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In a refrigerator one removes heat from a lower temperature and deposits to the surrounding at a higher temperature. In this process, mechanical work has to be done, which is provided by an electric motor. If the motor is of 1 kW power, and heat is transferred from −3°C to 27°C, find the heat taken out of the refrigerator per second assuming its efficiency is 50 % of a perfect engine. |
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Answer» Given : T1 = 27°C = 27 + 273 = 300 K T2 = −3°C = −3 + 273 = 270 K Efficiency, η = 1 – \(\frac{T_2}{T_1}\) = 1 - \(\frac{270}{300}\) = 1 – 0.9 = 0.1 = \(\frac{1}{10}\) Efficiency of refrigerator is 50% of percentage engine. η' = 50% of η = 0.5 ×\(\frac{1}{10}\)=\(\frac{1}{20}\) ∴ Coefficient of performance, \(β=\frac{Q_2}{W}=\frac{1-η'}{η'}\) \(β=\frac{1-\frac{1}{20}}{\frac{1}{20}}=\frac{1-0.05}{0.05}\) = \(\frac{0.95}{0.05}\) = 19 Q2 = 19% of work done by motor on refrigerator = 19 × 1 = 19 K J/S |
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