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Assuming the domestic refrigerator as reversible engine working between melting point of ice and the room temperature of 27°C, calculate the energy in Joule that must be supplied to freeze one kg of water. (Given melting point of ice = 0°C L = 80 cal g-1) |
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Answer» Given : T1 = 27 + 273 = 300 K T2 = 0 + 273 = 273 K m = 1 kg = 1000 g; L = 80 cal g-1 Heat to be removed, Q2 = mL = 1000 × 80 cal = 8 × 104 cal From the relation, \(\frac{Q_1}{Q_2}=\frac{T_1}{T_2},\) \(Q_1=\frac{T_1}{T_2}\times Q_2\) \(=\frac{300}{273}\times 8\times10^4\) = 87912.1 cal Energy required to be supplied, W = Q1 − Q2 = (87912.1 – 80,000) cal = 7912.1 cal = 7912.1 × 4.2 J = 33230.8 J |
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