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The hardest thing for me about studying thermodynamics, is connecting the formal theory with actual problem instances. Here is an example:

In Callen’s Thermodynamics and an Introduction to Thermostatics (second edition), Chapter 4, Example 1, the following scenario is described:

A particular system is constrained to constant mole number and volume, so that no work can be done on or by the system. [...]. Two such systems, with equal heat capacities, have initial temperatures $T_{10}$ and $T_{20}$, with $T_{10} < T_{20}$. An engine is to be designed to lift an elevator (i.e., to deliver work to a purely mechanical system), drawing energy from the two thermodynamic systems. What is the maximum work that can be so delivered?

What I don't understand here is: Why does the question mention two such systems? More generally, I don't understand how I'm supposed to envision this scenario.

I understand that the given system cannot do mechanical work, so it can only transfer (or receive) heat to another system, which I believe is the engine. Therefore, the engine needs to be able to do the mechanical work. What I imagine is: the given system is in contact with an engine. The engine is simply (for example) a gas in a cylinder with a movable piston. The gas in the engine is heated by the system, and therefore expands, and this moves a piston which gives us the required mechanical motion. I understand that this process can continue until the temperature of the given system equals the temperature of the gas in the engine.

So where does the second system enter? Why do we need it? Am I imagining the scenario correctly?

EDIT

Here is a synthesis of all the answers I received, in words that are easier for me to understand. I hope I got it right.

We are supposed to think about the engine as a rather limited entity, in at least these aspects:

  1. It has a small heat capacity. This means that, when it comes into contact with the high temperature system (which I'll call system $A$), a small amount of heat transferred to the engine is enough to raise its temperature to the level of system $A$. At this point, no more energy (heat) can be transferred to the engine from system $A$.

  2. The engine's ability to do work might be mechanically bounded. For example, if the work is performed by expanding gas in a piston, the piston might have a maximal expansion length, after which it cannot do anymore work (because the piston cannot mechanically move anymore).

It follows that, when running the process, the engine will "quickly" get from its initial state to a final state in which it cannot do anymore work - either because it cannot take in anymore heat from system $A$ (this heat is what drives the work to be done), or because it is mechanically "stuck".

However, more work can be done if the engine is brought back to its initial state, and then the entire process is repeated. But to bring the engine back to its initial state, the engine must dump heat to another system - a low temperature system (which I'll call system $B$) - so that the engine's temperature can return to the initial value. System $B$ allows the engine to work in cycles, and hence to perform more and more work.

As these cycles are repeated many times, system $A$'s temperature is slowly reduced, and system $B$'s temperature is slowly increased, until both temperature become equal. At this point, no more work can be done.

If the given scenario did not include system $B$, then we would need to consider more carefully the details of the engine, e.g its temperature and heat capacity. In this case, the engine would take heat from system $A$ until the temperatures are equal, while performing work. And we don't have repeating cycles.

