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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.
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.
In a refrigerator one removes heat from a lower temperature and deposits to the surroundings 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 1kW 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.
In a refrigerator one removes heat from a lower temperature and deposits to the surroundings 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 1kW 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.
Here, power of motor, `W= 1kW`
`T_(1)= 27^(@)C= (27+273)K= 300 K, T_(2)= -3^(@)C= (-3+273)K= 270K`
`eta = 1-(T_(2))/(T_(1))= 1-(270)/(300)= 1/(10)`.
Efficiency of refrigerator `= 0.5 eta= 1/(20)`
`COP= beta=(Q_(2))/W= (1-eta)/(eta)= (1-1/20)/(1//20)=19`
`Q_(2)= 19W = 19W :.` Heat taken out of refrigerator/sec `= Q_(2)= 19kW`
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