,
Marvin Sebourn

I mis-read the problem/ My aging eyes read the water as being 32
degrees as well as the ice.

so the people reading here can disregard your statement
hence you never should have clicked send
therefore your answer if you gave one or hinted at one, even if correct, is wrong, as you derived it improperly, tis why people need to show their work, lest someone with eyes issues such as yourself ...

Christopher may have an occasional "eye issue", but he never shows lack of courage, lack of truthfulness, or lack of sound reasoning. These laudable characteristics are highly desirable, especially when dealing with someone who lacks courage, truthfulness, or reasoning, such as...
Christopher had the courage to put forth an answer, unlike certain others here. And also the courage to admit a mistake.
Marvin Sebourn

Which asshole was this one?

[Intelligence]

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

[Marvin Sebourn]

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus. Post your resume' here, or CV. Let's see what you've done in life besides lie, hide and insult your betters, which means practically everybody.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

I could ask a child to count to 50, and it would take less time to solve the very simple problem. If you can't imagine a very simple solution, you have no feel for the problem.


So in speaking of your experience, why don't you show your resume' or your CV? Let's see what you've done in life besides insult people and play like you are superior to others here.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

I could ask a child

At least you are consistent.

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

I could ask a child

At least you are consistent.

Yes, I consistently ask the child, but again and again you refuse to answer. The child hides from posts by snipping them.

So I'll ask again: to dazzle us with your experience and expertise, why don't you post your resume' or your CV? What have you done in your life besides insult people and play superior, as an undisciplined child would do? Particulars?

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

I could ask a child

At least you are consistent.

Yes, I consistently ask the child

You have been harrassing some kid over your inability to be able to demonstrate knowledge you feign to possess?

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

I could ask a child

At least you are consistent.

Yes, I consistently ask the child

You have been harrassing some kid over your inability to be able to demonstrate knowledge you feign to possess?

No, I've been bothering some child who has shown absolutely nothing in the way of his purported knowledge and is unwilling to give even the simplest of his bona fides. Maybe his multiple roles as a sock puppy have him confused as to his identity. Possibly he is lined up to appear on Dr. Phil.

Where is your promised list of my sockpuppets, sockpuppy?

And finally, if you treat me with respect, I will reciprocate. You're not making any points, and I am boring people, even myself.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

I could ask a child

At least you are consistent.

Yes, I consistently ask the child

You have been harrassing some kid over your inability to be able to demonstrate knowledge you feign to possess?

No

You have failed to demonstrate you can show your work, the calculations required, for problems like this.

[Marvin Sebourn]

Post by Intelligence

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
It's easily a work-in-your-head problem, when you think about it and have a little background. The next times, with simple numerical values, should be solved within seconds.

Prove it.

Focus.

You have failed to show you know how to do the calculations and work on a question you yourself put here, yet tried to insinuate you did.

I could ask a child

At least you are consistent.

Yes, I consistently ask the child

You have been harrassing some kid over your inability to be able to demonstrate knowledge you feign to possess?

No

You have failed to demonstrate you can show your work, the calculations required, for problems like this.

You know only that I have only not shown you, puppy, to your satisfaction, and you have no knowledge of what I have shown others, so your statement is ONLY true concerning you, and may be FALSE in all other respects. In any case, we do not know-you have not shown-that you are qualified to make a decision in this case. Or any other case. As your postings show up here, that inability appears quite possible. I’m not here to teach you elementary school multiplication.

You have shown absolutely nothing in the way of your purported knowledge and you are unwilling to furnish the simplest of your bona fides. Maybe your multiple roles as a sock puppy have you confused as to your identity. (”Well, self, it’s Friday evening. Which sockpuppy am I this hour? “Intelligence”, or was it "Big Orange Trump?”)

Why do you hide from my posts?

Where is your promised list of my sockpuppets, sockpuppy?

You’re boring. You only seem to post here to get a bit of an adrenaline rush. Your average response time to my posts has been about nine minutes, so we know it's a bated breath scenario for you, puppy. I’m finished giving adrenaline rushes to you, sockpuppy, and my amusement at your foolishness wanes. Your foolishness has overloaded the "foolishness" circuits. Saturated them.

Bye.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn
I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp

it came to -.2 degrees

[Peter Pan]

Post by Intelligence

I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

It's such a simple problem there is no need to
approximate anything. Yet you got it wrong.

