Answer
$1.260\text{ kW}$
Work Step by Step
$P_1=140 \mathrm{kPa}$ sat. vapor
$ h_1=h_{g@140 \mathrm{kPa}}=239.19 \mathrm{~kJ} / \mathrm{kg}$
$s_1=s_{g @140 \mathrm{kPa}}=0.94467 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$
$\left.\begin{array}{l}P_2=0.8\ \mathrm{MPa} \\ s_2=s_1\end{array}\right\} h_2=275.40\ \mathrm{~kJ} / \mathrm{kg}$
$\left.\begin{array}{l}P_3=0.8\ \mathrm{MPa} \\ \text { sat. liquid }\end{array}\right\} h_3=h_{f @ 0.8 \mathrm{MPa}}=95.48 \mathrm{~kJ} / \mathrm{kg}$
$h_4 \equiv h_3=95.48 \mathrm{~kJ} / \mathrm{kg}$ (throttling)
The quality of the refrigerant at the end of the throttling process is $$
x_4=\left(\frac{h_4-h_f}{h_{f g}}\right)_{@ 140 \mathrm{kPa} }=\frac{95.48-27.06}{212.13}=\mathbf{0 . 3 2 2 5}
$$ (b) The COP of the refrigerator is determined from its definition, $$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}}=\frac{h_1-h_4}{h_2-h_1}=\frac{239.19-95.48}{275.40-239.19}=\mathbf{3 . 9 6 9}
$$ (c) The power input to the compressor is determined from $$
\dot{W}_{\text {in }}=\frac{\dot{Q}_L}{\mathrm{COP}_{\mathrm{R}}}=\frac{(300 / 60) \mathrm{kW}}{3.969}=1.260 \mathrm{~kW}
$$
