# Two body system ,entropy, max work

## Homework Statement

Two equal bodies with temperatures T1 and T2.
T1 > T2. The specific heat c does not depend on temperature.
What is the maximal work which can be done from this system?

I have to get equation for max work which includes c and temperatures.

## Homework Equations

This seems to be rather simple problem, but it has been very difficult for me so far to understand the concepts and judgments behind this problem. So could someone please help me?

## The Attempt at a Solution

I know that we have to make a system.Where one body is the heater and other is cooler.
So i know:
The heat from heater dQ1=T1dS1
and cooler dQ2=T2dS2
goes to reservoir where work is done
dW=dQ1-dQ2=-T1dS1-T2dS2

Offcourse i know that change in entropy is equal to
dS=dS1+dS2

The next step should be to establish that max work is equal to
dAmax=-(dT)dS

and further use the carno cycle efficiency coefficient
Do not have any idea how to do it keeping in mind the previous relations :(

Bystander
Homework Helper
Gold Member
and further use the carno cycle efficiency coefficient
Can you write this much for us?

Chestermiller
Mentor
If it all is carried out in carnot cycle, what is dS, and how are dS1 and dS2 related? What does this say about how dQ1 and dQ2 related?

Chet

Can you write this much for us?

dAmax=(1-T2/T1)dQ1

Chestermiller
Mentor
The easiest way to do this problem is to figure out how to determine the final equilibrium temperature of the two bodes. To do that, you start out by answering the questions I asked in post #3.

Chet

The easiest way to do this problem is to figure out how to determine the final equilibrium temperature of the two bodes. To do that, you start out by answering the questions I asked in post #3.

I think i got to the relation of work.
1) If the temperature of one body decreases then the temperature of other body increases.
dQ1=-dQ2
2) Entropy of the system increases
dS1+dS2>0

From 1)
T1dS1-T2dS2=0
S2=T1dS1/T2

and from 2)
dS1+T1dS1/T2>0
(T2-T1)dS1>0
dA>0

But here it is positive and there is no reason for me to think that this is the maximal work.
Maybe (T2-T1)dS1 is the maximal work because entropy increases?

Chestermiller
Mentor
None of this assessment is correct.

1. Q2 is not equal to -Q1. Otherwise, no work would be done.
2. The total entropy of the two bodies does not increase if the process is carried out reversibly. So ΔS1+ΔS2=0.

Here's what happens. After all the heat has been transferred and all the work has been done, the temperatures of the two bodes are going be the same. Call this final temperature Tf. Do you know how to determine the change in entropy of the hot body if its temperature decreases from T1 to Tf? If so, please write out your expression. Similarly, do you know how to determine the change in entropy of the cold body if its temperature increases from T2 to Tf. If so, please write out your expression. The sum of these two entropy changes will be equal to zero.

We can continue with the solution to the problem after you have had a chance to respond to my two questions above.

Chet

None of this assessment is correct.

1. Q2 is not equal to -Q1. Otherwise, no work would be done.
2. The total entropy of the two bodies does not increase if the process is carried out reversibly. So ΔS1+ΔS2=0.

Here's what happens. After all the heat has been transferred and all the work has been done, the temperatures of the two bodes are going be the same. Call this final temperature Tf. Do you know how to determine the change in entropy of the hot body if its temperature decreases from T1 to Tf? If so, please write out your expression. Similarly, do you know how to determine the change in entropy of the cold body if its temperature increases from T2 to Tf. If so, please write out your expression. The sum of these two entropy changes will be equal to zero.

We can continue with the solution to the problem after you have had a chance to respond to my two questions above.

Chet
Unfortunately i dont have ideas have to incorporate the final temperature in ΔS1+ΔS2=0.
I would write
dQ1/T1+dQ2/T2=0

Chestermiller
Mentor
Unfortunately i dont have ideas have to incorporate the final temperature in ΔS1+ΔS2=0.
I would write
dQ1/T1+dQ2/T2=0
OK. Let me get you started.

For reversibly heating or cooling each of the bodies,

##TdS=cdT##, or, dS = c d lnT

So, ##ΔS_1=c\ln{\left(\frac{T_f}{T_1}\right)}## and ##ΔS_2=c\ln{\left(\frac{T_f}{T_2}\right)}##

With these equations and the requirement that the combined entropy change of the two bodies is equal to zero, you should be able to determine Tf in terms of T1 and T2. What do you get?

Chet

a

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Never mind the previous post,its a mistake.

OK. Let me get you started.