[Intelligence]

Post by Peter Pan

Post by Intelligence

Post by Marvin Sebourn
I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

It's such a simple problem there is no need to
approximate anything.

It allows latitude.
Today I will show how to do it in an easier fashion, and the work involved, rather than the approx way I did last night; see, marvin wanted to copy off of me, so I waited aND WAITED and finally performed the approximation in a more difficult process then the now easier known to you way as you watched me; see, that is how you got your bachelor's degree, through academic dishonesty; which makes your credentials worthless; problem is, one needs a 4 year degree to be a federal agent, with a fraudulently obtained degree, you never should have been a federal agent;

[%]

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Marvin Sebourn
I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

It's such a simple problem there is no need to
approximate anything.

It allows latitude.
Today I will show how to do it in an easier fashion, and the work involved, rather than the approx way I did last night; see, marvin wanted to copy off of me, so I waited aND WAITED and finally performed the approximation in a more difficult process then the now easier known to you way as you watched me; see, that is how you got your bachelor's degree, through academic dishonesty; which makes your credentials worthless; problem is, one needs a 4 year degree to be a federal agent, with a fraudulently obtained degree, you never should have been a federal agent;

what should he be

[Intelligence]

Post by %

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Marvin Sebourn
I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

It's such a simple problem there is no need to
approximate anything.

It allows latitude.
Today I will show how to do it in an easier fashion, and the work involved, rather than the approx way I did last night; see, marvin wanted to copy off of me, so I waited aND WAITED and finally performed the approximation in a more difficult process then the now easier known to you way as you watched me; see, that is how you got your bachelor's degree, through academic dishonesty; which makes your credentials worthless; problem is, one needs a 4 year degree to be a federal agent, with a fraudulently obtained degree, you never should have been a federal agent;

what should he be

if I utilized a closer approximation, it would be closer to zero at -0.05 ...

[Peter Pan]

Post by Intelligence

Post by Peter Pan

Post by Intelligence

I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

It's such a simple problem there is no need to
approximate anything. Yet you got it wrong.

It allows latitude.
Today I will show how to do it in an easier fashion, and the work involved, rather than the approx way I did last night; see, marvin wanted to copy off of me, so I waited aND WAITED and finally performed the approximation in a more difficult process then the now easier known to you way as you watched me; see, that is how you got your bachelor's degree, through academic dishonesty; which makes your credentials worthless; problem is, one needs a 4 year degree to be a federal agent, with a fraudulently obtained degree, you never should have been a federal agent;

Too late. Once i'm a federal agent, it can't be undone.
You're busted. You're facing a major case of Fraudulent
Thermal Dynamics.

You say you're going to show us how to do the problem,
but you let your insanity derail you from your task.

[Intelligence]

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Marvin Sebourn
I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

It's such a simple problem there is no need to
approximate anything. Yet you got it wrong.

It allows latitude.
Today I will show how to do it in an easier fashion, and the work involved, rather than the approx way I did last night; see, marvin wanted to copy off of me, so I waited aND WAITED and finally performed the approximation in a more difficult process then the now easier known to you way as you watched me; see, that is how you got your bachelor's degree, through academic dishonesty; which makes your credentials worthless; problem is, one needs a 4 year degree to be a federal agent, with a fraudulently obtained degree, you never should have been a federal agent;

Too late. Once i'm a federal agent, it can't be undone.

20 year law for how long you can be a cop.

That ended for you last may.

Unfortunately some dumbass extended your career past the mandatory retirement

[Peter Pan]

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Intelligence

I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

It's such a simple problem there is no need to
approximate anything. Yet you got it wrong.

It allows latitude.
Today I will show how to do it in an easier fashion, and the work involved, rather than the approx way I did last night; see, marvin wanted to copy off of me, so I waited aND WAITED and finally performed the approximation in a more difficult process then the now easier known to you way as you watched me; see, that is how you got your bachelor's degree, through academic dishonesty; which makes your credentials worthless; problem is, one needs a 4 year degree to be a federal agent, with a fraudulently obtained degree, you never should have been a federal agent;

Too late. Once i'm a federal agent, it can't be undone.
You're busted. You're facing a major case of Fraudulent
Thermal Dynamics.

20 year law for how long you can be a cop.
That ended for you last may.
Unfortunately some dumbass extended your career past the mandatory retirement

No mandatory retirement age for me. It takes 20 years in
my specialty to learn how to handle troublemakers like
you.

