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Suppose we have three systems made up of the same amount of the same substance, $S_1, S_2$ and $S_3$. They start with temperatures $T_1$, $T_2$, $T_3$ such that $T_1>T_2>T_3$.

We can place the systems next to each other along a line: $S_1 | S_2 |S_3$, where $|$ indicates that the systems on either side of it are in contact.

Let us assume that this line of systems, taken as a whole, can be treated as isolated. Within the line itself, however, heat and work may be exchanged between any two systems in contact. The three systems therefore begin to get into thermodynamic equilibrium with each other.

At the very start of the process, $S_1$ loses heat $\delta Q_{12}$ to $S_2$, and $S_2$ loses heat $\delta Q_{23}$ to $S_3$. The systems' entropies change by $dS_1$, $dS_2$ and $dS_3$ respectively.

I would like to show that $dS_1 + dS_2 + dS_3 > 0$, using the Clausius inequality. In other words, I would like to show that the Second Law of Thermodynamics has as a consequence that the entropy for the whole line of systems increases as the line reaches internal thermodynamic equilibrium.

The Clausius inequality automatically gives $d S_1 \geq \frac{-\delta Q_{12}}{T_2}$ and $d S_3 \geq \frac{\delta Q_{23}}{T_2}$, because $S_1$ and $S_3$ are both only in contact with $S_2$, which is at temperature $T_2$.

But can I use the Clausius inequality to complete my argument and say: $d S_2 \geq \frac{\delta Q_{12}}{T_1} + \frac{-\delta Q_{23}}{T_3}$? Can/how would the Clausius inequality can be applied when a system is in contact with two reservoirs of different temperatures?

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You incorrectly applied the Clausius inequality to this problem. The temperatures that must be used in the denominators are the values at the interfaces between 1 and 2 and between 2 and 3 (see Moran et al, Fundamentals of Engineering Thermodynamics). So the correct application of the Clausius inequality should look like this: $$\Delta S_1\geq\int{\frac{dQ_{21}}{T_{12}}}$$ $$\Delta S_2\geq\int{\frac{dQ_{12}}{T_{12}}}+\int{\frac{dQ_{32}}{T_{23}}}$$ $$\Delta S_3\geq\int{\frac{dQ_{23}}{T_{23}}}$$where $T_{12}$ is the temperature at the interface between systems 1 and 2, and $T_{23}$ is the temperature at the interface between systems 2 and 3, with $$dQ_{12}=-dQ_{21}$$and$$dQ_{23}=-dQ_{32}$$So, when you add these up, you get (as expected) $$\Delta S_1+\Delta S_2+\Delta S_3\geq 0$$

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  • $\begingroup$ Hi Chet. If the three substances are thermal reservoirs, what will be the temperatures at the interfaces? The mean? $\endgroup$ – Bob D Oct 17 at 11:01
  • $\begingroup$ @BobD This is an instance in which the world of ideal reservoirs is in conflict with the real world. For real reservoirs that approach ideal behavior, there are temperature gradients near the boundaries (where entropy generation is taking place), and the temperature is continuous at the boundary, displaying a boundary value part-way between the bulk average temperatures of the reservoirs. In the case of an ideal reservoir in contact with, say, an ideal gas, all the temperature variation and entropy generation is assumed to take place within the gas, and the temp at the boundary is that of $\endgroup$ – Chet Miller Oct 17 at 15:06
  • $\begingroup$ the reservoir is the same as the bulk. No temperature gradients are allowed near the ideal reservoir boundary in this case. The reservoir is assumed to have infinite thermal conductivity compared to the gas. $\endgroup$ – Chet Miller Oct 17 at 15:08
  • $\begingroup$ Chet, thanks for the explanation. Would my answer be correct if I stated in the second assumption that the substances are "ideal" thermal reservoirs? $\endgroup$ – Bob D Oct 17 at 21:01
  • $\begingroup$ @BobD Actually, I disagree with several things in your answer, starting with with the first assumption. For the energy balances on the three reservoirs, I have $$\Delta U_1=Q_{21}$$ $$\Delta U_2=Q_{12}+Q_{32}$$ $$\Delta U_3=Q_{23}$$with $Q_{21}=-Q_{12}$ and $Q_{32}=-Q_{23}$ If we add the three changes in internal energy, we obtain (as expected) $$\Delta U_1+\Delta U_2+\Delta U_3=0$$But none of this implies that $Q_{12}=Q_{23}$. That would mean that $\Delta U_2=0$, which is certainly not a necessary condition. $\endgroup$ – Chet Miller Oct 18 at 1:10
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I don't know if this helps you "complete your argument", consider the following involving only heat transfer between the substances:

  1. For an isolated system, system energy must be conserved, so the heat exiting $S_1$ equals the heat entering $S_3$. Consequently

$Q_{12}$ = $Q_{23}$ = Q

  1. Assume $S_1$, $S_2$, and $S_3$ are thermal reservoirs, so that the heat transfers occur isothermally.

  2. The changes in entropy for each reservoir are

$$\Delta S_{1}=-\frac{Q}{T_1}$$ $$\Delta S_{2}=+\frac{Q}{T_2}-\frac{Q}{T_2}$$ $$\Delta S_{3}=+\frac{Q}{T_3}$$

  1. Total entropy change of the isolated system is

$$\Delta S_{Total}=\Delta S_{1}+\Delta S_{2}+\Delta S_{3}$$

$$\Delta S_{Total}=-\frac{Q}{T_1}+\frac{Q}{T_2}-\frac{Q}{T_2}+\frac{Q}{T_3}$$

$$\Delta S_{Total}=\frac{Q}{T_{3}}-\frac{Q}{T_{1}}$$

For all $T_{1}>T_3$, $\Delta S_{Total}>0$

If $T_{1}\to T_3$ (reversible heat transfer) $\Delta S_{Total}= 0$

Therefore $\Delta S_{Total}≥0$

Hope this helps.

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