A refrigerator takes heat from water at `0^(@)C` and transfer it to room at `27^(@)C` . If `100 kg` of water is converted in ice at `0^(@)C` then calculate the work done. (Latent heat of ice `3.4xx10^(5) J//kg`)

Explanation

Coefficient of performance `(COP) = (T_(2))/(T_(1)-T_(2)) = (273)/(300-273) = (273)/(27)`
` W= (Q_(2))/(COP) = (mL)/(COP) = (100xx 3.4xx10^(5))/(273//27) = (100 xx 3.4 xx 10^(5) xx27)/(273) = 3.36 xx 10^(7)J`