A Carnot freezer takes heat from water at 0,^ | vedclass

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(D) The heat removed from the water (sink) is given by $Q_{sink} = mL = 5 	imes 336 	imes 10^3 = 1.68 	imes 10^6\,J$.For a Carnot refrigerator,the

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$1.67 	imes 10^5\,J$

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(D) The heat removed from the water (sink) is given by $Q_{sink} = mL = 5 	imes 336 	imes 10^3 = 1.68 	imes 10^6\,J$.For a Carnot refrigerator,the coefficient of performance is given by $\beta = \frac{T_{sink}}{T_{source} - T_{sink}} = \frac{Q_{sink}}{W}$.Given temperatures: $T_{sink} = 0 + 273 = 273\,K$ and $T_{source} = 27 + 273 = 300\,K$.Substituting the values: $\frac{273}{300 - 273} = \frac{1.68 	imes 10^6}{W}$.$\frac{273}{27} = \frac{1.68 	imes 10^6}{W}$.$W = \frac{1.68 	imes 10^6 	imes 27}{273} \approx 1.6615 	imes 10^5\,J$.Rounding to the nearest option,the energy consumed is $1.67 	imes 10^5\,J$.

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