Answer
$Q_{31 \text { out }} =271.7\text{ kJ}$
$\eta_{\text {th }}=34.7\%$
Work Step by Step
$(b)$ The temperature at state 2 and the heat input are $$
\begin{aligned}
T_2=T_1\left(\frac{P_2}{P_1}\right)^{(k-1) / k}=(300 \mathrm{~K})\left(\frac{1000 \mathrm{kPa}}{100 \mathrm{kPa}}\right)^{0.4 / 1.4}=579.2 \mathrm{~K} \\
Q_{\text {in }}=m\left(h_3-h_2\right)=m c_p\left(T_3-T_2\right) \\
416 \mathrm{~kJ}=(0.5 \mathrm{~kg})(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(T_3-579.2\right) \longrightarrow T_3=1407.1 \mathrm{~K}
\end{aligned}
$$ Process $3-1$ is a straight line on the $P-v$ diagram, thus the $w_{31}$ is simply the area under the process curve, $$
\begin{aligned}
w_{31} & =\operatorname{area}=\frac{P_3+P_1}{2}\left(v_1-v_3\right)=\frac{P_3+P_1}{2}\left(\frac{R T_1}{P_1}-\frac{R T_3}{P_3}\right) \\
& =\left(\frac{1000+100 \mathrm{kPa}}{2}\right)\left(\frac{300 \mathrm{~K}}{100 \mathrm{kPa}}-\frac{1407.1 \mathrm{~K}}{1000 \mathrm{kPa}}\right)(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \\
& =251.4 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ Energy balance for process 3-1 gives $$
\begin{aligned}
E_{\text {in }}-E_{\text {out }} & =\Delta E_{\text {system }} \longrightarrow-Q_{31 \text { out }}-W_{31 \text { out }}=m\left(u_1-u_3\right) \\
Q_{31 \text { out }} & =-m w_{31 \text { out }}-m c_v\left(T_1-T_3\right)=-m\left[w_{31 \text { out }}+c_\nu\left(T_1-T_3\right)\right] \\
& =-(0.5 \mathrm{~kg})[251.4+(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})(300-1407.1) \mathrm{K}] \\
& =271.7 \mathrm{~kJ}
\end{aligned}
$$ (c) The thermal efficiency is then $$
\eta_{\text {th }}=1-\frac{Q_{\text {out }}}{Q_{\text {in }}}=1-\frac{271.7 \mathrm{~kJ}}{416 \mathrm{~kJ}}=0.347=34.7 \%
$$
