Answer
$T_{2}=437.5\text{ K}$
$Q_{out}=975.8\text{ kJ}$
$S_{gen}=3.92\text{ kJ/K}$
Work Step by Step
(a) The mass and the final temperature of $\mathrm{CO}_2$ may be determined from ideal gas equation
$
\begin{aligned}
& m=\frac{P_1 V}{R T_1}=\frac{(100 \mathrm{kPa})\left(0.8 \mathrm{~m}^3\right)}{\left(0.1889 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)(250 \mathrm{~K})}=1.694 \mathrm{~kg} \\
& T_2=\frac{P_2 V}{m R}=\frac{(175 \mathrm{kPa})\left(0.8 \mathrm{~m}^3\right)}{(1.694 \mathrm{~kg})\left(0.1889 \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}\right)}=437.5 \mathrm{~K}
\end{aligned}
$
(b) The amount of heat transfer may be determined from an energy balance on the system
$
\begin{aligned}
Q_{\text {out }} & =\dot{E}_{\mathrm{c}, \mathrm{in}} \Delta t-m c_v\left(T_2-T_1\right) \\
& =(0.5 \mathrm{~kW})(40 \times 60 \mathrm{~s})-(1.694 \mathrm{~kg})(0.706 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\\&(437.5-250) \mathrm{K}=\mathbf{9 7 5 . 8} \mathbf{~ k J}
\end{aligned}
$
(c) The entropy generation associated with this process may be obtained by calculating total entropy change, which is the sum of the entropy changes of $\mathrm{CO}_2$ and the surroundings
$
\begin{aligned}
S_{\text {gen }} & =\Delta S_{\text {co2 }}+\Delta S_{\text {surs }}=m\left(c_p \ln \frac{T_2}{T_1}-R \ln \frac{P_2}{P_1}\right)+\frac{Q_{\text {out }}}{T_{\text {sum }}} \\
& =(1.694 \mathrm{~kg})\left[(0.895 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}) \ln \frac{437.5 \mathrm{~K}}{250 \mathrm{~K}}-(0.1889 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K})\\ \ln \frac{175 \mathrm{kPa}}{100 \mathrm{kPa}}\right]+\frac{975.8 \mathrm{~kJ}}{300 \mathrm{~K}} \\
& =\mathbf{3 . 9 2}\ \mathrm{kJ} / \mathrm{K}
\end{aligned}
$
