Melting of 1 kg of ice: ΔS

Problem

One kilogram of ice at $0\ ^\circ$C melts reversibly. Calculate the entropy change of the ice (latent heat of fusion $\lambda_\text{fus} = 3{,}33\cdot 10^5$ J/kg).

Solution

Melting takes place at constant temperature $T = 273$ K, so all the heat absorbed is exchanged at that temperature. The heat needed for melting is $Q = m\lambda_\text{fus}$, and the entropy change is $\Delta S = \frac{Q}{T} = \frac{m\lambda_\text{fus}}{T} = \frac{1\cdot 3{,}33\cdot 10^5}{273} = 1{,}22\cdot 10^3\;\text{J/K}$ $\ev{\Delta S = +1{,}22\cdot 10^3\;\text{J/K}}$ The positive sign indicates that, going from the ordered solid state to the more disordered liquid state, the system increases its own entropy.

Collegamenti

Argomenti: Entropy and the second law Concetti: Entropy · Latent heat and changes of state · Temperature Competenze: Calorimetric equation · Entropy balance