# Homework Help: Two body system ,entropy, max work

1. Jan 9, 2015

### prehisto

1. The problem statement, all variables and given/known data
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.

2. Relevant 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?

3. 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 :(

2. Jan 9, 2015

### Bystander

Can you write this much for us?

3. Jan 9, 2015

### Staff: 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

4. Jan 9, 2015

### prehisto

dAmax=(1-T2/T1)dQ1

5. Jan 9, 2015

### Staff: 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

6. Jan 10, 2015

### prehisto

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?

7. Jan 10, 2015

### Staff: 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.

Now let me help you work through this problem.
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

8. Jan 11, 2015

### prehisto

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

9. Jan 11, 2015

### Staff: Mentor

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

10. Jan 11, 2015

### prehisto

a

Last edited: Jan 11, 2015
11. Jan 11, 2015

### prehisto

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
$T_f=\sqrt{T_1T_2/c}$

12. Jan 11, 2015

### Staff: Mentor

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

13. Jan 11, 2015

### prehisto

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

14. Jan 11, 2015

### Staff: Mentor

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

15. Jan 11, 2015

### prehisto

$dQ=cdT -> dQ_1 =cdT=c \int dT= c(T_f-T_1)$
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.
$dQ=c(T_1-T_2)$

16. Jan 11, 2015

### Staff: Mentor

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

Chet

17. Jan 11, 2015

### prehisto

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
$Q_1-Q_2=-c(T_f-T_1)-c(T_f-T_2)=-cT_f+cT_1-cT_f+cT_2=-2cT_f+c(T_1+T_2)$

18. Jan 11, 2015

### Staff: Mentor

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

19. Jan 11, 2015

### prehisto

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

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.

20. Jan 11, 2015

### Staff: Mentor

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

21. Jan 11, 2015

### Staff: Mentor

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

22. Jan 12, 2015

### prehisto

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.

Last edited: Jan 12, 2015
23. Jan 12, 2015

### Staff: Mentor

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

24. Jan 12, 2015

### prehisto

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

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