For reversibly heating or cooling each of the bodies,

##TdS=cdT##, or, dS = c d lnT

So, ##ΔS_1=c\ln{\left(\frac{T_f}{T_1}\right)}## and ##ΔS_2=c\ln{\left(\frac{T_f}{T_2}\right)}##

With these equations and the requirement that the combined entropy change of the two bodies is equal to zero, you should be able to determine Tf in terms of T1 and T2. What do you get?

Chet
##ΔS=c\ln{\left(\frac{T_f}{T_1}\right)} + c\ln{\left(\frac{T_f}{T_2}\right)} = c\ln{\left(\frac{T_f^2}{T_2 T_1}\right)} ##
## c\ln{\left(\frac{T_f^2}{T_2 T_1}\right)}=0##
which gives
## T_f=\sqrt{T_1T_2/c} ##

Chestermiller
Mentor
Never mind the previous post,its a mistake.

##ΔS=c\ln{\left(\frac{T_f}{T_1}\right)} + c\ln{\left(\frac{T_f}{T_2}\right)} = c\ln{\left(\frac{T_f^2}{T_2 T_1}\right)} ##
## c\ln{\left(\frac{T_f^2}{T_2 T_1}\right)}=0##
which gives
## T_f=\sqrt{T_1T_2/c} ##
Not quite. The c shouldn't be in there, should it? If you divide both sides of the equation by c first, your get ##\ln{\left(\frac{T_f^2}{T_2 T_1}\right)}=0##. So again, what is Tf?

Next, we're going to determine the heat lost by the hot body and the heat gained by the cold body.

Chet

Not quite. The c shouldn't be in there, should it? If you divide both sides of the equation by c first, your get ##\ln{\left(\frac{T_f^2}{T_2 T_1}\right)}=0##. So again, what is Tf?

Next, we're going to determine the heat lost by the hot body and the heat gained by the cold body.

Chet

Yes,my mistake.
## T_f=\sqrt{T_1T_2} ##
or
##T_f^2=T_1T_2 ##

Looking forward to it. This time i tink , i wont rash in to it

Chestermiller
Mentor
Yes,my mistake.
## T_f=\sqrt{T_1T_2} ##
or
##T_f^2=T_1T_2 ##

Looking forward to it. This time i tink , i wont rash in to it
OK. I find it an interesting result that the final temperature is the geometric mean of the two initial body temperatures.

Now that you know Tf, tell me, in terms of c, T1, and T2, how much heat is removed from the hot body Q1 between its initial and its final states? Also, in terms of c, T1, and T2, how much heat is added to the cold body Q2 between its initial and its final states? What is the net of Q1 and Q2?

Chet

Now that you know Tf, tell me, in terms of c, T1, and T2, how much heat is removed from the hot body Q1 between its initial and its final states? Also, in terms of c, T1, and T2, how much heat is added to the cold body Q2 between its initial and its final states? What is the net of Q1 and Q2?

Chet

## dQ=cdT -> dQ_1 =cdT=c \int dT= c(T_f-T_1) ##
then i can subsitute
## T_f^2=T_1T_2 ##
and obtain
## dQ_1= - c(\sqrt{T_1T_2}-T_1) ##
I suppose if heat is removed,it should go with negative sign

The same goes for the second body
##dQ_2 = c(T_f-T_2) ##
## dQ_2=c(\sqrt{T_1T_2}-T_2) ##

But in the case of net of Q1 and Q2 i obtain.
## dQ=c(T_1-T_2) ##

Chestermiller
Mentor
## dQ=cdT -> dQ_1 =cdT=c \int dT= c(T_f-T_1) ##
then i can subsitute
## T_f^2=T_1T_2 ##
and obtain
## dQ_1= - c(\sqrt{T_1T_2}-T_1) ##
I suppose if heat is removed,it should go with negative sign

The same goes for the second body
##dQ_2 = c(T_f-T_2) ##
## dQ_2=c(\sqrt{T_1T_2}-T_2) ##

But in the case of net of Q1 and Q2 i obtain.
## dQ=c(T_1-T_2) ##
Your equations for Q1 and Q2 are correct, but check your algebra. This final answer is wrong.

Chet

Your equations for Q1 and Q2 are correct, but check your algebra. This final answer is wrong.

