The temperature inside and outside a refrigerator are $273 \,K$ and $300 \,K$ respectively. Assuming that the refrigerator cycle is reversible, for every joule of work done, the heat delivered to the surrounding will be nearly ........ $J$
Medium
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(a)
$\eta=1-\frac{T_2}{T_1} ; \eta=1-\frac{273}{300}=\frac{9}{100}$
$\beta=\frac{1-\eta}{\eta}=\frac{100}{9}-1=\frac{91}{9}-11 \,J$
$\beta=\frac{Q}{W}$
For $W=1 \,J$
$Q=\beta$
$Q=11 \,J$
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