Answer
a) $ \eta =0.4 =40\%$
b) $P_H =400W$
c) $P_C = 666.67W$
Work Step by Step
a) We have
$\eta = 1-\frac{T_C}{T_H}$
$\eta =1- \frac{300}{500}$
$ \eta =0.4 =40\%$
b)
Now $ \eta =\frac{W}{Q_H}$
Dividing numerator and denominator on right side by $t$, we get
$0.4 = P_{out}/P_H$
where $P_H$ is rate of heat input
$ P_H = 0.4* P_{out}$
$P_H = 0.4* 1000W$
$P_H =400W$
c) For carnot engine we have
$\frac{Q_H}{Q_C} =\frac{T_H}{T_C}$
$\frac{Q_H}{Q_C} =\frac{500K}{300K}$
$ Q_H = 5/3Q_C$
Dividing both sides by $t$ we get
$P_H = 0.6 * P_C$
where $P_C $ is rate of heat output
$ 400W = 0.6 * P_C$
$P_C = 666.67W$
