Answer
$\frac{dQ_H}{dt} = 4664.17 W$
Work Step by Step
To find the rate of heat input, $\frac{dQ_H}{dt}$, we need to first identify the efficiency,
$\epsilon = \frac{T_H -T_L}{T_H} $
$\epsilon = \frac{373 K -333 K}{373K} $
$\epsilon = 0.1072$
The rate of heat input is
$\frac{dQ_H}{dt} = \frac{1}{\epsilon} \frac{dW}{dt} $
And $ \frac{dW}{dt} $ is the power which is 500W
$\frac{dQ_H}{dt} = \frac{1}{0.1072}500W $
$\frac{dQ_H}{dt} = 4664.17 W$
