A Carnot engine takes $6000 \,cal$ of heat from a reservoir at $627^{\circ} C$ and gives it to a sink at $27^{\circ} C$. The work done by the engine is ......... $kcal$
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(c)
$T_1=627+273=900\,K$
$Q_1=6000\,Cal$
$T_2=27+273=300\,K$
From Carnot Theorem $\frac{Q_1}{T_1}=\frac{Q_2}{T_2}$
$Q_2=\frac{T_2}{T_1} \times Q_1$
$=\frac{300}{900} \times 6000$
$=2000\,cal$
Work Done $=Q_1-Q_2$
$=6000-2000$
$=4000\,Cal$
$=4\,Kcal$
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