If minimum possible work is done by a refrigerator in converting 100 grams of water at $0^{\circ} \m

Question

If minimum possible work is done by a refrigerator in converting 100 grams of water at $0^{\circ} \mathrm{C}$ to ice, how much heat (in calories) is released to the surroundings at temperature $27^{\circ} \mathrm{C}$ (Latent heat of ice $=80 \mathrm{cal} /$ gram ) to the nearest integer?\

Professor PC Thomas Classes · Answer

(8791) : 100 g water at $0^{\circ} \mathrm{C}$ to ice, $T_{1}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$\ $L_{	ext {ice }}=80 \mathrm{cal} / \mathrm{g}$\ Heat energy extracted from the water,\ $Q_{2}=m L=100 	imes 80=8000 \mathrm{cal}$\ $T_{2}=0^{\circ} \mathrm{C}=273 \mathrm{~K} ; \quad \frac{Q_{1}}{Q_{2}}=\frac{T_{1}}{T_{2}}$\ $Q_{1}=Q_{2} \cdot \frac{300}{273}=1.099$\ $Q_{2}=8000 	imes 1.099=8791 \mathrm{cal}$\