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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)`. |
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Answer» 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` |
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