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I'm working on a problem that asks me the efficiency of a heat engine exchanging heat with an infinite cold reservoir and a finite hot reservoir of capacity $C$ that operates between the temperatures $T_H=10\,T_C$. Looking up at the solution, I found out that the book I'm reading it from makes use of the change of entropy being equal to 0, which is reasonable since it's a reversible process, but it states that $$\frac{Q_C}{T_C}=C\int_{T_H}^{T_C}\frac{dT}{T}$$being $\delta Q=CdT$, and I don't quite get why it integrates between precisely $T_H$ and $T_C$ and not between $T_H$ and some generic $T_f$. Is the efficiency of any reversible engine independent of its operating parameters? Can I take an arbitrary $T_f$ and still get the same result, thus justifying the $T_C$ in the integral? And if I can, is this possibility limited to this specific kind of cycle or can it be applied to every reversible one? (btw with this identities the efficiency comes out nicely as $\approx 0,744$). Thank you all :)

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  • $\begingroup$ Just so I understand the problem, should the limits for the integral be 10$T_C$ to $T_C$? $\endgroup$
    – Bob D
    Jul 30 at 14:34
  • $\begingroup$ @BobD Yup, from $T_H=10T_C$ to $T_C$ $\endgroup$ Jul 30 at 15:17
  • $\begingroup$ OK. BTW I agree with Chet Miller's answer. They are assuming that heat is extracted reversibly from the limited capacity high temperature source until its temperature goes down to that of the low temperature reservoir, which would result in the maximum possible work. $\endgroup$
    – Bob D
    Jul 30 at 15:30
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It seems to be referring to the maximum amount of work you can squeeze out of the system. During the operation, the temperature of the hot reservoir would be getting lower. To get the maximum amount out of this, you would have to run the system in a sequence of mini-cycles, where, in early cycles, the hot reservoir would be close to $T_H$, while, in later cycles, the hot reservoir would approach $T_C$. After the hot reservoir reaches $T_C$, of course, no more work can be squeezed out.

So the change in entropy of the hot reservoir over the entire sequence of cycles would be $$\Delta S_H=-\int_{T_C}^{T_H}{\frac{CdT}{T}}$$The change in entropy of the cold reservoir would be $$\Delta S_C=\frac{Q_C}{T_C}$$This would be enough information to determine the heat rejected to the cold reservoir $Q_C$ over the entire sequence of cycles. And from that you can determine the work and the efficiency.

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  • $\begingroup$ You just beat me to essentially the same answer! $\endgroup$
    – Bob D
    Jul 30 at 14:39
  • $\begingroup$ Thanks Chet, that makes sense. So basically with this setup the work that can be extracted is obviously limited. But the flowing of heat from the hot reservoir to the cold one reminds me a lot of a spontaneous and irreversible process. Then why would the total entropy be equal to 0? Being a state function, and being the final state different from the initial one, the entropy should have increased, shouldn't it? Hope I'm being clear and still thanks for the answer :) $\endgroup$ Jul 30 at 15:14
  • $\begingroup$ It’s not spontaneous if we break it up into a sequence of reversible mini-cycles in which we only transfer a differential amount of heat during each mini-cycle. $\endgroup$ Jul 30 at 15:28
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    $\begingroup$ @ChetMiller My bad, you're completely right, thanks for the answer it's been very useful $\endgroup$ Jul 30 at 15:40
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Good morning! Welcome to this forum. It is full of interesting and customized content.

Perhaps my answer is a bit late, but here it goes:

As the hot reservoir is finite, the engine will convert its stored heat into power out.

The stored energy till decrease along with the continuously extracted shaft power.

Carnot´s relationship with respect to reservoirs´temperature relates the maximum thermodynamic efficency.

This efficiency will approach to zero along with the continuously extracted work, as the higher temperature tends to the lower.

The machine will "dry out" of extractable work.

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