Answer
$\dot{W}_{\text {net, out }}=3373\text{ kW}$
Work Step by Step
Using variable specific heats for air,
$$
\begin{aligned}
& T_3=2000\ \mathrm{R} \longrightarrow h_3=504.71\ \mathrm{Btu} / \mathrm{lbm} \\
& T_4=1200\ \mathrm{R} \longrightarrow h_4=291.30\ \mathrm{Btu} / \mathrm{lbm} \\
& r_p=\frac{P_2}{P_1}=\frac{120}{15}=8 \\
& \dot{Q}_{\text {out }}=\dot{m}\left(h_4-h_1\right) \longrightarrow h_1=291.30-6400 / 40=131.30\ \mathrm{Btu} / \mathrm{lbm} \\
& \rightarrow P_{r_1}=1.474 \\
& P_{r_2}=\frac{P_2}{P_1} P_{r_1}=(8)(1.474)=11.79 \longrightarrow h_{2 s}=238.07\ \mathrm{Btw} / \mathrm{lbm} \\
& \dot{W}_{\mathrm{C} \text { in }}=\dot{m}\left(h_{2 s}-h_1\right) / \eta_C=(40 \mathrm{lbm} / \mathrm{s})(238.07-131.30) /(0.80)=5339\ \mathrm{Btu} / \mathrm{s} \\
& \dot{W}_{\mathrm{T}, \text { out }}=\dot{m}\left(h_3-h_4\right)=(40 \mathrm{lbm} / \mathrm{s})(504.71-291.30) \mathrm{Btu} / \mathrm{lbm}=8536\ \mathrm{Btu} / \mathrm{s} \\
& \dot{W}_{\text {net out }}=\dot{W}_{\mathrm{T}, \text { oat }}-\dot{W}_{\mathrm{C}, \text { in }}=8536-5339=3197\ \mathrm{Btu} / \mathrm{s}=\mathbf{3 3 7 3} \mathbf{k W} \\
&
\end{aligned}
$$