Post by Intelligence

Post by Peter Pan
You say you're going to show us how to do the problem,
but you let your insanity derail you from your task.

So you don't know how to do the problem. You don't even
know where to start. Why would anyone want to copy your
work? Even duke knows better.

[Intelligence]

Post by Peter Pan
No mandatory retirement age for me. It takes 20 years in
my specialty to learn how to handle troublemakers like
you.

only in your fantasies, moron.

I merely lack having the money i need to do what I need to in life.

You couldn't stop a paper bag surrounding your breathing apparatus

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn
I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

Somehow you combined 10 grams of ice at 0 degrees C with 40 grams of water at 20 C and got -0.2 degrees for the resultant? Which would mean the water was converted to ice?

I did am (sic) approximation of your intelligence and found it to be about .2 standard deviations below the average. But I was being kind.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn
I’m not here to teach you elementary school multiplication.

I did am approximation of your problem and I got a freezing temp
it came to -.2 degrees

Somehow you combined 10 grams of ice at 0 degrees C with 40 grams of water at 20 C and got -0.2 degrees for the resultant?

So tell me, since everybody knows you word is worthless, and can't be trusted from one statement to the next, of more interest, is why your word is worthless, that should be more interesting ...

[Peter Pan]

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer? Did he skip this whole
exercise?

[Intelligence]

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

[Peter Pan]

Post by Intelligence

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

Did you read the instructions? Or you don't know what
"Numerical answers only, please" means.

Or did you just want to copy the solution.

[Intelligence]

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

Did you read the instructions? Or you don't know what
"Numerical answers only, please" means.
Or did you just want to copy the solution.

i don't think you can plug in the equations, and do the conversions and calculations and cancellations

[Marvin Sebourn]

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

Did you read the instructions? Or you don't know what
"Numerical answers only, please" means.
Or did you just want to copy the solution.

i don't think you can plug in the equations, and do the conversions and calculations and cancellations

So do it yourself. Impress us all.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

Did you read the instructions? Or you don't know what
"Numerical answers only, please" means.
Or did you just want to copy the solution.

i don't think you can plug in the equations, and do the conversions and calculations and cancellations

So do it yourself.

I have the equations for the problem.
If you would comprehend how to do those, I would give you one that is a non zero answer, but as of yet, I do not think you can do the calculations to begin with.

[Marvin Sebourn]

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

Did you read the instructions? Or you don't know what
"Numerical answers only, please" means.
Or did you just want to copy the solution.

i don't think you can plug in the equations, and do the conversions and calculations and cancellations

So do it yourself.

I have the equations for the problem.

Don't over-complicate the problem I put forth. If you do, it tells me you have no feel or grasp of a very simple situation.

Post by Intelligence
If you would comprehend how to do those, I would give you one that is a non zero answer, but as of yet,

But as of yet, you don't know whether I do or not, and you, suspecting that I do know, are afraid to give me even a simple problem. The problem I cited was very simple. Yet you haven't shown you know how to solve it.

Post by Intelligence
I do not think you can do the calculations to begin with.

That's only an assertion. You offer nothing of evidence to back it up. That's not unusual for you. That's a convenient way for you to back out.

And you said you are here to prove me a fraud.
Prove I am a fraud.
Provide evidence.
Show your work.

If you do come up with a problem, I anticipate you will try and complicate it as much as possible. Keep it as simple as I did, "Intelligence".

Marvin Sebourn
***@aol.com

[t***@gmail.com]

Your failure to show the work on the problem you presented
is the evidence of you being a fraud

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

Did you read the instructions? Or you don't know what
"Numerical answers only, please" means.
Or did you just want to copy the solution.

i don't think you can plug in the equations, and do the conversions and calculations and cancellations

So do it yourself.

I have the equations for the problem.

Don't over-complicate the problem I put forth. If you do, it tells me you have no feel or grasp of a very simple situation.

Post by Intelligence
If you would comprehend how to do those, I would give you one that is a non zero answer, but as of yet,

But as of yet, you don't know whether I do or not, and you, suspecting that I do know, are afraid to give me even a simple problem. The problem I cited was very simple. Yet you haven't shown you know how to solve it.

Post by Intelligence
I do not think you can do the calculations to begin with.

