Question 2.46: A reversible engine with 45% thermal efficiency discharges 1...

\begin{aligned}T_2 &=30+273=303 \ K \\Q_2 &=1600 \ kJ / min \\\eta &=\frac{W}{Q_1}=\frac{W}{W+Q_2}=1-\frac{T_2}{T_1} \\0.45 &=\frac{W}{W+1600}=1-\left\lgroup\frac{303}{T_1}\right\rgroup \\T_1 &=\frac{303}{0.55}=550.9 \ K \text { or } 277.9^{\circ} C\end{aligned}

Also 0.45 W + 720 = W

\begin{aligned}W &=\frac{720}{0.55}=1309.09 \ kJ / min \\\text { Power } &=\frac{1309.09}{60}=21.82 \ kW\end{aligned}