## A Question of Equilibrium

It’s been quite a while since I’ve been able to find time to post any items in the cute problems folder, and I don’t have much time today either, but here’s a quickie. You may well find this a lot harder than it looks at first sight. At least I did!

*An isolated system consists of two identical components, each of constant heat capacity C, initially held at temperatures T _{1} and T_{2} respectively. What is the maximum amount of work that can be extracted from the system by allowing the two components to reach equilibrium with each other?*

As usual, answers through the comments box please. There is no prize, even if you’re right.

Follow @telescoper

March 30, 2017 at 5:55 pm

C (T1 + T2 – 2*sqrt (T1*T2)), I think.

A nice opportunity to brush up my thermodynamics.

March 30, 2017 at 8:03 pm

You don’t get marks unless you explain your working!

March 30, 2017 at 8:47 pm

Entropy dS=CdT/T so initial S = C*ln(T1)+C*ln(T2)

Final entropy is 2C*ln(Tf). Set equal to initial S and

get Tf=sqrt(T1*T2). So dE = C(T1+T2-2*sqrt(T1*T2))

and Dave Henley is right.

March 30, 2017 at 8:59 pm

Thanks. I was waiting till I got to a proper computer to type up my explanation (which wasn’t as succinct as yours), but you beat me to it.

March 31, 2017 at 10:38 am

Here is a longer piece of reasoning to get to the same result (ideal that’s what I would expect from students):

From the second law of thermodynamics we have that the total entropy of the system, considered isolated, satisfies:

Delta S = Delta S1 + Delta S2 >= 0

Delta S1 = Int_T1^Tf C dT/T = C ln (Tf / T1)

Delta S2 = Int_T2^Tf C dT/T = C ln (Tf / T2)

Overall this gives Delta S = 2C ln Tf – C ln T1T2 >= 0.

This inequality is equivalent to asking that Tf >= sqrt(T1T2).

Fair enough, one needs to determine Tf now.

From the first law of thermodynamics we have that:

Delta U = Q1 + Q2 + W.

Considering again the system as isolated this gives

Delta U = Q1 + Q2 + W = 0.

Then Q1 = C(Tf-T1) and Q2 = C(Tf-T2).

This gives that Tf = (T1+T2)/2 – W/(2C).

As expected if W = 0 we retrieve the usual Tf = (T1+T2)/2 while if W > 0 then Tf will have a lower value but which cannot go below sqrt(T1T2) because of the second law. This of course puts a bound on W by substituting Tf by its above expression:

W <= C[T1+T2 – 2 sqrt(T1T2)].

March 31, 2017 at 12:07 pm

For bonus points, design an apparatus that extracts this amount of work.

March 31, 2017 at 12:13 pm

If you could do that you wouldn’t worry about bonus points – you could make a fortune!

March 31, 2017 at 12:10 pm

The answers posted are correct. On reflection I should have made it clearer that the system is to be regarded as isolated, and have now edited the question to that effect.

If I were marking this in an exam, I would have insisted for full marks on some sort of preliminary reasoning. The point is that the maximum energy available for mechanical work is when the system operates reversibly, in which case the total change of entropy is zero. Then the calculation by Ned Wright follows.

“Total” here means “system and surroundings” so it is necessary that the system should be isolated. If an engine increases the entropy of its surroundings then it can perform even less mechanical work than if it were isolated…

When I did this problem myself many years ago I jumped to the assumption that it would require maximization of some function by differentiation, but that wasn’t a wise assumption…

March 31, 2017 at 1:39 pm

I did it a clunkier way than Ned Wright. Run a Carnot engine between T1 and T2 to extract a small amount of energy dE. The efficiency of a Carnot engine is 1-T1/T2, so

dE = -(1-T1/T2) C dT2

(T2 is the higher temperature, and dT2 is negative.) And energy conservation tells us that

dE = -C (dT1 + dT2)

These lead to

dT1 + dT2 = (1-T1/T2) dT2.

This cleans up to dT1/dT2 = -T1/T2, and the solution to this is T1 T2 = constant.

This is certainly the long way around compared to Ned ‘s way. Much simpler to say dS = dT1/T1 + dT2/T2 = 0. I’m annoyed that I didn’t see that line of attack.

March 31, 2017 at 1:43 pm

Yes, it’s a cute problem but no every solution is as cute!

March 31, 2017 at 4:13 pm

I think the reason you’re going around the houses a bit is because the Carnot engine’s efficiency is derived from the requirement on the entropy change.

If it’s any consolation, I didn’t find these other lines of reasoning terribly obvious – they were a bit terse for me, and I haven’t done any thermodynamics in years! I eventually settled on the following reasoning:

The two components will settle at some final temperature, Tf. Each system’s entropy will change by a calculation of:

dS = C dT/T

integrated from the starting temperature to the final temperature. If we are to extract the maximum work from the process, we need this process to be reversible (as you say, a Carnot engine) in which case the final entropy will not have changed – any entropy change in one component will have been compensated by a change in the other.

So doing those integrals for the two systems and making sure they sum to zero gives us Tf = sqrt(T1 T2). This tells us the final temperature of the two components.

Since we know the final temperature and C, we know how much energy was taken from the hot component: C (T1-Tf)

And dumped into the cold component: C(T2 – Tf).

That won’t sum to zero, because some of the energy has left the system as the work done. So taking the difference and substituting the final temperature gives Dave’s speedy answer.

April 1, 2017 at 4:22 pm

Yes, you’re right. That’s exactly why my solution is more roundabout than necessary. It happens that I lectured on the Carnot cycle not too long ago, so maybe it’s not surprising that that approach leapt to mind.

April 18, 2017 at 11:12 am

All good physicists should keep in shape by regularly reading the Carnot-cycle blog.

April 18, 2017 at 2:01 pm

A wise and noble comment Phillip. I salute you.