A Carnot's engine works as a refrigerator between $250\, K$ and $300\, K$. It receives $500\, cal$ heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is ..... $J$
JEE MAIN 2018, Medium
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Given: Temperature of cold body, $T_2= 250\,K$ temperature of hot body; $T_1 = 300\, K$ Heat received, $Q_2= 500\, cal$ work done, $W =$ ?
Effociency$ = 1 - \frac{{{T_2}}}{{{T_1}}} = \frac{W}{{{Q_2} + W}}$
$ \Rightarrow 1 - \frac{{250}}{{300}} = \frac{W}{{{Q_2} + W}}$
$W = \frac{{{Q_2}}}{5} = \frac{{500 \times 4.2}}{5}\,\,J = 420\,J$
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