Answer
$\eta_{\text {th }}=0.390$
Work Step by Step
From the steam tables (Tables A-4E, A-5E, and A-6E),$$
\begin{aligned}
h_1 & =h_{f @ 5 \text { psia }}=130.18\ \mathrm{Btu} / \mathrm{lbm} \\
v_1 & =v_{f @ 5 \text { psia }}=0.01641\ \mathrm{ft}^3 / \mathrm{lbm} \\
w_{\text {p,in }} & =v_1\left(P_2-P_1\right) \\
& =\left(0.01641 \mathrm{ft}^3 / \mathrm{lbm}\right)(2500-5) \mathrm{psia}\left(\frac{1 \mathrm{Btu}}{5.404 \mathrm{psia} \cdot \mathrm{ft}^3}\right) \\
& =7.58\ \mathrm{Btu} / \mathrm{lbm} \\
h_2 & =h_1+w_{\text {p,in }}=130.18+7.58=137.76\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ $$
\begin{aligned}
& \left.\begin{array}{l}
P_4=5\ \mathrm{psia} \\
x_4=0.80
\end{array}\right\} \begin{array}{l}
h_4=h_f+x_4 h_{f g}=130.18+(0.80)(1000.5)=930.58\ \mathrm{Btu} / \mathrm{lbm} \\
s_4=s_f+x_4 s_{f g}=0.23488+(0.80)(1.60894)=1.52203 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}
\end{array} \\
& \left.\begin{array}{l}
P_3=2500\ \mathrm{psia} \\
s_3=s_4=1.52203\ \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}
\end{array}\right\} \begin{array}{l}
h_3=1450.8\ \mathrm{Btu} / \mathrm{lbm} \\
T_3=\mathbf{9 8 9 . 2}^\circ \mathrm{F}
\end{array}
\end{aligned}
$$ Thus, $$
\begin{aligned}
q_{\text {in }} & =h_3-h_2=1450.8-137.76=1313.0\ \mathrm{Btu} / \mathrm{lbm} \\
q_{\text {out }} & =h_4-h_1=930.58-130.18=800.4\ \mathrm{Btu} / \mathrm{lbm}
\end{aligned}
$$ The thermal efficiency of the cycle is $$
\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{800.4}{1313.0}=\mathbf{0 . 3 9 0}
$$
