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

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Sol: T_{1} = 27 °C = 300 K , L_{ice} = 80 cal/g

Heat energy extracted from water , Q_{2} = m L

Q_{2} = 100 x 80 = 8000 cal

T_{2} = 0 °C = 273 K

$\displaystyle \frac{Q_1}{Q_2} = \frac{T_1}{T_2}$

$\displaystyle Q_1 = Q_2 \times \frac{T_1}{T_2} $

$\displaystyle Q_1 = 8000 \times \frac{300}{273} $

= 8791 cal