The temperatures inside and outside of a refr | vedclass

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(D) Given: Inside temperature $T_{2} = 260 \ K$,Outside temperature $T_{1} = 315 \ K$,Work done $W = 1 \ J$.For a reversible refrigerator,the coeffici

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$5$

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(D) Given: Inside temperature $T_{2} = 260 \ K$,Outside temperature $T_{1} = 315 \ K$,Work done $W = 1 \ J$.For a reversible refrigerator,the coefficient of performance $\beta$ is given by $\beta = \frac{Q_{2}}{W} = \frac{T_{2}}{T_{1} - T_{2}}$.Substituting the values: $\beta = \frac{260}{315 - 260} = \frac{260}{55} \approx 4.73$.Since $\beta = \frac{Q_{2}}{W}$,we have $Q_{2} = \beta 	imes W = 4.73 	imes 1 = 4.73 \ J$.The heat delivered to the surroundings $(Q_{1})$ is the sum of the heat extracted from the inside $(Q_{2})$ and the work done $(W)$: $Q_{1} = Q_{2} + W$.$Q_{1} = 4.73 \ J + 1 \ J = 5.73 \ J$.

The temperatures inside and outside of a refrigerator are $260 \ K$ and $315 \ K$ respectively. Assuming that the refrigerator cycle is reversible,calculate the heat delivered to the surroundings for every joule of work done. (in $.73$)

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