A Carnot engine takes $3\times {10}^6$  cal of heat from a reservoir at ${627}^{0}C$  and gives it to a sink at ${27}^{0}C$ . The work done by the engine is

$T_1=627+273=900K$

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

$T_2=27+273=300K$

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

$\Rightarrow Q_2=\frac{T_2}{T_1}\times Q_1=\frac{300}{900}\times 3\times {10}^6$

$=1\times {10}^6$  cal

Work done $=Q_1-Q_2$

$=3\times {10}^6-1\times {10}^6=2\times {10}^6$  cal

$=2\times 4.2\times {10}^{6}J=8.4\times {10}^{6}J$