Answer
$\eta_{\text {th }}=61.0\%$
$\dot{Q}_{\text {in }}=148\text{ kW}$
Work Step by Step
The definition of cycle thermal efficiency reduces to $$
\eta_{\text {th }}=1-\frac{1}{r^{k-1}}=1-\frac{1}{10.5^{1.4-1}}=0.6096=\mathbf{6 1 . 0} \%
$$ The rate of heat addition is then $$
\dot{Q}_{\text {in }}=\frac{\dot{W}_{\text {net }}}{\eta_{\text {th }}}=\frac{90 \mathrm{~kW}}{0.6096}=148 \mathrm{~kW}
$$
