Find the change in the entropy in the following process 100 gm of ice at 0°C melts when dropped in a bucket of water at 50°C (Assume temperature of water does not change)
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Gain of entropy of ice
$s_1 = \frac{\Delta Q}{T} = \frac{\mathrm{mL}}{T} = \frac{80 \times 100}{(0 + 273)} = \frac{8 \times 10^3}{273} \text{ cal/K}$
Loss of entropy of water $= S_2 = -\frac{\Delta Q}{T} = -\frac{\mathrm{mL}}{T}$ $= \frac{80 \times 100}{(273 + 50)} = \frac{8 \times 10^3}{323} \text{ cal/K}$ Total change of entropy $S_1 + S_2 = \frac{8 \times 10^3}{273} - \frac{8 \times 10^3}{323} = +4.5 \text{ cal/K}$
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