## Thermodynamics B/C

WhatScience?
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### Re: Thermodynamics B/C

I used to feel this way about Physics too, but once you understand the concepts involved formerly impossible questions appear quite simple, and the whole learning process becomes very fulfilling.
This is why I like physics more than bio.

I dropped A and P and Microbes because it was just too much memorization. Thermo at least feels more like a science, where everything has a logical solution.
Did you just...
imply...
that microbio...
isn’t a science...
I did NOT

When I say feels like a science, I mean the logic part.

This is just what I enjoy in a science, not what makes up a science.
"When you clean your room, you are increasing the total chaos of the universe" - Hank Green Crash Course (Entropy)

Events 2018

Thermodynamics, Potions and Poisons, Disease Detectives, Optics, and Towers

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### Re: Thermodynamics B/C

Okay, I'll ask the next one, and if someone else wants to give the rest of the answers for the other question, he/she can?

A mathematician homeowner measures temperatures in a temperature unit he created called S. This system has units on a logarithmic scale, so heating something by 1 degree Celsius would cause the temperature in S to multiply by some constant factor. This homeowner heats his house at a rate of 100 kW when it is 10 S out to maintain an inside temperature of 90 S. If it is 5 S outside, how much heat (in kJ) must he add to his house per minute to maintain an inside temperature of 405 S (assuming the same conditions as before).

Justin72835
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### Re: Thermodynamics B/C

Okay, I'll ask the next one, and if someone else wants to give the rest of the answers for the other question, he/she can?

A mathematician homeowner measures temperatures in a temperature unit he created called S. This system has units on a logarithmic scale, so heating something by 1 degree Celsius would cause the temperature in S to multiply by some constant factor. This homeowner heats his house at a rate of 100 kW when it is 10 S out to maintain an inside temperature of 90 S. If it is 5 S outside, how much heat (in kJ) must he add to his house per minute to maintain an inside temperature of 405 S (assuming the same conditions as before).
Pretty creative question! I'm not 100% sure on this one (my Algebra 2 skills are a bit rusty) but I'll give it a shot.

The rate of heat that must be supplied is proportional to the change in temperature:

$100kW=k(\Delta T)$

Since this equation works only if the scale is linear, we have to modify it a bit:

$100kW=k(\ln{T_f/T_i})=k(\ln{90/10})=k(\ln{9})$

If we apply this to the second scenario:

$H=k(\ln{T_f/T_i})=k(\ln{405/10})$

Cross-multiplying gives:

$H=100kW*\frac{k(\ln{405/10})}{k(\ln{90/10})}=168.5kW$

Then to get heat transferred per minute, just multiply by 60 and change the units a bit:

$168.5kW*60s=10107 kJ/minute$
"The fault, dear Brutus, is not in our stars,
But in ourselves, that we are underlings."

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### Re: Thermodynamics B/C

Okay, I'll ask the next one, and if someone else wants to give the rest of the answers for the other question, he/she can?

A mathematician homeowner measures temperatures in a temperature unit he created called S. This system has units on a logarithmic scale, so heating something by 1 degree Celsius would cause the temperature in S to multiply by some constant factor. This homeowner heats his house at a rate of 100 kW when it is 10 S out to maintain an inside temperature of 90 S. If it is 5 S outside, how much heat (in kJ) must he add to his house per minute to maintain an inside temperature of 405 S (assuming the same conditions as before).
Pretty creative question! I'm not 100% sure on this one (my Algebra 2 skills are a bit rusty) but I'll give it a shot.

The rate of heat that must be supplied is proportional to the change in temperature:

$100kW=k(\Delta T)$

Since this equation works only if the scale is linear, we have to modify it a bit:

$100kW=k(\ln{T_f/T_i})=k(\ln{90/10})=k(\ln{9})$

If we apply this to the second scenario:

$H=k(\ln{T_f/T_i})=k(\ln{405/10})$

Cross-multiplying gives:

$H=100kW*\frac{k(\ln{405/10})}{k(\ln{90/10})}=168.5kW$

Then to get heat transferred per minute, just multiply by 60 and change the units a bit:

$168.5kW*60s=10107 kJ/minute$
Overall right, but pay attention to the question details (I bolded the part of the question I'm referring to specifically). Your turn!

Justin72835
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### Re: Thermodynamics B/C

Overall right, but pay attention to the question details (I bolded the part of the question I'm referring to specifically). Your turn!
Oh yeah, looks like I completely missed that part of the question :/.

You have a well-insulated rigid container separated into two equal halves by a small removable membrane of negligible volume. On one side of the membrane, you have 3 moles of a monatomic gas at a temperature of 25K. On the other side, you have a vacuum. If the membrane separating the two halves was quickly removed, allowing the gas to expand into the other half of the container, calculate the change in entropy of the gas.

