# Irreversibility of heat transfer between two reservoirs

In a closed system the entropy change is given by $\Delta s=s_{gen}+\int \delta Q/T$. If $s_{gen}=0$, the state change is called reversible.

Why is a heat flux between a hot and a cold reservoirs called irreversible? I understand that $\Delta s>0$, but why should there be a positive $s_{gen}$?

## 2 Answers

The formula you have written applies to a system, with $S_{gen}$ being the entropy generated in the interior of the system, and $\int~dQ/T$ being the entropy flux across the system's boundary. To apply this equation to the example of heat exchange $Q$ between two heat reservoirs at temperatures $T_1$ and $T_2$, with $T_1>T_2$, you must first define your system. If you take the hot reservoir or cold reservoir as your system, then indeed there is no entropy generated within their interior, and only flux term remains (equal to $-Q/T_1$ or $+Q/T_2$ depending on your choice of system). However if you take both of them together as a system, then flux term will become zero (because heat flux is then internal) while entropy generation term will be non-zero (equal to $Q/T_2-Q/T_1$).

For any process to be labeled as irreversible, entropy change not of any particular system but that of the entire universe (system+surroundings) must be considered. Obviously, for the universe flux terms are always zero (there is nothing else to exchange heat with), and any increase in entropy is always $S_{gen}$.

P.S. Any isolated system is a universe unto itself.

If you had a conductive wall (with negligible thermal inertia) between the two reservoirs that was conducting heat from the ideal hot reservoir to the ideal cold reservoir, then all the entropy generation would have taken place within the wall, and none would have taken place within either of the ideal reservoirs. You could make this wall as thin and conductive as you wish, but the essence of the physical picture would not change: the ideal hot reservoir would all be at Th and the ideal cold reservoir would all be at Tc, and all the entropy generation would take place within the wall.

So, $$\Delta S_h=-Q/T_h$$ $$\Delta S_c=Q/T_c$$ $$\Delta S_w=Q/T_h-Q/T_c+s_{gen}=0$$and $$\Delta S_{universe}=\Delta S_h+\Delta S_c+\Delta S_w=s_{gen}=Q/T_c-Q/T_h$$

But now suppose you somehow placed the two reservoirs into direct contact with one another (without a wall). In this case, you could no longer have two ideal reservoirs at uniform temperatures, because the interface between them would have to be at a temperature somewhere between the hot and cold temperatures. So placing them into direct contact would invalidate the assumption of ideal reservoirs. In the case of real reservoirs with stirring, thermal boundary layers would develop in proximity to the interface between the two reservoirs; these thermal boundary layers would take the place of the wall. The bulk of the entropy generation would be taking place within these boundary layers. The remainder of the reservoirs would still be behaving as essentially ideal.