My question is related to "Temperature in the definition of entropy?". If a system changes its state from $A$ to $B$, we have that

$$ S(B) - S(A) \ge \int_A^B \frac{\delta Q}{T}, \tag{1}$$

where equality holds for reversible paths between $A$ and $B$. According to the linked question, $T$ refers to the temperature of the heat source, not to the temperature of the system.

In Fermi's "Thermodynamics" we find the following example.

"[...] we consider the exchange of heat by thermal conduction between two parts, $A_1$ and $A_2$, of a system. Let $T_1$ and $T_2$ be the temperatures of these two parts, respectively, and let $T_1 < T_2$. Since heat flows by conduction from the hotter body to the colder body, the body $A_2$ gives up a quantity of heat $Q$ which is absorbed by the body $A_1$."

Which is fine so far. He continues, however, with the following.

"Thus, the entropy of $A_1$ changes by an amount of $Q / T_1$, while that of $A_2$ changes by an amount $-Q/T_2$."

Which is, what I do not understand.

First, the reservoir of $A_1$ should be $A_2$ hence, the temperature of the reservoir should be $T_2$, but $T_1$ is used. The same holds for $A_2$.

Second, since the author talks about equality, this process should be reversible. But how could that be done? Clearly, just bringing the two subsystems in contact with each other does not yield a reversible process. One could connect the two systems by a metal rod. Then, at $A_1$ the rod would have temperature $T_1$ and at $A_2$ it would have temperature $T_2$. Thus, leading to the author's calculation, but that means that the author implicitly assumes that the heat is delivered to $A_1$ at temperature $T_1$ and extracted from $A_2$ at temperature $T_2$. Furthermore, it ignores whatever happens in this rod. Is this explanation correct? If so, is it clear from the text, that the heat has to be delivered at said temperatures? Am I missing something?

Third, assume we bring the two systems in contact. Then the reservoir of $A_1$ is at temperature $T_2$ and the sink of $A_2$ is at $T_1$. Thus, $$ S(A_1^\textrm{later}) - S(A_1^\textrm{earlier}) \ge Q / T_2 \quad \text{and} \quad S(A_2^\textrm{later}) - S(A_2^\textrm{earlier}) \ge -Q / T_1 \,. $$ Therefore, the change in total entropy $$ S(A^\textrm{later}) - S(A^\textrm{earlier}) = S(A_1^\textrm{later}) + S(A_2^\textrm{later}) - (S(A_1^\textrm{earlier}) + S(A_2^\textrm{earlier})) \ge Q / T_2 - Q / T_1 \,. $$ Since $T_1 < T_2$ the right-hand side of this inequality is negative. We know that the total entropy has to increase (or at least stay constant). Hence, while this result is not a contradiction (it is just a lower bound), it is a rather bad bound. Nevertheless, is it correct to apply the inequality (1) this way or am I violating any assumptions here?


2 Answers 2


This is a really good question, and, I might add, you are working with a really good book.

To understand what is happening here, you have to understand that the interface between the two reservoirs cannot be at both temperatures at the same time. There is only one temperature at the boundary of each reservoir, and that must be the temperature of the reservoir itself (since an ideal reservoir is assumed to have infinite thermal conductivity). So, to analyze this problem, you need to imagine a thin conductive medium between the two reservoirs, with each surface of the conductive medium in contact with one of the reservoir. So you have heat conduction through the medium as a result of the temperature gradient in the medium. The medium is assumed of have negligible heat capacity, but, because of the temperature gradient, all the entropy generation within the system takes place within the medium, and none of it takes place within the reservoirs themselves. This is because the reservoirs assumed to be ideal, such that their entropy changes are always equal to $Q/T_{res}$. If you do an entropy balance on the intervening medium, it is $$\Delta S_M=\frac{Q}{T_H}-\frac{Q}{T_C}+S_{gen}=0$$where $S_{gen}$ is the entropy generated within the conductive medium, and, from the above equation, it is equal to:$$S_{gen}=\frac{Q}{T_C}-\frac{Q}{T_H}$$The hotter reservoir delivers heat to the conductive medium at the temperature of the conductive medium at the boundary $T_H$, and its entropy change is $$\Delta S_H=-\frac{Q}{T_H}$$ Similarly, the colder reservoir receives heat form the conductive medium at the temperature of the conductive medium at the boundary $T_C$, and its entropy change is $$\Delta S_C=+\frac{Q}{T_C}$$So the entropy change of the combination of two reservoirs and intervening conductive medium is: $$\Delta S=\Delta S_H+\Delta S_M+\Delta S_C=S_{gen}=\frac{Q}{T_C}-\frac{Q}{T_H}$$

So the key to all this is recognizing that the interface between the two reservoirs cannot be at both temperatures at the same time, and that it is therefore necessary of include the thin conductive medium at their junction to make sense out of all this.

Incidentally, I thought I should also mention that the temperature T in the Clausius inequality is not necessarily a reservoir temperature at the system boundary. The system obviously doesn't know what it is in contact with at its boundary, and can't tell the difference between a reservoir and anything else in the surroundings that it is contact with. So, in reality, T is simply the temperature at the boundary with the surroundings through which the heat Q is flowing. Fermi neglects to make this important point in his text.

  • $\begingroup$ Thank you. This explanation helps a lot. What I do not understand is, why is $\Delta S_M$ equal to zero? $\endgroup$
    – H. Rittich
    May 8, 2020 at 8:23
  • 1
    $\begingroup$ Entropy is a physical property of a material (such as the intervening medium), irrespective of any process that the material has suffered. For a solid like the intervening medium, its entropy change is given by $\Delta S=MC\ln{(T_f/T_i)}$, where M is its mass and C is its heat capacity. Since we have specified that the mass and heat capacity of the medium are negligible, so also must the entropy change of the medium (assuming that its initial and final temperatures are somewhere between the two reservoir temperatures). $\endgroup$ May 8, 2020 at 11:09
  • $\begingroup$ Thanks. That helps. $\endgroup$
    – H. Rittich
    May 8, 2020 at 12:31

If the only thing that is happening is heat exchange (i.e., no work), then $dU = dq$ by the First Law. In this case, heat behaves like a state function (it is in fact equivalent to a state function). Which means that it does not matter whether the actual heat exchange is reversible or not -- we can proceed as if the process were reversible (because $dU = dq_\textrm{irrev} = dq_\textrm{rev}$). For reversible processes we can take $T$ to be the temperature of the system in the Clausius inequality. In fact, if the temperature of the body is well-defined (as in a quasi-static process), we can always use the temperature of the body in the Clausius inequality even if the process is irreversible.

Another way of looking at it is to go with your example of a connecting rod. There is a temperature gradient in the rod so that the temperature changes smoothly between $T_2$ on one end to $T_1$ on the other. Take that scenario to the limit, where the rod is an infinitesimally thin wire, so that whatever is happening in the wire is negligible compared to the heat transfer process between bodies $A_1$ and $A_2$. This is a reversible heat exchange, as the heat is transferred through an infinitesimal temperature difference at any point in the wire, and Fermi's result follows.

See for example Denbigh Principles of Chemical Equilibrium page 42, including the footnote.


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