A Carnot engine takes 5000kcal of heat from a reservoir at 727 ° C...

Given,

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

${T}_{2}\left(Sink\right) = {127}^{°}C = \left(127 + 273\right)K = 400K$

${T}_{1}\left(Source\right) = {727}^{°}C = \left(727 + 273\right)K = 1000K$

The efficiency of the Carnot cycle is given by,

$η = 1 - \dfrac{{T}_{sink}}{{T}_{source}} = \dfrac{W}{Q}$

$⇒ \left(1 - \dfrac{400}{1000}\right) = \dfrac{6}{10} = \dfrac{W}{5000}$$⇒ W = 3000kcal$

$= 3 \times \times 4 . 2 \times {10}^{6}J = 12 . 6 \times {10}^{6}J$.