That's only an assertion. You offer nothing of evidence to back it up. That's not unusual for you. That's a convenient way for you to back out.
And you said you are here to prove me a fraud.
Prove I am a fraud.
Provide evidence.
Show your work.
If you do come up with a problem, I anticipate you will try and complicate it as much as possible. Keep it as simple as I did, "Intelligence".
Marvin Sebourn

[Marvin Sebourn]

On Tuesday, February 12, 2019 at 10:17:55 AM UTC-6,

Post by t***@gmail.com
Your failure to show the work on the problem you presented
is the evidence of you being a fraud

Wow, a top post which is not true, with the transparent aim of encouraging the reader to ignore my post! (Apologies for my top post reply, regular reply continued at end)

Post by t***@gmail.com

Post by Marvin Sebourn

Post by Intelligence

Post by Marvin Sebourn

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

Hey, where's MX's answer?

Are you gonna show your work?

Did you read the instructions? Or you don't know what
"Numerical answers only, please" means.
Or did you just want to copy the solution.

i don't think you can plug in the equations, and do the conversions and calculations and cancellations

So do it yourself.

I have the equations for the problem.

Don't over-complicate the problem I put forth. If you do, it tells me you have no feel or grasp of a very simple situation.

Post by Intelligence
If you would comprehend how to do those, I would give you one that is a non zero answer, but as of yet,

But as of yet, you don't know whether I do or not, and you, suspecting that I do know, are afraid to give me even a simple problem. The problem I cited was very simple. Yet you haven't shown you know how to solve it.

Post by Intelligence
I do not think you can do the calculations to begin with.

That's only an assertion. You offer nothing of evidence to back it up. That's not unusual for you. That's a convenient way for you to back out.
And you said you are here to prove me a fraud.
Prove I am a fraud.
Provide evidence.
Show your work.
If you do come up with a problem, I anticipate you will try and complicate it as much as possible. Keep it as simple as I did, "Intelligence".
Marvin Sebourn

Chameleon:
"Your failure to show your work..."

Show my work? You didn't even show your answer!

You're afraid of exposing yourself here.

Your change of garb reminds me of a chameleon. You both have a common characteristic. I'm not surprised since someone pointed out that both you and the chameleon are types of lizards.

And I add this seriously-the work on the problem is very basic. The only involved mass is H20. All figures are rounded for ease of computation. The individual masses are immaterial as long as we maintain the same ratio of water to ice at 4 to 1. If you look at it correctly, you can, mentally, determine the equilibrium temperature within fifteen seconds or so, "in your head".

So far you have proven nothing about me being a fraud, and shown nothing of your knowledge of the problem. And it's been nearly 48 hours for you to consider the problem. At least you are consistent in your failing.

But-I can be serious, and civil. Why don't you try that?

Marvin Sebourn
***@aol.com

[t***@gmail.com]

Post by Marvin Sebourn
On Tuesday, February 12, 2019 at 10:17:55 AM UTC-6,

Post by t***@gmail.com
Your failure to show the work on the problem you presented
is the evidence of you being a fraud

Wow, a top post which is not true

Where ever one replies to a post has zero bearing on the veracity of said post content.

You have failed to show you know how to do the calculations required for your original question, one in which you hinted within a day would be proffered.

[Marvin Sebourn]

Post by t***@gmail.com

Post by Marvin Sebourn
On Tuesday, February 12, 2019 at 10:17:55 AM UTC-6,

Post by t***@gmail.com
Your failure to show the work on the problem you presented
is the evidence of you being a fraud

Wow, a top post which is not true

Where ever one replies to a post has zero bearing on the veracity of said post content.

Except that a top post can detract attention from the preceding post, as the chameleon's did.

Post by t***@gmail.com
You have failed to show you know how to do the calculations required for your original question,

You have failed to show you know the answer to a question very basically framed.

Post by t***@gmail.com
one in which you hinted within a day would be proffered.

You have poor reading comprehension. I wrote: "Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February."

I requested that the explanation of how a solution was obtained not be posted until after about 24 hours.

Does what I wrote say that I would provide calculations after 24 hours? No.

Does what I wrote say that anyone else would provide calculations after 48 hours? No, of course not.

But if someone did explain the (quite simple) calculations I requested for them to wait approximately 24 hours.

Since you are so eager, I invite you to provide calculations. That is, of course, after you have shown the correct answer. Go. You have my permission. Do not pass GO, do not collect $200...

Marvin Sebourn
***@aol.com

[t***@gmail.com]

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
On Tuesday, February 12, 2019 at 10:17:55 AM UTC-6,

Post by t***@gmail.com
Your failure to show the work on the problem you presented
is the evidence of you being a fraud

Wow, a top post which is not true

Where ever one replies to a post has zero bearing on the veracity of said post content.

