Answer
$T_{3}=1332 K$
$\dot{Q}_{\text {in }}=212\text{ kW}$
Work Step by Step
We begin by using the process types to fix the temperatures of the states. $$
\begin{aligned}
& T_2=T_1\left(\frac{\boldsymbol{v}_1}{\boldsymbol{v}_2}\right)^{k-1}=T_1 r^{k-1}=(330 \mathrm{~K})(17)^{1.4-1}=1025 \mathrm{~K} \\
& T_3=T_2\left(\frac{v_3}{v_2}\right)=T_2 r_c=(1025 \mathrm{~K})(1.3)=1332 \mathrm{~K}
\end{aligned} $$ Combining the first law as applied to the various processes with the process equations gives $$
\eta_{\mathrm{th}}=1-\frac{1}{r^{k-1}} \frac{r_c^k-1}{k\left(r_c-1\right)}=1-\frac{1}{17^{1.4-1}} \frac{1.3^{1.4}-1}{1.4(1.3-1)}=0.6597
$$ According to the definition of the thermal efficiency, $$
\dot{Q}_{\text {in }}=\frac{\dot{W}_{\text {est }}}{\eta_{\text {ti }}}=\frac{140 \mathrm{~kW}}{0.6597}=212 \mathrm{~kW}
$$
