Change in Entropy — Question 4

Question 4

Heating through a phase transition requires three entropy contributions

A .100 kg sample starts as ice at 253 K and ends as liquid water at 293 K, at fixed pressure. Use melting temperature 273 K, $c_i=2100$ J/(kg K), $c_w=4200$ J/(kg K), and latent heat $L_f=334000$ J/kg. Treat all values as constant in the stated ranges.

1. Calculate the entropy changes for warming ice, melting, and warming water.

2. Find total heat and total entropy change. Explain why the latent-heat term is divided by a fixed temperature rather than integrated over a temperature interval.

Question 4 - Solution

Heating through a phase transition requires three entropy contributions

(a) Method and calculation

$$\Delta S_i=mc_i\ln(273/253)\approx15.9773\,\mathrm{J/K}.$$$$\Delta S_m=mL_f/273\approx122.344\,\mathrm{J/K},\quad\Delta S_w=mc_w\ln(293/273)\approx29.6943\,\mathrm{J/K}.$$

(b) Interpretation and further calculation

$$\Delta S_{\mathrm{total}}\approx168.016\,\mathrm{J/K}.$$$$Q=mc_i(20)+mL_f+mc_w(20)=4200+33400+8400=46000\,\mathrm J.$$An equilibrium phase change at fixed pressure proceeds at its coexistence temperature, so $\int\delta Q_{\mathrm{rev}}/T=Q_{\mathrm{latent}}/T_m$.

Physical check

The large entropy of melting reflects latent heat transfer without a temperature rise.