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A Carnot's engine works as a refrigerator between $250 \mathrm{~K}$ and $300 \mathrm{~K}$. It receives $500 \mathrm{cal}$ heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:
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The correct answer is:
$420 \mathrm{~J}$
$420 \mathrm{~J}$
Given: Temperature of cold body, $\mathrm{T}_2=250$ $\mathrm{K}$ temperature of hot body; $\mathrm{T}_1=300 \mathrm{~K}$ Heat received, $Q_2=500 \mathrm{cal}$ work done, $\mathrm{W}=$ ?
Efficiency $=1-\frac{T_2}{T_1}=\frac{W}{Q_2+W}$
$\Rightarrow 1-\frac{250}{300}=\frac{W}{Q_2+W}$
$\mathrm{~W}=\frac{Q_2}{5}=\frac{500 \times 4.2}{5} J=420 \mathrm{~J}$
Efficiency $=1-\frac{T_2}{T_1}=\frac{W}{Q_2+W}$
$\Rightarrow 1-\frac{250}{300}=\frac{W}{Q_2+W}$
$\mathrm{~W}=\frac{Q_2}{5}=\frac{500 \times 4.2}{5} J=420 \mathrm{~J}$
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