If minimum possible work is done by a refrigerator in converting 100 grams of water at 00C to ice, how much heat (in calories) is released to the surroundings at temperature 270C (Latent heat of ice = 80 cal/gram) to the nearest integer ?

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answer is 8791.00.

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Detailed Solution

Qabs=mLQabs=100(80)calQabs=8000calNow, CoP=Qabsω=QabsQrej-Qabs=TsourceTsink-Tsource=273300-273⇒8000Qrej-8000=10.11⇒Qrej=8791.21⇒Qrej≈8791

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