A refrigerator is driven by a 1HP motor having an efficiency of 80%. The refrigerator works between `0^(@)C and 38^(@)C`. Calculate the time required by the refrigerator to freeze 1 litre of water, heat of fusion of ice is `2.26xx10^(5)Jkg^(-1)`.

Explanation

Available power `=(80)/(100)xx746.6=592.8W`
Heat draw from 1 litre of water `(m=1kg)`
`Q_(2)=mL=1xx2.26xx10^(5)J`
i.e., `Q_(2)=2.26xx10^(5)J` (cold reservoir)
`T_(1)=38+273=311K=T_(2)=273K`
We know that `(Q_(1))/(Q_(2))=(T_(1))/(T_(2)) " " Q_(1)=(T_(1))/(T_(2))xxQ_(2)=(311)/(273)xx2.26xx10^(5)`
`Q_(1)=2.574xx10^(5)J`
Heat rejected to the hot surroundings. `=2.574xx10^(5)J`
The chemical work to be by the compressor is
`W=Q_(1)-Q_(2)=(2.574.226)J`
`=0.314xx10^(5)J`
but `W=Pt`
i.e., `t=(W)/(P)=(0.314xx10^(5))/(592.8)`
`t~~53.0s`