Except that a top post can detract attention from the preceding post, as the chameleon's did.

Post by t***@gmail.com
You have failed to show you know how to do the calculations required for your original question,

You have failed to show you know the answer to a question very basically framed.

Post by t***@gmail.com
one in which you hinted within a day would be proffered.

You have poor reading comprehension. I wrote: "Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February."
I requested that the explanation of how a solution was obtained not be posted until after about 24 hours.
Does what I wrote say that I would provide calculations after 24 hours? No.

Why don't you just say you couldn't show the work for the calculations?

Post by Marvin Sebourn
Does what I wrote say that anyone else would provide calculations after 48 hours? No, of course not.
But if someone did explain the (quite simple) calculations I requested for them to wait approximately 24 hours.

so you say it is simple

[Marvin Sebourn]

Post by t***@gmail.com

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
On Tuesday, February 12, 2019 at 10:17:55 AM UTC-6,

Post by t***@gmail.com
Your failure to show the work on the problem you presented
is the evidence of you being a fraud

Wow, a top post which is not true

Where ever one replies to a post has zero bearing on the veracity of said post content.

Except that a top post can detract attention from the preceding post, as the chameleon's did.

Post by t***@gmail.com
You have failed to show you know how to do the calculations required for your original question,

You have failed to show you know the answer to a question very basically framed.

Post by t***@gmail.com
one in which you hinted within a day would be proffered.

You have poor reading comprehension. I wrote: "Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February."
I requested that the explanation of how a solution was obtained not be posted until after about 24 hours.
Does what I wrote say that I would provide calculations after 24 hours? No.

Why don't you just say you couldn't show the work for the calculations?

Because what you suggest is not true. I can.

Because, relating to you, I adhere to the Pearls/Swine admonition.

Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
Does what I wrote say that anyone else would provide calculations after 48 hours? No, of course not.
But if someone did explain the (quite simple) calculations I requested for them to wait approximately 24 hours.

so you say it is simple

[t***@gmail.com]

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn

Post by t***@gmail.com

Post by Marvin Sebourn
On Tuesday, February 12, 2019 at 10:17:55 AM UTC-6,

Post by t***@gmail.com
Your failure to show the work on the problem you presented
is the evidence of you being a fraud

Wow, a top post which is not true

Where ever one replies to a post has zero bearing on the veracity of said post content.

Except that a top post can detract attention from the preceding post, as the chameleon's did.

Post by t***@gmail.com
You have failed to show you know how to do the calculations required for your original question,

You have failed to show you know the answer to a question very basically framed.

Post by t***@gmail.com
one in which you hinted within a day would be proffered.

You have poor reading comprehension. I wrote: "Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February."
I requested that the explanation of how a solution was obtained not be posted until after about 24 hours.
Does what I wrote say that I would provide calculations after 24 hours? No.

Why don't you just say you couldn't show the work for the calculations?

Because what you suggest is not true. I can.

Prove it.

[Cloud Hobbit]

Post by t***@gmail.com
You have failed to show you know how to do the calculations required for your original question, one in which you hinted within a day would be proffered.

And you have failed to provide an answer.

You should have received enough information about Marvin to know that he is quite capable of solving the problem in question.

You're just violating the one commandment.

Don't be a dick.

[Intelligence]

Post by Cloud Hobbit

Post by t***@gmail.com
You have failed to show you know how to do the calculations required for your original question, one in which you hinted within a day would be proffered.

And you have failed to provide an answer.

All you did is find a 'clever' way to apparently guess the answer in the wrong way. Do you know how easy it would be to do that? Example: you could think in your head 'if I don't know, just put zero'. Or, 'If I do not understand the question, just put zero'. Or if you wanted to suggest writing 32 in anywhere for whatever reason as an answer, it might be accepted. Facts are, the answers proffered would be numerically equivalent to your grade score, on a scale of 100, IMO.

[t***@gmail.com]

Post by Intelligence

Post by Cloud Hobbit

Post by t***@gmail.com
You have failed to show you know how to do the calculations required for your original question, one in which you hinted within a day would be proffered.

And you have failed to provide an answer.