Lior
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  • \You are supposed to assume that you can transfer heat fiom the hot system to an engine and lesser heat from the engine to the cold system so that, in the end, the hot system and the cold system are at the same final temperature. – Chet Miller Feb 14 '23 at 13:27
  • "and lesser heat from the engine to the cold system" - why ? why isn't it enough to only "transfer heat fiom the hot system to an engine"? – Lior Feb 14 '23 at 14:30
  • Because the engine won’t be be a it’s original state after you’re done. And they expect you to get the engine and its original state when you’re done. – Chet Miller Feb 14 '23 at 15:47
  • why is this expected, though? is it implicitly assumed? (I conclude from the answers that it probably is, and that this is a condition for maximum work) – Lior Feb 14 '23 at 19:50
  • here are a few (heated) exchanges I have had over the years with @Bob_D and others regarding isothermal work: https://physics.stackexchange.com/q/707625/ and https://physics.stackexchange.com/q/747886/ and https://physics.stackexchange.com/q/708159/ – hyportnex Feb 14 '23 at 20:28
  • and https://physics.stackexchange.com/q/746242/ and https://physics.stackexchange.com/q/736525/ and here https://chat.stackexchange.com/transcript/141379/2022/12/18 – hyportnex Feb 14 '23 at 20:30
  • @Lior That doesn't make sense. You previously stated neither reversible processes or the Carnot cycle have been introduced, but now you are saying the deleted text refers to entropy S, a second law and reversible process concept. What gives? – Bob D Feb 14 '23 at 21:42
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    @hyportnex "Heated" exchanges. NICE!!! – Bob D Feb 14 '23 at 21:47
  • No. They should have mentioned that the engine state does not change from beginning to end state. – Chet Miller Feb 14 '23 at 21:55
  • @BobD I stand by what I said. In Callen, reversible processes and Carnot cycles are only introduced later. Maybe this is a nonstandard treatment. – Lior Feb 14 '23 at 22:08
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    @Lior You don't have to defend yourself. I simply don't understand how Callen can introduce an equation involving entropy, without having introduced the second law. That's all. – Bob D Feb 14 '23 at 22:12
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    Wow, what a horribly composed exercise/scenario (even when re-including the omission)! It introduces “a particular system” and then suddenly says that there are two of these systems. Instead of giving an overview of what is being described in the first sentence, it is given in the penultimate one. These two systems are described completely abstractly, but then are supposed to do a very specific work (lifting an elevator). More general, there is a strong verbosity and abstraction dissonance over the entire text. What is $10$ in “$T_{10}$” referring to? – Wrzlprmft Feb 15 '23 at 07:12
  • … The engine itself is not introduced as a system (although it is). There are no specifications on the engine whatsoever, e.g., that it should be returned to its original state (which is necessary to have any specific answer on this). – Wrzlprmft Feb 15 '23 at 07:33
  • @Wrzlprmft That's how I feel whenever I try to study thermodynamics... and Callen is a widely recommended reference (for example https://physics.stackexchange.com/a/36326/23677 and https://physics.stackexchange.com/a/45328/23677) – Lior Feb 15 '23 at 10:45
  • The restatement isn't accurate. (1) The engine doesn't need to have a certain mechanical bound; it can expand arbitrarily. But it does need to be returned to its original state, because Callen says it doesn't accumulate entropy. (2) The subsequent unconventional redefinition of "engine" as something containing the cold reservoir is only going to produce confusion moving forward. It's ill-advised to redefine consensus terms during the learning process. (3) The analysis is also wrong, as the "engine plus cold reservoir" is no longer a system in internal equilibrium, so entropy will be generated. – Chemomechanics Feb 15 '23 at 18:17

4 Answers4

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What I don't understand here is - why does the question mention two such systems? More generally, I don't understand how I'm supposed to envision this scenario.

Simply think of the two systems as a heat source and heat sink between which a heat engine can operate. Heat is taken from the high temperature system, partially converted to work, with the remaining energy transferred to the cold temperature system per the second law, until the two systems reach thermal equilibrium.

The maximum amount of work that can then be obtained using the two systems is that produced by a Carnot heat engine cycle operating between the systems. But since the systems are not thermal reservoirs, an infinite number of Carnot cycles need to be carried out between the two systems, with each subsequent cycle producing less work than the prior cycle, until the two systems reach the same temperature and no further work can be obtained.

why isn't one system enough? why do we need that "the remaining energy transferred to the cold temperature system" ? why can't the engine take heat from the high temperature system, convert some of it to work, and that is that ?

That would violate the Kelvin-Planck statement of the second law, which is:

No heat engine can operate in a cycle while transferring energy with a single heat reservoir

Hope this helps.

