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I am reading the second edition of Callen's Thermodynamics and am a little confused about an example he presents. The example is outlined below and can be found on page 101.

In the simplest case we consider the transfer of heat $\delta Q$ from one system of at temperature $T$ to another at the same temperature. Such a process is reversible, the increase in entropy of the recipient subsystem $\delta Q/T$ being exactly counterbalanced by the decrease in entropy $-\delta Q /T$ of the donor system.

Something I did not consider while writing my first thread about the same example (but a different question), that I have since then started thinking about, is how such a transfer would even be possible. If the systems start at identical temperatures, and we move an infinitesimal quantity of heat $\delta Q$ between them, they will no longer be in equilibrium since the temperatures are now infinitesimally different from each other.

Callen shows earlier in his book that a characterization of two simple systems that can exchange energy being in equilibrium is that the temperatures have to be precisely equal, that is, $T^{(1)}=T^{(2)}$. This is the state where the composite system has maximum entropy. So in the example outlined above, if we were to move $\delta Q$ of heat between the systems, we would change the temperatures, so would we not also decrease entropy since $T^{(1)}\ne T^{(2)}$? Callen says that total entropy is not affected but I don't understand how this can be.

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    $\begingroup$ Supplying heat does not necessarily change the temperature, because, for example, the system might expand. $\endgroup$ Commented 9 hours ago
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    $\begingroup$ @BobD I gave you a physical example which supports my point that in general, temperature difference is not necessary for the heat transfer. Also the whole Carnot theory is based on this. Surveys of textbooks don't matter in this respect, only arguments based on facts and established theories(classical thermodynamics). Textbooks authors can take various approaches, better or worse, and one of them may be focusing on particulars, such as the heat conduction equation/Newton's law of cooling, without giving the general concept of heat in classical thermodynamics clearly. $\endgroup$ Commented 6 hours ago
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    $\begingroup$ @BobD Definition of heat in classical thermodynamics does not address mechanism of how/why it occurs. It does not matter; the theory is not a causal story. Temperature difference may be and often is regarded as the cause of heat transfer in practice, but that is irrelevant here. There need not be any cause for heat transfer, the theory works perfectly well without it. $\endgroup$ Commented 4 hours ago
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    $\begingroup$ @BobD no, definition of heat in classical thermodynamics does not have to include any temperature differences. On the contrary, the whole theory is based on the idea that heat transfer is possible between bodies of the same temperature, and is used in Carnot/Clausius arguments. You're just completely off on this point. Some engineering/undergraduate sources may be wrong too, if they elevate particular features of heat such as Newton's law of cooling/Fourier's law into definition of heat, and thus mislead about the general thermodynamics concept; textbooks contain errors, that's nothing new. $\endgroup$ Commented 4 hours ago
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    $\begingroup$ @BobD yes, the Carnot engine with its working medium and the reservoir it is connected to having the same temperature is an idealization. That does not mean it is wrong or that heat definition the whole theory is based on should be altered to contain temperature differences. That would introduce an unnecessary specialized concept(not all heat transfers happen in presence of temperature difference) and complicate the Carnot/Clausius arguments. $\endgroup$ Commented 3 hours ago

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The physics as presented in Callen is more than good enough, but you might be confused about the rigorous argumentation that is devoid of the rigorous mathematics.

One trick to this is to be extremely precise with the mathematics. Fill in the rigorous mathematics, if you will. This is done in Ian Ford's textbook on statistical thermodynamics.

You can easily show for the case of the classical ideal gas, that the entropy generation is quadratic in the temperature difference. This means that, after summing an infinitely many linear temperature steps, the resulting entropy generation is still too small, i.e. sum/integrates to zero. Because of that, pretending that there is no temperature difference is a safe thing to pretend.