All you did is find a 'clever' way to apparently guess the answer in the wrong way. Do you know how easy it would be to do that? Example: you could think in your head 'if I don't know, just put zero'. Or, 'If I do not understand the question, just put zero'. Or if you wanted to suggest writing 32 in anywhere for whatever reason as an answer, it might be accepted. Facts are, the answers proffered would be numerically equivalent to your grade score, on a scale of 100, IMO.

or, maybe you thought it was multiple choice so you picked from temperatures already offered, or if you had a fifty fifty, if they removed the bottom two choices, no way in theory your answer given would be amiss, even though figured out wrongfully; [i am the other guy with that email, currently intelligence or ***@gmail.com]

[Cloud Hobbit]

Post by Intelligence
All you did is find a 'clever' way to apparently guess the answer in the wrong way. Do you know how easy it would be to do that? Example: you could think in your head 'if I don't know, just put zero'. Or, 'If I do not understand the question, just put zero'. Or if you wanted to suggest writing 32 in anywhere for whatever reason as an answer, it might be accepted. Facts are, the answers proffered would be numerically equivalent to your grade score, on a scale of 100, IMO.

Still don't have an answer, do you.

[Intelligence]

And I add this seriously-the work on the problem is very basic. The only involved mass is H20. All figures are rounded for ease of computation. If you look at it correctly, you can, mentally, determine the equilibrium temperature within fifteen seconds or so, "in your head".

You have a strange definition of correctly.

[Intelligence]

Post by Peter Pan

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

0° C.
Duke has the *real* answer sitting by his computer.

you flunked twice

[ed wolf]

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

I did not remember how to do it and had to look it up, found everything
in today´s ISO units like 333,5 J/g for melting Ice at 0°C, stared at
it, and went back to your nice old figures in Calories and °C, no need
for a calculator, indeed. 0°C is right, isn´t it?
Do you really use Calories as a unit in Physics, these days?
And a question for the you: if your bottle had no air and is completely
airtight while the ice melted, at what temperature will this water start
to boil? My guess is when it comes to the boil it is better for Pernod
than for tea. No hurry,am not sure I can work this out at all.
Maybe someone intelligent above average around here?
regards
ed

[Marvin Sebourn]

Post by ed wolf

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

I did not remember how to do it and had to look it up, found everything
in today´s ISO units like 333,5 J/g for melting Ice at 0°C, stared at
it, and went back to your nice old figures in Calories and °C, no need
for a calculator, indeed. 0°C is right, isn´t it?

Yes, Ed.

Post by ed wolf
Do you really use Calories as a unit in Physics, these days?

Probably everybody else does. I used calories because it avoided the 4.188 (?) J / Gram, but also as a whippersnapper I grew up with Centigrade and Kilocycles, which was in my daily vocabulary of use as a ham and a Navy Electronic Tech.

Post by ed wolf
And a question for the you: if your bottle had no air and is completely
airtight while the ice melted, at what temperature will this water start
to boil? My guess is when it comes to the boil it is better for Pernod
than for tea. No hurry,am not sure I can work this out at all.
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

Post by ed wolf
regards
ed

Regards also (73) from
Marvin

Marvin Sebourn
***@aol.com

[Intelligence]

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

[ed wolf]

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

Post by Intelligence
And a question for you: if your bottle had no air in it and is completely
airtight (and rigid) while the ice melted, at what temperature will this
water start to boil? My guess is when it comes to the boil it is better
for Pernod than for tea.

Maybe the conditions are not quite clear, English not being my first
language, but feel free to ask if in daubt. Personally, I am a welder
and engineer and find it too much serious work to do this in my tired
little spare time.

BTW:
You kept asking how long the equilibrium would need to settle, IMHO for
that one would have to know the shape of the ice(slush, cubes,
a sphere?) and the shape of the bottle,and is it static, shaken or
stirred? If you had any of those answers, would that help you?
Without knowing these conditions nothing can be calculated, as I am
sure you know.
Ask Marvin Sebourn about that, it is his bottle, after all.
And , for a change, dazzle us with brilliance.
Ed Wolf

[t***@gmail.com]

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

[ed wolf]

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

[Peter Pan]

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear from Mr. Yost (or his sockpuppets) on that subject.

Rapid footsteps fading into the distance, crickets, etc.

[t***@gmail.com]

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

[Peter Pan]

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory, that's too much to ask?

You made Ed's point for him.

The shame is, you could have possibly shown yourself to
be kinda right about the non-zero temp, but for reasons
(273+ of them) you can't understand.

Then why get involved in the first place? Oh yeah...
wasn't someone just talking about insane liars.