Note that because the gas expanded by itself (called free expansion), no work was done by the gas. Also, because the container was insulated and the gas expanded in just a fraction of a second, you can assume that no heat was transferred. Lastly, because of the last two conditions, you know that the gas experienced no change in temeprature.
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But in ourselves, that we are underlings."

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### Re: Thermodynamics B/C

Overall right, but pay attention to the question details (I bolded the part of the question I'm referring to specifically). Your turn!
Oh yeah, looks like I completely missed that part of the question :/.

You have a well-insulated rigid container separated into two equal halves by a small removable membrane of negligible volume. On one side of the membrane, you have 3 moles of a monatomic gas at a temperature of 25K. On the other side, you have a vacuum. If the membrane separating the two halves was quickly removed, allowing the gas to expand into the other half of the container, calculate the change in entropy of the gas.

Note that because the gas expanded by itself (called free expansion), no work was done by the gas. Also, because the container was insulated and the gas expanded in just a fraction of a second, you can assume that no heat was transferred. Lastly, because of the last two conditions, you know that the gas experienced no change in temeprature.
S = klnW
The number of states doubles for each particle, so overall, there are 2^N times W in the new container.
S = kln(2^N * W) = k * (Nln2 + lnW) = kNln2 + klnW
Change in S = kNln2 = 3 mol * R * ln 2 = [b]17.29 J/K[/b]

As far as I know, the temperature isn't relevant as long as it remains constant.

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### Re: Thermodynamics B/C

Overall right, but pay attention to the question details (I bolded the part of the question I'm referring to specifically). Your turn!
Oh yeah, looks like I completely missed that part of the question :/.

You have a well-insulated rigid container separated into two equal halves by a small removable membrane of negligible volume. On one side of the membrane, you have 3 moles of a monatomic gas at a temperature of 25K. On the other side, you have a vacuum. If the membrane separating the two halves was quickly removed, allowing the gas to expand into the other half of the container, calculate the change in entropy of the gas.

Note that because the gas expanded by itself (called free expansion), no work was done by the gas. Also, because the container was insulated and the gas expanded in just a fraction of a second, you can assume that no heat was transferred. Lastly, because of the last two conditions, you know that the gas experienced no change in temeprature.
S = klnW
The number of states doubles for each particle, so overall, there are 2^N times W in the new container.
S = kln(2^N * W) = k * (Nln2 + lnW) = kNln2 + klnW
Change in S = kNln2 = 3 mol * R * ln 2 = [b]17.29 J/K[/b]

As far as I know, the temperature isn't relevant as long as it remains constant.
Correct. Dang, that's a really cool way of doing that problem. I was thinking that you could do it this way:

$\Delta S = \int_{V_i}^{V_f}\frac{PdV}{T} = \int_{V_i}^{V_f}\frac{nRdV}{V} = nR\ln_{V_f/V_i}$

But I think the method you used was much more elegant. Your turn!

Edit: Can you explain how you knew that there were 2^N times W microstates in the new container? Thanks.
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### Re: Thermodynamics B/C

Can you explain how you knew that there were 2^N times W microstates in the new container? Thanks.
Each particle has two times the number of places it can be, thus 2 times the number of microstates for each particle, thus 2^N (because of how probability works).

Lord Kelvin collaborates with a Russian scientist but notices there is a discrepancy in their data! Lord Kelvin measured his temperature in Kelvin, and the Russian scientist measured his temperature in Réaumur! How can Lord Kelvin figure out the temperature in Kelvin based on this Russian scientist's measurements?

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### Re: Thermodynamics B/C

Can you explain how you knew that there were 2^N times W microstates in the new container? Thanks.
Each particle has two times the number of places it can be, thus 2 times the number of microstates for each particle, thus 2^N (because of how probability works).

Lord Kelvin collaborates with a Russian scientist but notices there is a discrepancy in their data! Lord Kelvin measured his temperature in Kelvin, and the Russian scientist measured his temperature in Réaumur! How can Lord Kelvin figure out the temperature in Kelvin based on this Russian scientist's measurements?
I'm pretty sure that for this problem you would just use an equation to convert between the two temperatures.

For converting from Kelvin to Réaumur: R = (K - 273.15)*0.8

For converting from Réaumur to Kelvin: K = 1.25R + 273.15
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### Re: Thermodynamics B/C

I'm pretty sure that for this problem you would just use an equation to convert between the two temperatures.

For converting from Kelvin to Réaumur: R = (K - 273.15)*0.8

For converting from Réaumur to Kelvin: K = 1.25R + 273.15