A reservoir is at 827 ° C and Carnot's engine takes a thousand...

Given, ${Q}_{1} = {10}^{6}cal$

${T}_{1} = 827 ° C = ( 827 + 273 ) = 1100K$,

${T}_{2} = 27 ° C = ( 27 + 273 ) = 300K$.

As, $\dfrac{Q1}{T1} = \dfrac{Q2}{T2}$

$∴ {Q}_{2} = \dfrac{T2}{T1}{Q}_{1} = \dfrac{300}{1100} \times {10}^{6}= 2. 72 \times {10}^{5}cal$
work done, $W = {Q}_{1} - {Q}_{2} = {10}^{6} - 2. 72 \times {10}^{5} = 7. 27 \times {10}^{5}cal$

The efficiency of the engine,
$η = \left(1 - \dfrac{T2}{T1}\right) \times 100$

$⇒ η = \left(1 - \dfrac{300}{1100}\right) \times 100$

$= 72. 72$