A refrigerator is driven by `1000 W` electric motor having an efficiency of `60%`. The refrigerator is considerd as a reversible heat engine operating between `273K and 303K`. Calculate the time required by it to freeze `32.5 kg` of water at `0^(@)C`. Heat losses may be rejected. Latent heat of fusion of ice `= 336xx10^(3)J kg^(-1)`.

Explanation

Power of motor `= 1000 W`
Efficiency of motor `= 60%`
`:.` Actual power availiable `= (60)/(100)xx1000W`
`=600W`
If the refrigerator take `t` seconds to freeze `32.5 kg` of water, then energy supplied `= 600xxt "joule"`.
Heat drawn from sink of freeze `32.5 kg` of water
`Q_(2)= mL= 32.5xx336xx10^(3)J`
`=1.092xx10^(7)J`
As `(Q_(2))/(Q_(1))= (T_(2))/(T_(1))`
`:. Q_(1)= (T_(1))/(T_(2))xxQ_(2)= (303)/(273)xx1.092xx10^(7)J`
`=1.212xx10^(7)J`
As `W=Q_(1)-Q_(2)`
`:. W= (1.212-1.092) 10^(7)J= 0.12xx10^(7)J`
As energy supplied = work done
`:. 600 t= 0.12xx10^(7)`
`t=(0.12xx10^(7))/(600)= 2000s= 33 min 20 s`