[Intelligence]

Post by Peter Pan

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory

No, give the answer to the problem.

[Peter Pan]

Post by Intelligence

Post by Peter Pan

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory, that's too much to ask?
You made Ed's point for him.
The shame is, you could have possibly shown yourself to
be kinda right about the non-zero temp, but for reasons
(273+ of them) you can't understand.
Then why get involved in the first place? Oh yeah...
wasn't someone just talking about insane liars.

No, give the answer to the problem.

Nope. It's classified information.
You don't have a need to know.

[Intelligence]

Post by Intelligence

Post by Peter Pan

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory, that's too much to ask?

No, give the answer to the problem.

Nope.

That is what was wrongfully demanded, that the equations not give the proper answer.

[Peter Pan]

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory, that's too much to ask?

No, give the answer to the problem.

Nope. It's classified information.
You don't have a need to know.

That is what was wrongfully demanded, that the equations not give the proper answer.

If you wrongfully demanded equations that do not give the
proper answer, it's easy to see why Marvin gave up on
you.

Maybe Earl could share with you some mugged G equations.

[Intelligence]

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by Intelligence

Post by Peter Pan

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from 0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory, that's too much to ask?

No, give the answer to the problem.

Nope. It's classified information.
You don't have a need to know.

That is what was wrongfully demanded, that the equations not give the proper answer.

If you wrongfully demanded equations that do not give the
proper answer

no the other person was, tis why i didn't give them

[Ted]

Post by Peter Pan

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from
0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory, that's too much to ask?
You made Ed's point for him.
The shame is, you could have possibly shown yourself to
be kinda right about the non-zero temp, but for reasons
(273+ of them) you can't understand.

LOL.

Post by Peter Pan
Then why get involved in the first place? Oh yeah...
wasn't someone just talking about insane liars.

LOL.

[Intelligence]

Post by Peter Pan

Post by t***@gmail.com

Post by Peter Pan

Post by ed wolf

Post by t***@gmail.com

Post by ed wolf

Post by Intelligence

Post by ed wolf
No hurry,am not sure I can work this out at all.

Post by ed wolf
Maybe someone intelligent above average around here?

I certainly don't know, Ed, but I welcome your post with wisdom AND wit.

not from you two, eh?

Here is what you carelessly snipped, it might prove a problem where
intelligence is needed, or experience with such calculations.
Since the melting 0°C Ice did not provoke you to any explanation how the
average Joe or Jane would solve it, and what your more intelligent way

You want me to give you the equations?

If they are correct, follow the conditions described, and differ from
0°C as a result, yes please.
ed wolf

Well, since you took him up, that's the last you'll ever
hear

He put 3 qualifications on his response
so essentially he did not want me to give him the equations I certainly have

So if he requires the equations to be correct, relevant,
and supporting your theory, that's too much to ask?
You made Ed's point for him.
The shame is, you could have possibly shown yourself to
be kinda right about the non-zero temp, but for reasons
(273+ of them) you can't understand.

LOL.

So you got that wrong so you flunked

Post by Peter Pan
Then why get involved in the first place? Oh yeah...
wasn't someone just talking about insane liars.

LOL.

[Amazing Answers]

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

And then a protocell pops out?

[Intelligence]

Post by Amazing Answers

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

And then a protocell pops out?

look, if this guy can't even do a simple calculation involving one ice cube ...

[Marvin Sebourn]

Post by Amazing Answers

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

And then a protocell pops out?

Yes. At 9:48 am this morning the Amazing Answers protocell pops out.


Marvin Sebourn
***@aol.com

[Intelligence]

Post by Amazing Answers

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

And then a protocell pops out?

Yes.

I do not know what you two dingbats problem is, abiogenesis is a non issue. You two want to argue past each other over something you know little about is pointless.
The issue of abiogenesis on earth is irrelevant in the big picture, but abiogenesis somewhere in place and time, is it viable; A few months ago I looked up a vital angle involving sugar molecules, it was viable.

[Marvin Sebourn]

Post by Intelligence

Post by Amazing Answers

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

And then a protocell pops out?

Yes.

I do not know what you two dingbats problem is, abiogenesis is a non issue. You two want to argue past each other over something you know little about is pointless.
The issue of abiogenesis on earth is irrelevant in the big picture, but abiogenesis somewhere in place and time, is it viable; A few months ago I looked up a vital angle involving sugar molecules, it was viable.

