Answer
$2.03$
Work Step by Step
$$
\begin{aligned}
& \left.\begin{array}{l}
P_4=120\ \mathrm{kPa} \\
x_4=0.34
\end{array}\right\} h_4=95.41 \mathrm{~kJ} / \mathrm{kg} \\
& h_3=h_4 \\
& \left.\begin{array}{l}
h_3=95.41 \mathrm{~kJ} / \mathrm{kg} \\
x_3=0 \text { (sat_liq.) }
\end{array}\right\} P_3=799\ \mathrm{kPa} \equiv 800\ \mathrm{kPa} \\
& P_2=P_3 \\
& \left.\begin{array}{l}
P_2=799\ \mathrm{kPa} \\
T_2=70^{\circ} \mathrm{C}
\end{array}\right\} h_2=306.92 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_1=P_4=120\ \mathrm{kPa} \\
x_1=1 \text { (sat. vap.) }
\end{array}\right\} h_1=236.99 \mathrm{~kJ} / \mathrm{kg}
\end{aligned}
$$ The mass flow rate of the refrigerant is determined from $$
\dot{m}=\frac{\dot{W}_{\text {in }}}{h_2-h_1}=\frac{0.45 \mathrm{~kW}}{(306.92-236.99) \mathrm{kJ} / \mathrm{kg}}=0.006435 \mathrm{~kg} / \mathrm{s}
$$ (c) The refrigeration load and the COP are
$$
\begin{aligned}
\dot{Q}_L & =\dot{m}\left(h_1-h_4\right) \\
& =(0.06435 \mathrm{~kg} / \mathrm{s})(236.99-95.41) \mathrm{kJ} / \mathrm{kg} \\
& =0.9111 \mathrm{~kW} \\
\text { COP } & =\frac{\dot{Q}_L}{\dot{W}_{i n}}=\frac{0.9111 \mathrm{~kW}}{0.45 \mathrm{~kW}}=2.03
\end{aligned}
$$
