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Question

If the minimum possible work is done by a refrigerator in converting 100 grams of water at 0 C to ice, how much heat (in calories) is released to the surroundings at temperature 27 C (Latent heat of ice = 80 Cal/gram) to the nearest integer?

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Solution

Heat absorbed is given by,
Q2=mL=80×100=8000 Cal

Temperature of ice, T2=0 C=273 K

Temperature of surrounding, T1=273+27=300 K

Coefficient of performance
α=WQ2=Q1Q2Q2

As minimum work is done, this can be considered as ideal refrigerator,

Q1Q2Q2=T1T2T2

Q180008000=27273

Q1=8791 Cal

Hence, 8791 is the correct answer.

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