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 Cal/gram) to the nearest integer ?
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 Cal/gram) to the nearest integer ?
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Minimum work means a reversible (Carnot) refrigerator, so \(Q_h/Q_c = T_h/T_c\). Heat removed from the water is \(mL\); use absolute temperatures.
Updated On: Nov 5, 2026
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Correct Answer: 8791
Solution and Explanation
Heat extracted from water to freeze it: \(Q_c = mL = 100 \times 80 = 8000\) cal.
Minimum work corresponds to a reversible (Carnot) refrigerator working between \(T_c = 273\) K (water at \(0^\circ\mathrm{C}\)) and \(T_h = 273+27 = 300\) K (surroundings).
For a Carnot cycle \(\frac{Q_h}{T_h} = \frac{Q_c}{T_c}\), so \(Q_h = Q_c \cdot \frac{T_h}{T_c} = 8000 \times \frac{300}{273}\).
\(Q_h = 8791.2\) cal \(\approx 8791\) cal.
Answer: 8791
Minimum work corresponds to a reversible (Carnot) refrigerator working between \(T_c = 273\) K (water at \(0^\circ\mathrm{C}\)) and \(T_h = 273+27 = 300\) K (surroundings).
For a Carnot cycle \(\frac{Q_h}{T_h} = \frac{Q_c}{T_c}\), so \(Q_h = Q_c \cdot \frac{T_h}{T_c} = 8000 \times \frac{300}{273}\).
\(Q_h = 8791.2\) cal \(\approx 8791\) cal.
Answer: 8791
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