Ice melting: entropy of the universe

Problem

An ice cube melts spontaneously in a glass of water at room temperature. Does the entropy of the universe in this process increase, decrease or stay unchanged? Justify the answer by referring to the second law of thermodynamics.

Solution

Let us take as the thermodynamic “universe” the set formed by the ice and the surrounding room-temperature water: this set constitutes an isolated system, which exchanges neither heat nor work with the outside.

The melting of the ice in the warmer water happens spontaneously, and every spontaneous process in an isolated system is irreversible. The second law then requires the total entropy to increase: $\ev{\Delta S_\text{tot} > 0}$ so the entropy of the universe INCREASES.

In terms of the balance: the ice absorbs heat and its entropy grows by an amount $\Delta S_\text{ice} = +\int \dd Q / T_\text{low}$, evaluated at the low temperatures of the ice. The surrounding water releases the same amount of heat and its entropy decreases by $\Delta S_\text{water} = -\int \dd Q / T_\text{high}$, but evaluated at the higher temperatures of the environment. Since the same heat is released at a high $T$ and received at a lower $T$, the positive term prevails in magnitude over the negative one: the net balance is positive.

Linked atoms

Topics: Entropia e secondo principio Concepts: Entropia · Secondo principio della termodinamica Skills: Bilancio entropico