Qualitative: cooling coffee

Problem

A cup of boiling-hot coffee left on the table cools down to room temperature, but the reverse process (the cold coffee spontaneously warming up by drawing heat from the air) never happens. Explain in five lines why the first process increases the entropy of the universe and the second would decrease it.

Solution

As it cools, the coffee releases heat $Q$ to the air. The coffee is at a high temperature $T_\text{coffee}$, the air at a low temperature $T_\text{air}$: the same amount of heat makes the coffee lose an entropy $Q/T_\text{coffee}$ smaller than the entropy $Q/T_\text{air}$ gained by the air. Hence $\Delta S_\text{univ}=\frac{Q}{T_\text{air}}-\frac{Q}{T_\text{coffee}}>0 \qquad (T_\text{air}<T_\text{coffee})$ the direct process is allowed and spontaneous.

The reverse process would transfer heat from the cold air to the hot coffee: the signs flip and we would have $\Delta S_\text{univ}<0$, forbidden by the second law. Microscopically it would correspond to an extremely improbable conspiracy of air molecules, statistically impossible: that is why it is never observed. $\ev{\text{direct }\Delta S_\text{univ}>0,\ \text{reverse }\Delta S_\text{univ}<0}$

Linked atoms

Topics: Entropia e secondo principio Concepts: Entropia · Secondo principio della termodinamica Skills: Bilancio entropico · Interpretazione micro-macro