Bob D
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  • why isn't one system enough? why do we need that "the remaining energy transferred to the cold temperature system" ? why can't the engine take heat from the high temperature system, convert some of it to work, and that is that ? – Lior Feb 14 '23 at 14:22
  • @Lior See update to my answer – Bob D Feb 14 '23 at 14:27
  • thanks, you added the word "cycle". but how is this related to the question, which does not mention any cycles? – Lior Feb 14 '23 at 14:28
  • @Lior You can produce work by taking in heat from the hot system alone (without the cold system) but it will not be the maximum amount of work that can be produced. This is what happens in the reversible isothermal expansion of the Carnot Cycle. But for it to be reversible, you can only take a tiny amount of heat and produce a tiny amount of work at a time, leaving energy available in the hot system to do more work. – Bob D Feb 14 '23 at 14:39
  • To do more work you need to return to, as they say, the "well" by completing a cycle. The only way to complete the Carnot cycle and produce net work is to reject some heat to the colder system. – Bob D Feb 14 '23 at 14:39
  • I think this is a good example as to what happens when it is taught that "isothermal heat" transport can work. If Lior had been taught that entropy transport can only work when entropy is moved through a temperature gradient and that is $\Delta W= (\theta_{high} - \theta_{low})\Delta S$ he would have never had to ask this question; not that there is anything wrong with the question itself. And I am saying this as somebody who likes Callen's book. – hyportnex Feb 14 '23 at 18:52
  • @BobD thanks, but I cannot accept this as an answer yet. In Callen's book, at this point, neither reversible processes, nor Carnot cycles etc., were introduced yet. I feel that your answer does not exactly address my confusion, although I cannot articulate why. – Lior Feb 14 '23 at 19:20
  • @BobD Maybe this question will help: Would the example in Callen's book even make sense if it talked about only one system? would it make sense if it talked about three systems? Also, as I ask in the main post: Am I imagining the scenario correctly? – Lior Feb 14 '23 at 19:22
  • @hyportnex Maybe this is true, but frankly I don't know what "'isothermal heat' transport" is. I think that my question is more basic: I'm trying to understand how to imagine the scenario described, what is the roll of the different systems, etc. – Lior Feb 14 '23 at 19:24
  • @Lior OK. In that case, at this point in Callen's book has he introduced the first law? – Bob D Feb 14 '23 at 20:05
  • @BobD yes (although it is not called by that name) – Lior Feb 14 '23 at 21:38
  • @Lior By what name did they call it? – Bob D Feb 14 '23 at 21:43
  • No name. It is not listed as a postulate. What Callen says is "Consider some specified process that takes a system from the initial state $A$ to the final state $B$. We wish to know the amount of energy transferred to the system in the form of work and the amount transferred in the form of heat in that particular process. The work is easily measured by the method of mechanics. Furthermore, the total energy difference $U_B-U_A$ is measurable by the procedures discussed in Section 1.7. Subtracting the work from the total energy difference gives us the heat flux in the specified process." – Lior Feb 14 '23 at 22:21
  • @Lior OK. That's all first law related "stuff". – Bob D Feb 14 '23 at 22:37
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There are two factors that haven’t been mentioned yet in the existing (very good) answers: one is an implicit assumption, and one imposes in conjunction a fundamental necessity for two systems/thermal reservoirs (as opposed to only one such system).

  1. Callen’s engine is implicitly assumed to not contain an additional hot or cold reservoir. That is, it is implicitly assumed to have negligible heat capacity. Why? Because it's stated elsewhere in the problem that a net entropy change is allowed in the reservoirs only—not the engine.

    (An alternate assumption, already mentioned, is that the engine is simply returned in the same state it was provided. In that case, for this particular problem with a final temperature of $T_f=\sqrt{T_{10}T_{20}}$ for both reservoirs, calculated here, any heat engine with a non-negligible heat capacity would require that exact initial temperature at the beginning. It's simpler to assume a negligible heat capacity.)

    This is why the responses tend to discuss cyclic operation of the engine: It’s the only reasonable way to extract the maximum work, as tasked. The reader is expected to realize that the engine must collect some thermal energy, output work (among other required actions, specified below), collect some thermal energy, output work, etc. In other words, it cannot reasonably consume the available energy content of the hot reservoir in a single step, as it itself doesn’t have the capacity.

    If one cycle extracts work $W$, then $n$ cycles extract up to $nW$ total work—depending on how large the reservoirs are—which is better. Of course, finite thermal reservoirs will be gradually depleted with continued operation; they themselves are not cyclically regenerated. The final state of a common temperature between the two finite reservoirs is asymptotically approached in the ideal case as the number of cycles increases.

    (Is this a frustrating mix of idealizations—minimal temperature gradients to avoid generating entropy, for example, and a negligible heat capacity—and the reality that an actual heat engine must have a finite size and be driven by finite temperature gradients? Yes; and this is common to thermodynamics thought experiments involving thermodynamic reversibility and Carnot engines. The Carnot engine can't be built—all real processes are thermodynamically irreversible—yet we discuss it constantly. Physical approximations to it would have a power output of essentially zero. When Callen refers to the "maximum work" but a "common [final] temperature," the thermodynamics practitioner must immediately conceive of a quite unusual combination of very large hot and cold reservoirs connected by a very small heat engine, transferring minuscule heat and work very slowly over an arbitrarily large accumulation of cycles. Why go to this cognitive effort in assembling an improbable tower of assumptions, idealizations, and impractical assemblies? The payoff is not having to incorporate entropy generation calculations; entropy transfer calculations through heating and cooling are sophisticated enough at this stage.)

