The temperature inside and outside of refrigerator are $260\, K$ and $315\, K$ respectively. Assuming that the refrigerator cycle is reversible, calculate the heat delivered to surroundings for every joule of work done.
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$\mathrm{T}_{2}=260 \mathrm{K} . \mathrm{T}_{1}=315 \mathrm{K} \quad ; \mathrm{W}=1$ joule
Coefficient of performance of Carnot
refrigerator $\beta=\frac{\mathrm{Q}_{2}}{\mathrm{W}}=\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}-\mathrm{T}_{2}}$
$\therefore \frac{\mathrm{Q}_{2}}{1}=\frac{260}{315-260}=\frac{260}{55} \Rightarrow$
$\mathrm{Q}_{2}=\frac{260}{55}=4.73 \mathrm{J}$
$\mathrm{Q}_{1}=\mathrm{Q}_{2}+\mathrm{W}=5.73$
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