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 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

Given, temperature of the source is `27^(@)` C
`rArr" "T_(1)=(27+273)K=300K`
Temperature of sink `" "T_(2)=(-3+273)K=270K`
Efficiency of a perfect heat engine is given by
`" "eta=1-(T_(2))/(T_(1))=1-(270)/(300)=(1)/(10)`
Efficiency of refrigerator is 50% of a perfect engine
`therefore" "eta'=0.5xxeta=(1)/(2)eta=(1)/(20)`
`therefore` Coeffcient of performance of the refrigerator
`beta=(Q_(2))/(W)=(1-eta')/(eta')`
`=(1-(1//20))/((1//20))=(19//20)/(1//20)=19`
`rArr" "Q_(2)=betaW=19W" "(becausebeta=(Q_(2))/(W))`
Therefore, heat is taken out of the refrigerator at a rate of 19 kJ per second.