  2. Work doesn’t transfer entropy, but heat does, and entropy can’t be destroyed. This is the fundamental constraint of heat engines. Callen takes care to note that the reservoirs can’t provide work or matter—they can only heat or cool other things. This means that any energy obtained from the hot reservoir brings along entropy, and since entropy can’t be accumulated in cyclic engine operation, it must be dumped in a second location: the cold reservoir. This requirement holds for all heat engines (and does not constrain other engine types).

These two aspects seem to directly address many of your follow-up comments/questions.

(Edit.) A final note: I think you may be trying to ask the question Can a single thermal reservoir produce work in a single step? Yes, if a gradient other than a temperature gradient is available. For example, abandoning Callen's restriction of constant reservoir volume, you can let the reservoir expand slowly into a surrounding vacuum forever, until its volume is arbitrarily large, its temperature is arbitrarily low, and its thermal energy has been converted essentially entirely into work. This specific example is a pressure-difference engine, not a heat engine. And so the Second Law doesn't prohibit this operation, as the entropy increase from expansion counteracts the entropy decrease from cooling. But you can only do it once; external energy is required for recompression. And it's not what Callen aimed to teach in this example problem, although it's independently an instructive thermodynamics thought problem.

Chemomechanics
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  • I see. So it is implicitly assumed that the engine returns to its initial state. Which means that any energy it gained, needs to be removed, partly by heat. Indeed, Callen mentions that "The change in total entropy will occur entirely in the two thermal systems", which hints that the engine returns to its initial state. I still don't quite know why cycles are required to extract maximum work, but given the assumption that the engine is cyclic, at least now I understand why we need two subsystems, and the calculation makes more sense. – Lior Feb 14 '23 at 19:46
  • I'll probably accept this as the answer, I'll first wait to see if there are any additional answers or discussions here in the near future. – Lior Feb 14 '23 at 19:47
  • I edited my answer to clarify quantitatively why cycling produces more total work. – Chemomechanics Feb 14 '23 at 19:51
  • I don't feel like the clarification explains the point. We could say the same argument about any process, I think: "If one non-cyclic process extracts work $W$, then $n$ repetitions of it extract up to $nW$". (I understand that repeating a process requires bringing it back to its initial state. But, the two thermal reservoirs do not return to their initial state - their temperature changes in the process. So how can the entire process be cyclic if these systems are not?) – Lior Feb 14 '23 at 22:32
  • A reservoir is not a process. The engine, not the reservoirs, operates cyclically. If the reservoirs aren't effectively infinite, the temperature difference progressively decreases, as does the efficiency. See also this question. – Chemomechanics Feb 14 '23 at 22:39
  • In our scenario described in Callen, a reservoir is not mentioned (explicitly). Moreover, in the solution, Callen says that "The two thermal systems will be left at some common temperature". Is this final state (of having a common temperature) the end of a single cycle, or the end of many cycles? – Lior Feb 14 '23 at 23:06
  • I'm using "reservoir" (consistently with the consensus meaning) because it's less ambiguous than "system," which could also describe the engine Callen mentions. The final state of a common temperature between the two reservoirs is asymptotically approached in the ideal case as the number of cycles increases. – Chemomechanics Feb 14 '23 at 23:28
  • Do you agree with the following statement? Instead of talking about two thermal systems, Callen equivalently could have talked about a single one - the high temperature system; and the engine itself is the low temperature system. (e.g. the engine contains gas in a piston), with a high heat capacity. in this formulation, the engine does not perform cycles, but work is produced until the engine's temperature becomes equal to the given system's temperature. The analysis of this case would be identical (in particular, the maximum work calculation). – Lior Feb 15 '23 at 12:14
  • Please ask this as a separate question. The space in comment fields isn’t enough to specify the required details. – Chemomechanics Feb 15 '23 at 15:23
  • Ok, although this is the essence of the current question. I would hope that my comment would be addressed in the answer space, not the comments space.... – Lior Feb 15 '23 at 17:10
  • You're asking me to alter Callen's discussion to fit some scenario that you have in mind, but the comments are not an appropriate place for me to try to tease out, at length, all the details you haven't mentioned: Is the engine's heat capacity much larger than the reservoir's heat capacity, or about the same? Since the engine can expand to an arbitrarily large volume, why have you set its final temperature to the "given system's temperature" when it could produce more work by continuing to expand? And so on, and so on. This site simply isn't intended for 50 back-and-forth comments (so far). – Chemomechanics Feb 15 '23 at 18:06
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I assume a heat engine cycle operating in a cycle. The two systems are a source and sink for heat, respectively, for a heat engine. Based on the second law of thermodynamics, heat cannot be converted completely into work; therefore, not all the source heat can be converted into work and some heat must be rejected.

