JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Heat & Thermodynamics

A refrigerator doing the least possible work runs reversibly, like a Carnot machine, between the freezing point $T_1=273\text{K}$ and the surroundings $T_2=300\text{K}$. The heat drawn from the water as it turns to ice is $Q_1=mL=100\times 80=8000\text{cal}$. For a reversible cycle $\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$, so the heat delivered to the surroundings is $Q_2=Q_1\frac{T_2}{T_1}=8000\times \frac{300}{273}\approx 8791\text{cal}$.