A Carnot engine takes 5000   kcal of heat from a reservoir at 727 ° C and gives heat to a sink at…

Given,

 $Q\left(\mathrm{Heat}\right)=5000 \mathrm{kcal}=5\times {10}^6 \mathrm{cal}$

$T_2\left(\mathrm{Sink}\right)={127}^{°}\mathrm{C}=\left(127+273\right)\mathrm{K}=400 \mathrm{K}$

$T_1\left(\mathrm{Source}\right)={727}^{°}\mathrm{C}=\left(727+273\right)\mathrm{K}=1000 \mathrm{K}$

The efficiency of the Carnot cycle is given by,

$\eta =1-\frac{T_{\mathrm{sink}}}{T_{\mathrm{source}}}=\frac{W}{Q}$

$\Rightarrow \left(1-\frac{400}{1000}\right)=\frac{6}{10}=\frac{W}{5000}$$\Rightarrow W=3000 \mathrm{kcal}$

$=3\times \times 4.2\times {10}^{6 }\mathrm{J}=12.6\times {10}^{6 }\mathrm{J}$.