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?
(8791)
Given,
Heat absorbed, $Q_{2}=m L=80 \times 100=8000 \mathrm{Cal}$
Temperature of ice, $T_{2}=273 \mathrm{~K}$
Temperature of surrounding,
$T_{1}=273+27=300 \mathrm{~K}$
Efficiency $=\frac{w}{Q_{2}}=\frac{Q_{1}-Q_{2}}{Q_{2}}=\frac{T_{1}-T_{2}}{T_{2}}=\frac{300-273}{273}$
$\Rightarrow \frac{Q_{1}-8000}{8000}=\frac{27}{273} \Rightarrow Q_{1}=8791 \mathrm{Cal}$
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