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If you suppose to put in contact two ideal heat sources at the temperatures $T_{1}$ and $T_2$ with $T_{1}<T_{2}$ you obtain a strange thermodynamics process. If you consider just one of the two sources as a system (for example the one at the temperature $T_{1}$), its temperature stays constant during the whole process, so you can think it does a reversible process (it is always in equilibrium because the temperature is constant and well defined). But if you compute the entropy of the universe: $$\Delta S_{source1}+\Delta S_{source2}= Q[\frac{1}{T_{1}}-\frac{1}{T_{2}}]=\Delta S_{u}>0$$ which means that there must be an irreversible process inside the universe. So the system or the environment must do an irreversible process, but since they both do a reversible process (the enviromet does the same process of the system), it is born a paradox. I think that the paradox is due to the fact that ideal sources of heat don't exist, is it true? In the real world the two temperature can't stay constant.

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    $\begingroup$ No, the transfer of heat through a finite temperature difference is always irreversible. The sources are not in thermal equilibrium with each other. $\endgroup$
    – knzhou
    Jul 29, 2018 at 9:17
  • $\begingroup$ Yes, but you can consider just one source as our tdm system. In this case the system is always at the equilibrium because its temperature stays constant. So the system does a reversible process. @knzhou $\endgroup$
    – Landau
    Jul 29, 2018 at 11:41
  • $\begingroup$ I edit the question, hope it will be more clear. $\endgroup$
    – Landau
    Jul 29, 2018 at 11:43
  • $\begingroup$ That's simply not true: a process is not guaranteed to be reversible if all the pieces of a system were individually in equilibrium. $\endgroup$
    – knzhou
    Jul 29, 2018 at 13:17
  • $\begingroup$ I have defined the system only as one of the two reservoire. In this very particular case my system is in equilibrium despite it is in contact with another body with a different temperature. @knzhou $\endgroup$
    – Landau
    Jul 29, 2018 at 16:43

2 Answers 2

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There are several aspects to what you are asking, so let's break it down into bite sized pieces.

A key characteristic of an ideal reservoir is that its mass times heat capacity is virtually infinite. So let's first focus on a single reservoir and determine what happens if its mass times heat capacity, rather than being infinite, is very large, but finite. In such a case, the change in entropy of the reservoir, when it receives and amount of heat Q is given by:

$$\Delta S=mC\ln{(T_f/T_i)}\tag{1}$$ with $$Q=mC(T_f-T_i)\tag{2}$$If we combine these two equations to eliminate the final temperature $T_f$, we obtain: $$\Delta S=mC\ln{\left[1+\frac{Q}{mCT_i}\right]}\tag{3}$$If we next expand the logarithmic term in this equation in a Taylor series, we obtain: $$\Delta S=\frac{Q}{T_i}\left[1-\frac{1}{2}\frac{Q}{mCT_i}+...\right]\tag{4}$$ In the limit of mC becoming infinite, this approaches just $Q/T_i$. So, even though the temperature of the reservoir has changed only infinitesimally, because of its infinite heat capacity times mass, this has enabled it to experience a finite entropy change.

But this is not the whole story. We can also get an idea of the amount of entropy generation that takes place when we put a hot body in contact with a cold body and allow them to exchange heat. But, first let's consider a single body at temperature $T_i$, and we change the temperature at its boundary to a new constant (higher) value of $T_B$, and then allow a finite amount of heat transfer Q to take place, after which we allow the body to re-equilibrate thermally to a new temperature $T_f$. The relationship between this final temperature and the heat transferred Q is again given by Eqn. 2. Now, according to the Clausius inequality, the change in entropy of the body and the heat transferred at the boundary is given by $$\Delta S=mC{(T_f/T_i)}>\frac{Q}{T_B}$$We can change this into an equality by writing:$$\Delta S=mC{(T_f/T_i)}=\frac{Q}{T_B}+\sigma$$where $\sigma$ is the amount of entropy generated in the transition from the initial- to the final state of the system. If we again use Eqn. 2 to eliminate the final temperature from the equations and take the limit as mC becomes infinite, we obtain generated in the process: $$\sigma=\frac{Q}{T_i}-\frac{Q}{T_B}>0$$The entropy generated is positive because Q is positive and $T_B>T_i$.

