Question

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

Answer 1

8791

Answer 2

1. Calculate the heat absorbed from the cold reservoir ($Q_C$) as the latent heat of fusion: $Q_C = \text{mass} \times \text{latent heat of fusion} = 100 \, \text{g} \times 80 \, \text{Cal/g} = 8000 \, \text{Cal}$.
2. Convert temperatures to Kelvin: $T_C = 0^\circ C = 273 \, K$ and $T_H = 27^\circ C = 300 \, K$.
3. For minimum work (ideal Carnot refrigerator), the heat released to the hot reservoir ($Q_H$) is given by: $Q_H = Q_C \times \frac{T_H}{T_C}$
4. Substitute values and calculate: $Q_H = 8000 \, \text{Cal} \times \frac{300 \, K}{273 \, K} = 8000 \times \frac{100}{91} \, \text{Cal} \approx 8791.2 \, \text{Cal}$.
5. Rounding to the nearest integer gives 8791 Cal.