One of the thermodynamics textbooks by Sonntag and Van Wylen provides a clear discussion of heat engines related to the second law.

John Darby
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  • Heat cannot be converted completely into work with an engine operating in a cycle. – Chet Miller Feb 14 '23 at 13:29
  • Yes, I had assumed a cycle. I added this to my answer. Thanks for the clarification. – John Darby Feb 14 '23 at 13:51
  • how is "cycle" relevant here? the question doesn't mention this term. why can't the engine take heat from the source, convert some of it to work, and that is that ? – – Lior Feb 14 '23 at 14:26
  • For a non-cycle. Some, but not all, of the heat can be converted to work. You can add heat and extract work, but the internal energy of the system would increase since all the heat added cannot be converted into work. – John Darby Feb 14 '23 at 15:13
  • thanks, but I cannot accept this as an answer yet. In Callen's book, at this point, cycles were not introduced yet. I feel that your answer does not exactly address my confusion. Maybe these additional questions will clarify my misunderstanding: Would the example in Callen's book even make sense if it talked about only one system? would it make sense if it talked about three systems? Also, as I ask in the main post: Am I imagining the scenario correctly? – Lior Feb 14 '23 at 19:26
  • If this is not a cycle, I think it means what the comment by @Chet Miller says. My earlier comment saying heat added to a system cannot be converted completely to work, is also why for a cycle the system must reject heat to return to its original state (no change internal energy). In my opinion the problem in the book is not clearly stated. – John Darby Feb 14 '23 at 22:27
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More generally, I don't understand how I'm supposed to envision this scenario.

I want to focus on this part. The scenario as described mashes abstract and specific in a very weird way, so I would argue that it cannot really be envisioned at all. (Compare it to envisioning: “An object with momentum $p$ hits John Cleese inelastically.”) So, here are two ways of envisioning it:

The abstract way (as in thermodynamic textbooks)

  1. The high-temperature system transfers heat to a piston, being the main part of your engine.
  2. The gas in the piston expands, doing some work.
  3. The piston transfers heat to the low-temperature system thereby contracting. This may require some work.
  4. Go to Step 1 until the high- and low-temperature systems have thermally equilibrated.

A crucial point here is that the piston starts at some temperature, has a finite heat capacity, and maximum expansion length. Therefore, in Step 2, you cannot just transfer an arbitrary amount of heat and convert it to work. Also, you usually want to return the engine to its original state at the end of the process – which requires a way to cool down the engine. This is why you need the second system. Otherwise your piston would expand once, leaving it in a useless state, and that’s it.

(All of this isn’t mentioned in the quoted scenario, which makes the maximum work infinite, e.g., if you have an infinite piston that is initially filled with an ideal gas at 0 K and does not need to be returned to its original state.)

The practical way

Let’s take the example of a combustion engine:

  • The high-temperature system is not a thermodynamic system at all but chemical energy stored in a fuel mixture.
  • The low-temperature system is the surrounding.
  1. A gas-mixture in a piston is ignited. This transfers energy from the high-temperature “system” to the piston.
  2. The gas in the piston expands, doing some work.
  3. The waste gas is exhausted into the surrounding (heating it up) and exchanged with fresh fuel mixture. This requires some work that is either stored in the engine (e.g., as angular momentum) or provided by a second piston operating in a different cycle.
  4. Go to Step 1.

This may seem like a completely different thing, yet for the thermodynamic budget and efficiency considerations, it makes little difference if you replace Step 3 with something like adiabatic cooling.

Wrzlprmft
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  • Thanks for the great answer. Can you please briefly explain the significance of the piston having a finite heat capacity? – Lior Feb 15 '23 at 10:58
  • or confirm my understanding: if the heat capacity were infinite, then the piston would be able to absorb unlimited amounts of heat without equilibrating with the high-temperature system, and so would be able to produce unbounded amount of work? (if the expansion length is unlimited as well) – Lior Feb 15 '23 at 11:04
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    @Lior: The amount of work would not be unbounded, because energy conservation still applies, but since the piston never equilibrates with the high-temperature system, you would never need to reset it to its original position and in particular not need the low-temperature system to cool it. – Wrzlprmft Feb 15 '23 at 11:47