Answer
See explanation
Work Step by Step
(a) Process 1-2: isentropic compression. $$
T_2=T_1\left(\frac{\boldsymbol{v}_1}{\boldsymbol{v}_2}\right)^{k-1}=(300 \mathrm{~K})(14)^{0.4}=862 \mathrm{~K}
$$ Process $2-x$, $x-3$: heat addition, $$
\begin{aligned}
q_{\text {in }} & =q_{2-x, \text { in }}+q_{3-x, \text { in }}=\left(u_x-u_2\right)+\left(h_3-h_x\right) \\
& =c_v\left(T_x-T_2\right)+c_p\left(T_3-T_x\right) \\
1520.4 \mathrm{~kJ} / \mathrm{kg} & =(0.718 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\left(T_x-862\right)+(1.005 \mathrm{~kJ} / \mathrm{kg} \\\cdot \mathrm{K})\left(2200-T_x\right)
\end{aligned}
$$ Solving for $T_x$ we get $T_x=250 \mathrm{~K}$ which is impossible. Therefore, constant specific heats at room temperature turned out to be an unreasonable assumption in this case because of the very high temperatures involved.
