A Carnot engine takes $3\times {10}^6$ cal of heat from a reservoir at $627°C$ and gives it to a sink at $27°\mathrm{C}$ . Find the work done by the engine.

$T_1=627+273=900K$

$T_2=27+273=300K$

$Q_1=3\times {10}^{6}cal$

$\eta =\left(1-\frac{T_2}{T_1}\right)=\left(1-\frac{Q_2}{Q_1}\right)$

$\therefore \frac{Q_2}{Q_1}=\frac{T_2}{T_1}$

$\therefore Q_2=\frac{T_2}{T_1}\times Q_1$

$=\frac{3\times {10}^6\times 300}{900}={10}^{6}cal$

Work done by the engine,

$W=Q_1-Q_2=3\times {10}^6-{10}^{6}cal$

$=2\times {10}^{8}cal$