Heat from hot to cold: proof

Problem

Two thermostats at temperatures $T_1 > T_2$ are put in thermal contact and exchange an amount of heat $Q$. Starting from the formulas $\Delta S_\text{term} = Q/T_\text{term}$, show that the condition $\Delta S_\text{tot} \ge 0$ implies that heat $Q$ flows from the hot thermostat to the cold one. Specify the limiting case where $\Delta S_\text{tot} = 0$.

Solution

Let $Q$ be the heat that passes from the hot thermostat to the cold one. The hot thermostat gives up $Q$, so it receives $-Q$ and its entropy changes by $-Q/T_1$; the cold thermostat receives $+Q$ and its entropy changes by $+Q/T_2$. The total change is $\Delta S_\text{tot} = -\frac{Q}{T_1} + \frac{Q}{T_2} = Q\,\frac{T_1 - T_2}{T_1 T_2}$

Since $T_1 > T_2$ and absolute temperatures are positive, the factor $\dfrac{T_1 - T_2}{T_1 T_2}$ is strictly positive. So the sign of $\Delta S_\text{tot}$ coincides with the sign of $Q$: $\Delta S_\text{tot} \ge 0 \iff Q \ge 0$ The second law, by requiring $\Delta S_\text{tot} \ge 0$, forces $Q \ge 0$: heat indeed flows from the hot thermostat to the cold one, never spontaneously the other way round.

Limiting case: $\ev{\Delta S_\text{tot} = 0}$ holds only if $Q = 0$ (no exchange) or in the limit $T_1 \to T_2$, i.e. reversible contact between bodies at infinitesimally close temperatures.

Collegamenti

Argomenti: Entropy and the second law Concetti: Entropy · Second law of thermodynamics · Temperature Competenze: Entropy balance · Limiting-case and thought-experiment analysis