ΔS of ice melting

Problem

$100\;\text{g}$ of ice at $0\;\text{°C}$ melts completely (latent heat of fusion $L_f=334\;\text{kJ/kg}$). Calculate $\Delta S$ of the ice.

Solution

Melting occurs at constant temperature $T=273\;\text{K}$, so the heat absorbed is $Q=mL_f$ and the entropy change is $\Delta S=\frac{Q}{T}=\frac{mL_f}{T}$. Substituting $m=0{,}1\;\text{kg}$ and $L_f=334000\;\text{J/kg}$: $\Delta S=\frac{mL_f}{T}=\frac{0{,}1\cdot 334000}{273}=122\;\text{J/K}$ $\ev{\Delta S\approx 122\;\text{J/K}}$

Links

Topics: Entropy and the second law Concepts: Entropy · Temperature Skills: Entropy balance · Calorimetric equation