If we had done this same analysis for heat transferred out of a hot body, Q would have been negative and the boundary temperature would have been less than the initial reservoir temperature. So, in this case again, the amount of entropy generated would have been positive.

If we have two identical cubes of material, one at temperature $T_H$ and the other of temperature $T_C$ and we place them into contact with one another, the temperature at the interface between the two bodies cannot be $T_H$ and $T_C$ simultaneously. Our understanding of heat conduction tells us that, during the time that the cubes are in contact, the interface temperature will be the average of the two initial temperatures. So if, during the time the bodies are in contact, an amount of heat Q is transferred, the entropy generated in each of the two bodies up to the final state where they each have equilibrated thermally will be given, in the limit of infinite mC, by:

$$\sigma_H=-\frac{Q}{T_H}+\frac{2Q}{(T_H+T_C)}=\frac{Q}{T_H}\frac{(T_H-T_C)}{(T_H+T_C)}$$ $$\sigma_C=\frac{Q}{T_C}-\frac{2Q}{(T_H+T_C)}=\frac{Q}{T_C}\frac{(T_H-T_C)}{(T_H+T_C)}$$ So positive entropy is generated in both cubes during the irreversible transfer of heat from the hot cube to the cold cube.

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A heat reservoir in contact with a system is a boundary condition on the system. One of the requirements in any consistent analysis is that the boundary conditions not be contradictory. You can analyze two idealized voltage sources (i.e., idealized batteries) connected via a resistor but if the resistor has zero resistance then the sources are directly connected to each other and that problem cannot be solved within the confines of Kirchhoff's equations that do not allow such connections. A voltage source is a boundary condition on the circuit and having two different voltage sources between two different points are contradictory prescriptions on those same points. It is not the physics, here described by Kirchhoff's 1st and 2nd (nodal and loop) laws, but your idealized boundary conditions that are contradictory.

If you still wish to analyze the problem then you must introduce some resistance of the connecting wires, or internal resistance of the batteries, or sparks, radiation, or whatever. The same goes for connecting two heat sources. All real heat sources have finite capacities, their surfaces are not ideal, the emissivity is not constant, the temperature changes during connection and it is not uniform throughout, etc. If you ignore these when directly connecting them you get contradictions in your calculation when in fact your idealization is in error.

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  • $\begingroup$ So you think that the paradox exists just because bodies with $C\rightarrow\infty$ don't exist ? $\endgroup$
    – Landau
    Jul 29, 2018 at 16:45
  • $\begingroup$ That's not right. The fact that $\Delta S > 0$ when heat flows from hot to cold isn't a paradox or a "contradiction" at all. It's just basic thermodynamics. $\endgroup$
    – knzhou
    Jul 29, 2018 at 16:58
  • $\begingroup$ @knzhou Indeed, it is basic thermodynamics and is not a paradox that when heat flows form hot to cold $\Delta S >0$ but it is wrong to say that since the process involves only thermostats (infinite heat capacities) so their processes must be reversible as Landau stated "but since they both do a reversible process (the environment does the same process of the system), it is born a paradox". The "paradox" is there because of the inconsistent boundary conditions. $\endgroup$
    – hyportnex
    Jul 29, 2018 at 17:23
  • $\begingroup$ @Chester_Miller showed in detail that if you assume finite heat sources your "paradox" goes away; it is all about idealization. Either you consider only more realistic sources and then you have a much more complicated but realistic situation (finite size, finite conductivity, etc.), or you restrict your sources, i.e., idealize their properties and use them as boundary conditions but then you must have rules how to couple them and thus avoid inconsistent assumptions. $\endgroup$
    – hyportnex
    Jul 29, 2018 at 17:31

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