Answer
$ W_{\text {net,out }}=63.8\text{ kJ}$
Work Step by Step
The minimum pressure in the cycle is $P_3$ and the maximum pressure is $P_1$. $$
\frac{T_2}{T_3}=\left(\frac{P_2}{P_3}\right)^{(k-1) / k}
$$ or $$
P_2=P_3\left(\frac{T_2}{T_3}\right)^{k /(k-1)}=(20\ \mathrm{kPa})\left(\frac{1100 \mathrm{~K}}{300 \mathrm{~K}}\right)^{1.4 / 0.4}=1888\ \mathrm{kPa}
$$ The heat input is determined from $$
\begin{aligned}
& s_2-s_1=c_p \ln {\frac{T_2}{T_1}}^{\nexists 0}-R \ln \frac{P_2}{P_1}=-(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \frac{1888 \mathrm{kPa}}{3000 \mathrm{kPa}}=0.1329 \mathrm{~kJ} / \\
& Q_{\text {in }}=m T_H\left(s_2-s_1\right)=(0.6 \mathrm{~kg})(1100 \mathrm{~K})(0.1329 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})=87.73 \mathrm{~kJ}
\end{aligned}
$$ Then, $$
\begin{aligned}
& \eta_{\text {th }}=1-\frac{T_L}{T_H}=1-\frac{300 \mathrm{~K}}{1100 \mathrm{~K}}=0.7273=72.7 \% \\
& W_{\text {net,out }}=\eta_{\text {th }} Q_{\text {in }}=(0.7273)(87.73 \mathrm{~kJ})=\mathbf{6 3 . 8} \mathbf{k J}
\end{aligned}
$$
