Answer
$\dot{m}=984.6 \text{ kg/s}$
Work Step by Step
Using constant specific heats, $$
\begin{aligned}
T_{2 s} & =T_1\left(\frac{P_2}{P_1}\right)^{(k-1) / k}=(310 \mathrm{~K})(8)^{0.4 / 1.4}=561.5 \mathrm{~K} \\
T_{4 s} & =T_3\left(\frac{P_4}{P_3}\right)^{(k-1) / k}=(900 \mathrm{~K})\left(\frac{1}{8}\right)^{0.4 / 1.4}=496.8 \mathrm{~K} \\
w_{\text {net, out }} & =w_{\mathrm{T}, \text { out }}-w_{\mathrm{C}, \text { is }}=\eta_T c_p\left(T_3-T_{4 s}\right)-c_p\left(T_{2 s}-T_1\right) / \eta_C \\
& =(1.005 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})[(0.86)(900-496.8)-(561.5-310)(0.80)] \mathrm{K} \\
& =32.5 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ and $$
\dot{m}=\frac{\dot{W}_{\text {net,out }}}{w_{\text {net,out }}}=\frac{32,000 \mathrm{~kJ} / \mathrm{s}}{32.5 \mathrm{~kJ} / \mathrm{kg}}=984.6 \mathrm{~kg} / \mathbf{s}
$$