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  • $\begingroup$ What I am confused about is how such a process, as described above, would even be permissable given Callen's postulates. Callen's second postulate implies that two subsystems are in thermal equilibrium if and only if their temperatures are exactly equal to each other. If such is the case, then the system is in a state of maximum entropy. If we make any changes to any of the extensive variables (such as moving an infinitesimal quantity of heat), should we not have $dS < 0$, since we the extensive paramters are no longer such as to maximize the entropy? $\endgroup$ Commented 9 hours ago
  • $\begingroup$ That's why my answer is the one that is much more helpful; you need to realise that the mathematical modelling that we can rigorously write down are idealisations of actual physical processes. $\endgroup$ Commented 8 hours ago
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how such a transfer would even be possible.

In the theory(classical thermodynamics), such reversible transfer is possible because it is an axiom or a consequence of other axioms. In the situation considered, the bodies are assumed to have constant temperature, unaffected by the heat transfer.

This axiom is not unrealistic, when we recall experience with heat transfers in the world. It is true that if we have two containers with gas of constant volume, then if one gas loses heat, its temperature decreases. But for given amount of heat, this decrease can be made as small as desired, by using lots of gas with high enough heat capacity. And the decrease can be made zero, if the system giving off heat is compressed just right.

Temperature of both bodies with constant volume can be also maintained by another body with the same temperature, with immense heat capacity; this way, the change of temperature due to finite heat transfer can be made arbitrarily small.

Or the bodies can be phase mixtures, e.g., liquid water with ice floating in it. When such system gives off/accepts heat, its temperature does not change at all. The lost heat comes from the liquid part that freezes (giving off latent heat of water freezing) while no change of temperature occurs, and this heat may go into the other system, where it melts some ice, again without change of temperature.

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  • $\begingroup$ Ah, I mistakenly thought the system discussed were closed (by Callen’s terminology. I think this means isolated in usual terminology). If such were the case, then the heat transfer discussed could not be the case, right? That is, if we had a closed system consisting of two simple subsystems of identical temperature, then such a heat flow would not be permitted, right? $\endgroup$ Commented 14 hours ago
  • $\begingroup$ @Anna Heat flow can be prevented by an insulating wall, but if that is not present, then heat flow would be permitted even then. The change of temperature can be made arbitrarily small by using a large enough system. Arbitrarily small is then equivalent to 0, and this limiting case is what that part of theory describes. $\endgroup$ Commented 14 hours ago
  • $\begingroup$ Perhaps I am not understanding the mathematical formalism that is being used here. The way I am thinking about these "infinitesimals" is in the formalism of hyperreal numbers. As such, even if the heat capacity of each system were infinite, the temperatures would still change given an infinitesimal exchange of heat, albeit infinitesimally. Is this the wrong theory of infinitesimals for thermodynamics? $\endgroup$ Commented 11 hours ago
  • $\begingroup$ @Anna With heat capacity $C$, $dT = \frac{dQ}{C}$. If $C$ is infinite, then $dT=0$. $\endgroup$ Commented 8 hours ago
  • $\begingroup$ @Anna Infinite heat capacity occurs, e.g., in the case of phase mixture of liquid water and ice. This system does not change its temperature at all when giving off heat. A gas in a container has finite $C$, and does change its temperature. But the theory, when talking about reversible transfer of heat, refers to the limiting case $C = \infty$, and this ideal case is a very good approximation to a real heat transfer between two real gases, if the heat transfer is very small, and their temperature change only negligibly. $\endgroup$ Commented 8 hours ago
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To solve the apparent difficulty, we have to recall Callen's definition of heat. His definition is the same as that proposed by Caratheodory and used by Planck and Born at the beginning of the last century:

The exchanged head $Q$ is given by $$ Q = \Delta U - W, $$ where $W$ is the work exchanged in the transformation and $\Delta U$ the change of internal energy defined as the work of an adiabatic transformation between the same initial and final states. In other words, the heat transferred is the difference between the adiabatic work and the actual work on the chosen transformation.

