In a carnot engine, the temperature of reservoir is 527 ° C and...

Efficiency of Carnot engine is $η = 1 - \dfrac{{T}_{L}}{{T}_{H}} = \dfrac{W}{Q}$, where, $Q$ is heat absorbed.

Here, ${T}_{H} = 527 + 273 = 800K$

Now, $η = 1 - \dfrac{200}{800} = \dfrac{12}{Q}$

$∴ \dfrac{3}{4} = \dfrac{12000}{Q}$

$∴ Q = 16 \times {10}^{6}J$