Answer
$
\eta_{\text {th }}=0.375$
Work Step by Step
From the steam tables (Tables A-4, A-5, and A-6),
$$
\begin{aligned}
& P_1=P_{\text {sat } @ 40^\circ \mathrm{C}}=7385\ \mathrm{kPa} \\
& P_2=P_{\text {sat } @ 300^\circ \mathrm{C}}=8588\ \mathrm{kPa} \\
& h_1=h_{f @ 40^\circ \mathrm{C}}=167.53 \mathrm{~kJ} / \mathrm{kg} \\
& v_1=v_{f @ 40^{\circ} \mathrm{C}}=0.001008 \mathrm{~m}^3 / \mathrm{kg} \\
& w_{\mathrm{p}, \text { in }}=\boldsymbol{v}_1\left(P_2-P_1\right) \\
& \begin{array}{l}
=\left(0.001008 \mathrm{~m}^3 / \mathrm{kg}\right)(8588-7.385) \mathrm{kPa}\left(\frac{1 \mathrm{~kJ}}{1 \mathrm{kPa} \cdot \mathrm{m}^3}\right) \\
=8.65 \mathrm{~kJ} / \mathrm{kg}
\end{array} \\
& =8.65 \mathrm{~kJ} / \mathrm{kg} \\
& h_2=h_1+w_{\text {p,in }}=167.53+8.65=176.18 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
T_3=300^{\circ} \mathrm{C} \\
x_3=1
\end{array}\right\} \begin{array}{l}
h_3=2749.6 \mathrm{~kJ} / \mathrm{kg} \\
s_3=5.7059 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array} \\
& \left.\begin{array}{l}
T_4=40^{\circ} \mathrm{C} \\
s_4=s_3
\end{array}\right\} \begin{array}{l}
x_4=\frac{s_4-s_f}{s_{f g}}=\frac{5.7059-0.5724}{7.6832}=0.6681 \\
h_4=h_f+x_4 h_{f g}=167.53+(0.6681)(2406.0)\\=1775\text{ kJ/kg}
\end{array}
\end{aligned}
$$ Thus, $$
\begin{aligned}
w_{\mathrm{T}, \text { out }} & =h_3-h_4=2749.6-1775.1=\mathbf{9 7 4 . 5}\ \mathbf{k J} / \mathbf{k g} \\
q_{\text {in }} & =h_3-h_2=2749.6-176.18=\mathbf{2 5 7 3 . 4 ~k J} / \mathbf{k g} \\
q_{\text {out }} & =h_4-h_1=1775.1-167.53=1607.6\ \mathrm{~kJ} / \mathbf{k g}
\end{aligned}
$$ The thermal efficiency of the cycle is $$
\eta_{\text {th }}=1-\frac{q_{\text {out }}}{q_{\text {in }}}=1-\frac{1607.6}{2573.4}=\mathbf{0 . 3 7 5}
$$