In such a definition, there is no explicit reference to macroscopic temperature differences. Of course, at the microscopic level, we may have a mechanism based on local differences of temperature, but at the macroscopic (thermodynamic) level, we do not need to invent weird mechanisms to explain the heat transfer under an isothermal transformation. That's the main advantage of the Caratheodory approach to heat.

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  • $\begingroup$ Hello. I am having some trouble understanding how your answer relates to my question. Are you saying that the definition of heat makes no mention of macroscopic temperature differences, so that heat could move regardless of if there is a difference in temperature? $\endgroup$ Commented 7 hours ago
  • $\begingroup$ @Anna Indeed, you've got it. It's similar to how in mechanics, for a motion of a body to happen, non-zero net force is not needed. It is is needed in usual cases due to friction, but it is not needed in principle, as evidenced by experiments with reduced friction, or celestial motions or Earth satellite motions. $\endgroup$ Commented 7 hours ago
  • $\begingroup$ @Anna Yes, if you go back to Callen's definition of heat, you won't find any explicit reference to temperature. $\endgroup$ Commented 6 hours ago
  • $\begingroup$ @JánLalinský, so to be clear, a heat exchange occuring does NOT necessarily mean that temperatures will be affected? That is, as this heat is transferred, we can imagine it being done in such a way that the temperatures don't change at all, not even infinitesimally, such that $T^{(1)}=T^{(2)}$ always and entropy doesn't change? $\endgroup$ Commented 2 hours ago
  • $\begingroup$ @Anna exactly, that's how a reversible heat transfer is described in classical thermodynamics, and this is the basis on which the usual formulations of the Carnot/Clausius arguments stand. In addition to the mechanical example, maybe an electric example is ever closer: a transfer of charge(current) from one circuit element terminal to another element terminal connected to it does not imply there is a potential gradient between those terminals. People are confused by this because they think in terms of Ohm's law, but Ohm's law does not hold in general. $\endgroup$ Commented 1 hour ago
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What we consider as equilibrium is just a state close enough to equilibrium, as it takes an infinite time to reach equilibrium. Classically. As if we consider quantum effects, there would be fluctuations from equilibrium even then. These fluctuation are infinitely bigger than infinitesimal values.

A difference between system states means a finite value of the difference of some parameters. But infinitesimal value is smaller than ANY given value by definition (all in absolute values). Therefore there is no difference.

Experimentally, if we measure system parameters, and if their values do not differ from those expected in equilibrium significantly in statistical terms, we consider such a system being in equilibrium. Such evaluation has very wide margins, if we consider infinitesimal values.

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  • $\begingroup$ Doesn't that mean that the criterion of thermal equilibrium, $T^{(1)}=T^{(2)}$, is incorrect even in theory? Of course if were to make real measurements, we could not measure any difference between two temperatures that are infinitesimally close to each other. However, the theory demands that the temperatures be precisely equal at equilibrium, does it not? $\endgroup$ Commented 10 hours ago
  • $\begingroup$ The theory demands that the temperature difference is smaller than any given number. That is not the same as zero. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ But Callen derives, on page. 43, that for two systems in thermal equilibrium, it must be the case that $T^{(1)}=T^{(2)}$. Indeed, he does not make any remark that these may differ in any way, not even infinitesimally. That seems like an extremely important detail to include if it were the case. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ Well, equilibrium does not exist. Not even theoretically. Just its theoretical and practical approximation. $\endgroup$ Commented 10 hours ago
  • $\begingroup$ In Callen's framework equilibria do exist by one of the postulates. I am asking my questions with regards to his book and his postulates, not reality. I apologize if this was not clear in my original question. $\endgroup$ Commented 10 hours ago
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The theory of thermodynamics is based on the concept of the thermodynamic limit, where we may assume infinitely large systems and heat baths at fixed temperatures with infinite heat capacity. Then it is allowed for systems to exchange heat without changing temperature. Since $$\Delta Q = C \Delta T$$ where $C$ is the heat capacity, which is proportional to the system size, we have $$\Delta T = \Delta Q / C \to 0$$ when the system size goes to infinity.

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