Worked exercise — why heat flows from hot to cold

Problem

A thermostat 1 at $T_1 = 400\;\text{K}$ and one 2 at $T_2 = 300\;\text{K}$ are put in contact. $Q$ joules flow from the first to the second. Show that this is compatible with the second law only if $Q \geq 0$ (i.e. from hot to cold).

Solution

Setup. Let $Q$ be the heat that passes from thermostat 1 to thermostat 2. Thermostat 1 gives up $Q$, so it receives $-Q$; thermostat 2 receives $+Q$. For each we use the thermostat formula $\Delta S = Q_\text{ricevuto}/T$.

Entropy of the two thermostats: $\Delta S_1 = -\frac{Q}{400}, \qquad \Delta S_2 = +\frac{Q}{300}$

Total entropy: $\Delta S_\text{tot} = -\frac{Q}{400} + \frac{Q}{300} = Q\left(\frac{1}{300}-\frac{1}{400}\right) = \frac{Q}{1200}$

Conclusion. The second law requires $\Delta S_\text{tot} \geq 0$, i.e. $Q/1200 \geq 0$, and therefore: $\ev{Q \geq 0}$

Heat can only flow from the hot thermostat to the cold one. This is a genuine proof of the Clausius statement (“heat does not pass spontaneously from cold to hot”) starting from the entropic formulation of the second law alone. The factor $\left(\frac{1}{300}-\frac{1}{400}\right)$ is positive precisely because $T_2 < T_1$: it is the difference between the temperatures that fixes the allowed direction.

Collegamenti

Argomenti: Entropy and the second law Concetti: Second law of thermodynamics · Entropy Competenze: Entropy balance