# Callen's Maximum Work Theorem: why doesn't heat lost equal heat absorbed (and likewise for work)?

This question was, effectively, asked here (please refer to that question for additional context); however, I don't think the given answer is correct (or at least complete) despite my having added a bounty and having had a productive discussion with the "answerer" there. In particular, I don't believe the usage of reversible there is what Callen has in mind; Callen is referring to a process which is reversible in the sense of the entire composite system (i.e. isentropic), not with respect to a particular subsystem. I have since reread the section of Callen and still cannot convince myself of what's going on.

In Chapter 4.5 of his famous textbook, Callen gives his "Maximum Work Theorem"; effectively, it says that for a composite system (and process defined thereon) composed of three subsystems: (1) a primary subsystem which is taken from some initial state to some final state (2) a "reversible heat source" (RHS) which can only exchange heat and (3) a "reversible work source" (RWS) which can only exchange work. The theorem then states that

"for all processes leading from the specified initial state to the specified final state of the primary system, the delivery of work is maximum (and the delivery of heat is minimum) for a reversible process. Furthermore the delivery of work (and of heat) is identical for every reversible process."

The "reversible" caveats above are perhaps a bit of a misnomer; they are just intended to indicate that every process is quasistatic with respect to said subsystem.

As shown in Callen, one can show that in an infinitesimal process, the maximum work delivered to the RWS is given by $$dW_{RWS} = \left(1 - \frac{T_{RHS}}{T}\right)(-dQ) - dW$$ where unsubscripted quantities refer to the primary subsystem. My question is why the following logic is wrong: By the definitions of the two "reversible sources", any heat lost by the primary subsystem must be exchanged with RHS, and any work must be exchanged with the RWS. Therefore, $$dQ_{RHS} = -dQ$$ and $$dW_{RWS} = -dW$$.

This clearly contradicts $$dW_{RWS} = \left(1 - \frac{T_{RHS}}{T}\right)(-dQ) - dW$$. But what is wrong with the logic above? Certainly, we must have $$-dW - dQ = -dU = dQ_{RHS} + dW_{RWS}$$ (indeed this is part of Callen's derivation), but clearly I am wrong and it's not true that $$dQ_{RHS} = -dQ$$ and $$dW_{RWS} = -dW$$.

To be clear, I am not disputing that the derivation given by Callen is wrong; I follow it through completely. I am asking why my particular logic above fails, since I think any answer to this "paradox" may offer some insight (to me anyway) into how heat is converted to work.

My question here is different than the linked one precisely because that answer does not (it seems to me) explain why my logic above is wrong directly. I am not asking for an explanation of Callen's derivation, but why my (wrong) derivation is wrong.

• any heat lost by the primary subsystem must be exchanged with RHS, and any work must be exchanged with the RWS. Therefore, dQRHS=−dQ and dWRWS=−dW” The second statement doesn’t follow from the first. Only the sum is fixed. This is explicitly discussed in the last two paragraphs here. Jun 16 at 0:44
• Does this answer your question? Understanding Callen's eq. (4.9) and the maximum work theorem Jun 16 at 0:44
• Hi! Thank you again for helping me with this. It's really bothered me the last couple weeks here. I guess that I understand that it must indeed only be the sum that is fixed, but I'm afraid I can't understand how your last two paragraphs there resolve my problem. I get that the first sentence you give doesn't imply the latter, but I don't understand why the latter doesn't hold in and of itself. @Chemomechanics
– EE18
Jun 16 at 0:56
• Even in a reversible (with respect to the overall composite system) process, we don't have $Q = Q_{RHS}$ and likewise for work. I also want to note that when Callen doesn't use subscripts he is referring to the primary subsystem, not to certain (real) processes.
– EE18
Jun 16 at 0:56
• Does this answer your question? Understanding Callen's eq. (4.9) and the maximum work theorem Jun 16 at 4:39

Let's go through the derivation to see why your logic is wrong.

The setup: system undergoes change from state 1 to 2 during which it exchanges heat with a bath at $$T_\text{RHS}$$. What is the maximum amount of work that can be produced, or minimum amount that must be consumed, as a result of the change of state?

Answer I will assume closed system for simplicity, we can repeat the steps with an open system. By first law $$\Delta U_{12} = Q + W$$ By Second Law: $$S_\text{gen} = \Delta S_{12} - \frac{Q}{T_\text{RHS}} \geq 0 .$$ Solve the second law for $$Q$$, substitute into the first law and solve for work: $$W = \underbrace{\Delta U_{12} - T_\text{RHS}\Delta S_{12}}_{W_\text{RHS}} + T_\text{RHS} S_\text{gen}$$ or $$\boxed{ W = W_\text{RHS} + T_\text{RHS} S_\text{gen} \geq W_\text{RHS} .\vphantom{\sum_1^1} }$$ This is essentially Gallen's result. I don't have the book in front of me but if you take all processes to be reversible except for heat transfer from $$T_\text{RHS}$$ to $$T$$, then $$S_\text{gen}$$ is Gallen's term with $$dQ$$. (Not sure why Gallen complicates things with different signs for the work.)

Resolution 1 From the boxed equation is clear that $$W\neq W_\text{RHS}$$ because it is possible to add work to the system ($$W$$) that contributes to the entropy generation rather than to $$W_\text{RHS}$$.

Resolution 2 To put this differently, to obtain $$dW_\text{RHS}=-dW$$ in Gallen's equation you need to run a tiny Carnot cycle between $$T$$ and $$T_\text{RHS}$$ to extract work that you are losing because you are allowing heat to pass between two finite temperatures.

• I see; so you are saying that the resolution is that Callen is failing to mention there are "extra auxiliary systems" (e.g. the working fluid of the Carnot cycle you allude to) at play here?
– EE18
Jun 23 at 0:39
• @EE18 Not exactly. Gallen's point is that if you set $T=T_\text{RHS}$ then the irreversibility vanishes and $dW_\text{RHS}=-dW$. My point is that the term $(1-T_\text{RHS}/T)(-dQ)$ represents lost work due to irreversibility: to maximize your work you should either set the two temperatures equal to each other, or keep them as they are but add a Carnot. In either case you will maximize the work and the maximum value will be the same. Jun 23 at 0:58
• Is it lost work though? One can have $dQ <0$ (extraction of heat) as in an engine? I am not sure that your first part is what Callen is getting at?
– EE18
Jun 23 at 1:24
• (To be clear, I think I agree with your resolution 2.)
– EE18
Jun 23 at 1:30
• @Yes, in my last comment I am referring to Resolution 2. Part of you problem may be Gallen himself, his is not my favorite book, I think he over-explains things. You should understand thermodynamics in you own way, not someone else's, but this of course means that you should tolerate your teacher's explanations (Gallen's) until you form a point of view of you own. Also, don't waste you reputation points in bounties -- people will still try to give good answers. I will look into the discussion you linked. Jun 23 at 11:24