If minimum possible work is done by a refrigerator in converting $100\; grams$ 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 $=80 Cal / gram$ ) to the nearest integer?
JEE MAIN 2020, Medium
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$W + Q _{1}= Q _{2}$
$W = Q _{2}- Q _{1}$
$C.O.P. =\frac{ Q _{1}}{ W }=\frac{ Q _{1}}{ Q _{2}- Q _{1}}=\frac{273}{300-273}=\frac{ Q _{1}}{ W }$
$W =\frac{27}{273} \times 80 \times 100 \times 4.2$
$Q _{2}= W+Q_{1}$
$Q _{2}=\frac{27}{273} \times 80 \times 100 \times 4.2+80 \times 100 \times 4.2$
$Q _{2}=\frac{300}{273} \times 80 \times 100=8791.2 cal$
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