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 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.

Explanation

Temperature of source is `27^@C`
`therefore T_1=27+273`=300 K
Temperature of sink `T_2=-3+273`=270 K
The efficiency of a perfect engine ,
`eta=1-T_2/T_1`
`therefore eta=1-270/300=30/300=0.1`
The efficiency of a refrigerator ,
`eta.`=50% efficiency of a perfect engine,
`=0.1xx50/100`
=0.05
Coefficient of performance of a refrigerator ,
`beta=Q_2/W=(1-eta.)/(eta.)`
`therefore beta=(1-0.05)/0.05 =0.95/0.05`=19
`therefore Q_2=betaW`
=19 x 1 kW
=19 kW
`=19 "kJ"/s`
Therefore , heat is taken out of the refrigerator at a rate of 19 kJ per second.