If minimum possible work is done by a refrige | vedclass

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(C) For a refrigerator,the coefficient of performance is given by $COP = \frac{T_1}{T_2 - T_1} = \frac{Q_1}{W}$,where $T_1$ is the temperature of the

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$8791$

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(C) For a refrigerator,the coefficient of performance is given by $COP = \frac{T_1}{T_2 - T_1} = \frac{Q_1}{W}$,where $T_1$ is the temperature of the cold reservoir and $T_2$ is the temperature of the hot reservoir.Given: $T_1 = 0^{\circ} C = 273 \; K$,$T_2 = 27^{\circ} C = 300 \; K$.The heat to be removed from the water is $Q_1 = m 	imes L = 100 \; g 	imes 80 \; cal/g = 8000 \; cal$.Using the relation $\frac{Q_1}{W} = \frac{T_1}{T_2 - T_1}$,we have $W = Q_1 	imes \frac{T_2 - T_1}{T_1} = 8000 	imes \frac{300 - 273}{273} = 8000 	imes \frac{27}{273} \approx 791.21 \; cal$.The heat rejected to the surrounding is $Q_2 = Q_1 + W = 8000 + 791.21 = 8791.21 \; cal$.Rounding to the nearest integer,we get $8791 \; cal$.

If minimum possible work is done by a refrigerator in converting $100 \; g$ of water at $0^{\circ} C$ to ice,how much heat (in calories) is released to the surrounding at temperature $27^{\circ} C$ (Latent heat of ice $= 80 \; cal/g$)? Round to the nearest integer.

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