Q 11-11-144JEE MainJEE Main 2020 (3 Sep, Shift 2)Medium
If minimum possible work is done by a refrigerator in converting $100$ grams of water at $0^\circ\text{C}$ to ice, how much heat (in calories) is released to the surroundings at temperature $27^\circ\text{C}$ (Latent heat of ice $= 80\ \text{cal/gram}$) to the nearest integer?
Numerical value type. Enter your answer.
Answer: 8791
Heat removed from the water: $Q_2 = 100\times80 = 8000$ cal at $T_2 = 273$ K.
Minimum work means a reversible (Carnot) refrigerator, for which $\dfrac{Q_1}{Q_2} = \dfrac{T_1}{T_2}$:
$$Q_1 = 8000\times\frac{300}{273} \approx 8791\ \text{cal}$$
Solution by Sreeraj P, M.Sc Physics