You are the only one posting about abiogenesis. AmAns made some silly quip and I replied suggesting that he is at the level of a protocell. That's all.

And then you take off, hiding from your posts, hiding from the snips you have made, and acting as if you are superior, arguing with first grade logic.

I'm still waiting for you to list my sockpuppets. Remember from a couple of posts ago when you suggested I had socks, or is your short term memory going also? Do yo remember that you snipped:

"This "discussion" is boring me. I don't think that our interchanges bring you any credit. Time to bring down the curtain on your show. Responses such as your "how long is a typical class period at school, which is how long somebody would have for a test, obviously" show the lack of any argument from you. Your reply is not only way off topic, it is irrational.

Please list my sockpuppets here:

Or anywhere if you reply."


So list my sockpuppets, sockpuppy.

You're too scared. Cowardly. Because you'd again be shown to be wrong. Because I run no socks.

And you don't know how good I am at math. All I need here is to be able to solve a very simple problem, that you haven't even given the answer to.

You may consider yourself an American Srinivasa Ramanujan, a tragic undiscovered genius. But you're just another troll who wastes his time when he could be doing something of greater substance. Almost anything you could do would be of greater substance.

And concerning Ramanujan, I'm currently reading "The Man Who Knew Infinity" on Kindle. Try it.

List my sockpuppets, cowardly sockpuppy.

Marvin Sebourn
***@aol.com

[Intelligence]

Post by Marvin Sebourn

Post by Intelligence

On Wednesday, February 13, 2019 at 9:48:32 AM UTC-6, Amazing Answers

Post by Amazing Answers
And then a protocell pops out?

Yes.

I do not know what you two dingbats problem is, abiogenesis is a non issue. You two want to argue past each other over something you know little about is pointless.
The issue of abiogenesis on earth is irrelevant in the big picture, but abiogenesis somewhere in place and time, is it viable; A few months ago I looked up a vital angle involving sugar molecules, it was viable.

You are the only one posting about abiogenesis.

So you are completely insane and a liar.

[Christopher A. Lee]

On Wed, 13 Feb 2019 08:02:23 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Amazing Answers

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

And then a protocell pops out?

Yes. At 9:48 am this morning the Amazing Answers protocell pops out.

What's wrong with these morons?

Do they imagine we make these things up, the way theists make up ad
hoc nonsense to "support" their bullshit?

Do they imagine the scientists whose researched results we cite, are
lying about it the way theists do?

[Marvin Sebourn]

Post by Christopher A. Lee
On Wed, 13 Feb 2019 08:02:23 -0800 (PST), Marvin Sebourn

Post by Marvin Sebourn

Post by Amazing Answers

Post by Marvin Sebourn
We start with pure H2O.
10 grams of ice at 0 degrees Celsius, or 32 degrees Fahrenheit.
40 grams of water at 20 degrees Celsius, or 68 degrees Fahrenheit.
Both ice and water are in separate, perfectly insulated containers with no void for air.
The ice and the water are then combined into one single, perfectly insulated container with no void for air.
The specific heat of water = 1 calorie per gram
The heat of fusion of H2O = 80 calories per gram. I know the exact heat of fusion is more nearly 79.7 calories per gram, but for ease of calculation we will use 80 calories per gram.
Atmospheric pressure is normal, and constant at 14.7 psi.
When the water and the ice reach thermal equilibrium, what is the temperature of the combined mass?
Numerical answers only, please, at this time. Please no “Here’s how to do it” for about 24 hours, or after 8pm CST on Monday, 11 February.
Thanks
Marvin Sebourn

And then a protocell pops out?

Yes. At 9:48 am this morning the Amazing Answers protocell pops out.

What's wrong with these morons?
Do they imagine we make these things up, the way theists make up ad
hoc nonsense to "support" their bullshit?

I think some do imagine we make things up, Christopher, but I also think that they know what we say, but act differently. Is that cognitive dissonance? And the tension of CD shows up frequently-for example, Intelligence or Cody or whatever.

Intelligence is amusing. He speaks of my ignorance yet is afraid to answer most questions. He suggests I run socks but refuses to name any, giving some lame excuse. Maybe I should ask him to post his resume' or CV and make comparisons with what he has done in his life, other than anger and hate, exaggeration and insult-what have we seen here?

Marvin Sebourn

Post by Christopher A. Lee
Do they imagine the scientists whose researched results we cite, are
lying about it the way theists do?