1) In case of net Q, i have got
## Q=Q_1+Q_2=-c(T_f-T_1)+c(T_f-T_2)=-cT_f+cT_1+cT_f-cT_2=c(T_1-T_2) ##
2) in cace of work dW=dQ
## Q_1-Q_2=-c(T_f-T_1)-c(T_f-T_2)=-cT_f+cT_1-cT_f+cT_2=-2cT_f+c(T_1+T_2) ##

Chestermiller
Mentor
1) In case of net Q, i have got
## Q=Q_1+Q_2=-c(T_f-T_1)+c(T_f-T_2)=-cT_f+cT_1+cT_f-cT_2=c(T_1-T_2) ##
2) in cace of work dW=dQ
## Q_1-Q_2=-c(T_f-T_1)-c(T_f-T_2)=-cT_f+cT_1-cT_f+cT_2=-2cT_f+c(T_1+T_2) ##
The net Q is not Q1 + Q2 (your first equation). The net Q is Q1-Q2 (your second equation), and, yes, that is equal to W. Now, in your equation for Q, if you factor out the c and combine the terms, you get ## W=Q =Q_1-Q_2=(T_1-2T_f+T_2) ##
Now substitute your equation for Tf into the term in parenthesis, and you will get a perfect square. In terms of T1 and T2, what is that perfect square?

Chet

Now substitute your
equation for Tf into the term in parenthesis, and you will get a perfect square. In terms of T1 and T2, what is that perfect square?

## (T_1-2T_f+T_2)=T_1-2\sqrt{(T_1T_2)}+T_2=(\sqrt{T_1}-\sqrt{T_2})^2 ##

yes,i got it. As far as i can see now its mathematical operations which shows us relation for work in other form.
But where it leads us as far as physical interpretation.

Chestermiller
Mentor
## (T_1-2T_f+T_2)=T_1-2\sqrt{(T_1T_2)}+T_2=(\sqrt{T_1}-\sqrt{T_2})^2 ##

yes,i got it. As far as i can see now its mathematical operations which shows us relation for work in other form.
But where it leads us as far as physical interpretation.
In what we've done, we've looked at the problem from 20000 ft and examined the overall response, by just focusing on the initial and the final states. However, you are very perceptive to ask for a physical interpretation. In particular, I think you are asking, "how do we design a process in detail to achieve this maximum work?" That would help you greatly in physical understanding. Is that what you want?

Chet

Chestermiller
Mentor
## (T_1-2T_f+T_2)=T_1-2\sqrt{(T_1T_2)}+T_2=(\sqrt{T_1}-\sqrt{T_2})^2 ##

yes,i got it. As far as i can see now its mathematical operations which shows us relation for work in other form.
But where it leads us as far as physical interpretation.
In what we've done, we've looked at the problem from 20000 ft and examined the overall response, by just focusing on the initial and the final states. However, you are very perceptive to ask for a physical interpretation. In particular, I think you are asking, "how do we design a process in detail to achieve this maximum work?" That would help you greatly in physical understanding. Is that what you want?

Chet

That would help you greatly in physical understanding. Is that what you want?
Yes. I want to understand why our result is for maximum work.

i suppose that our result correspond to the maximal work because its a reversible process thus the net entropy change is zero ( no heat is given to environment which leads to increase of entropy) . With what we started in the first place.

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Chestermiller
Mentor
Yes. I want to understand why our result is for maximum work.

i suppose that our result correspond to the maximal work because its a reversible process thus the net entropy change is zero ( no heat is given to environment which leads to increase of entropy) . With what we started in the first place.
Your overall assessment is correct. But maybe you won't be satisfied until you dream an actual reversible process where you can bring about these reversible changes in the temperatures of the bodies and do the corresponding amount of reversible work (using some working fluid, for example). So, what are your thoughts on a process for accomplishing this. Here are some considerations:

1. It's a one shot deal, so the working fluid(s) do not need to be returned to its (their) original state(s) (i.e., the working fluid does not have to go through a cycle).
2. You don't need to use only one working fluid. You can use more than one.
3. The heat transferred to the working fluid(s) must equal the net heat transferred from the bodies.
4. The entropy change of the working fluid(s) must be equal to zero (since the entropy change of the bodies is zero and the process is reversible).
5. The working fluid(s) can be compressed or expanded by adding or removing tiny weights to or from a piston (or pistons).
6. The working fluid(s) can be an ideal gas (or gases).

Chet

Your overall assessment is correct. But maybe you won't be satisfied until you dream an actual reversible process where you can bring about these reversible changes in the temperatures of the bodies and do the corresponding amount of reversible work (using some working fluid, for example). So, what are your thoughts on a process for accomplishing this. Here are some considerations:

1. It's a one shot deal, so the working fluid(s) do not need to be returned to its (their) original state(s) (i.e., the working fluid does not have to go through a cycle).
2. You don't need to use only one working fluid. You can use more than one.
3. The heat transferred to the working fluid(s) must equal the net heat transferred from the bodies.
4. The entropy change of the working fluid(s) must be equal to zero (since the entropy change of the bodies is zero and the process is reversible).
5. The working fluid(s) can be compressed or expanded by adding or removing tiny weights to or from a piston (or pistons).
6. The working fluid(s) can be an ideal gas (or gases).

Chet
Thank you for you help with this problem. It was very helpful!