For a problem I have been working on, I faced some difficulties in understanding some parts. The problem is as follows:

A body of finite mass is originally at temperature $T_2$, higher than that of a heat reservoir at a temperature $T_1$. An engine operates in infinitesimal cycles between the body and the reservoir until it lowers the temperature of the body from $T_2 to T_1$. Prove that the maximum work obtainable from the engine is $W_{max}=Q - T_1(S_2 - S_1)$, where $S_1-S_2$ is the entropy change in the body and Q is the heat extracted from the body by the engine.

This is how I start:

(1)Firstly, I know that the engine is performing a cycle, so change in internal energy associated within the engine is 0, i.e. $dU=0$, and I also know that the change in entropy as applied to a reversible engine(a Carnot Engine) is $dS=\frac{dQ}{T} \implies dQ=TdS$

(2) Now for any heat engine, we know that the work done, W, by the engine is $W=Q_2-Q_1$ where $Q_2,Q_1$ represent the heat extracted from the higher temperature reservoir and heat rejected to the lower temperature reservoir respectively.

(3)Also, when the body's temperature is lowered from $T_2 to T_1$ the engine will top operating and the work output will be 0 because otherwise we would violate Kelvin-Planck's statement of 2nd law of thermodynamics. At this state, $W=Q'_1-Q_1=0 \implies Q'_1=Q_1$ here $Q'_1$ is the heat we extract from the body when it has arrived at the temperature $T_1$

(4) Now, as the body's temperature decreases to $T_1$, there is an entropy change in the body of amount $S_1-S_2$. Now, at $T_1$ the heat extracted from the body would be $$Q'_1=-T_1(S_1-S_2) \implies Q'_1=T_1(S_2-S_1)$$. from (3) i get that $Q_1=Q'_1=T_1(S_2-S_1)$

(5) Using the expression in (4) with the one in (2) I get $W=Q_2 - T_1(S_2 - S_1)$ and this is basically $W=Q - T_1(S_2 - S_1)$ because $Q_2=Q$ is the heat extracted from the engine.

The expression for work done in (5) represents the maximum work extractable because the engine was operated in infinitesimal cycles between the two temperatures this implies that the engine is reversible, and from Carnots's theorem we know that the efficiency of reversible engines is maximum.

Is this approach alright? Any places I need to change/modify? The way i have done it, gives a better feel for the underlying Physics(atleast to me) than the one given below:

https://en.wikipedia.org/wiki/Principle_of_maximum_work ;I found this online but i did not completely understand it at this point. Is my solution wrong then?


1 Answer 1


I can't really understand what you did, but I would have done this differently. For maximum work, there must be no entropy generation in the system, so the sum of the entropy change of the body plus the entropy change of the reservoir must be equal to zero. So, $$(S_2-S_1)-\frac{Q_1}{T_1}=0$$where $Q_1$ is the heat extracted by the reservoir. This gives $$Q_1=T_1(S_2-S_1)$$So the maximum work is just $$W_{max}=Q-Q_1=Q-T_1(S_2-S_1)$$

If this is the same thing you did, then it is correct.


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