Answer
$\eta_{\text {th }}=57.5\%$
$\dot{Q}_{\text {in }}=157\text{ kW}$
Work Step by Step
The definition of cycle thermal efficiency reduces to $$
\eta_{\text {th }}=1-\frac{1}{r^{k-1}}=1-\frac{1}{8.5^{1.4-1}}=0.5752=\mathbf{5 7 . 5} \%
$$ The rate of heat addition is then $$
\dot{Q}_{\text {in }}=\frac{\dot{W}_{\text {net }}}{\eta_{\text {th }}}=\frac{90 \mathrm{~kW}}{0.5752}=157\ \mathbf{k W}
$$
