Answer
$q_{\text {in }, 23} =217.4Btu/lbm$
$\eta_{\text {th }}=26.8\%$
Work Step by Step
(b) The temperature at state 2 and the heat input are $$
\begin{aligned}
q_{\text {in }, 12} & =u_2-u_1=c_v\left(T_2-T_1\right) \\
300 \mathrm{Btu} / \mathrm{lbm} & =(0.171 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})\left(T_2-540\right) \mathrm{R} \\
T_2 & =2294 \mathrm{R} \\
\frac{P_2 v_2}{T_2} & =\frac{P_1 v_1}{T_1} \longrightarrow P_2=\frac{T_2}{T_1} P_1=\frac{2294 \mathrm{R}}{540 \mathrm{R}}(14.7 \mathrm{psia})=62.46 \mathrm{psia} \\
q_{\text {in }, 23} & =h_3-h_2=c_P\left(T_3-T_2\right)=(0.24 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(3200-2294) \mathrm{R}=217.4\ \text{Btu/lbm}
\end{aligned}
$$ Process 3-4 is isentropic: $$
\begin{aligned}
T_4 & =T_3\left(\frac{P_4}{P_3}\right)^{(k-1) / k}=(3200 \mathrm{R})\left(\frac{14.7 \mathrm{psia}}{62.46 \mathrm{psia}}\right)^{0.4 / 1.4}=2117 \mathrm{R} \\
q_{\text {in }} & =q_{\text {in }, 12}+q_{\text {in }, 23}=300+217.4=\mathbf{5 1 7 . 4} \mathrm{Btu} / \mathrm{lbm} \\
q_{\text {out }} & =h_4-h_1=c_p\left(T_4-T_1\right)=(0.240 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R})(2117-540)=378.5\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ (c) The thermal efficiency is then $$
\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{378.5 \mathrm{Btu} / \mathrm{lbm}}{517.4 \mathrm{Btu} / \mathrm{lbm}}=\mathbf{2 6 . 8} \%
$$
