Answer
a) $\dot{Q}_L=9.416\text{ kW}$, $\dot{W}_{\text {in }}=3.627\text{ kW}$
b) $74.1\%$
c) $\mathrm{COP}_{\mathrm{R}}=2.60$
Work Step by Step
(a) From the refrigerant tables (Tables A-12 and A-13),$$
\begin{aligned}
& \left.\begin{array}{l}
P_1=0.20\ \mathrm{MPa} \\
T_1=-5^{\circ} \mathrm{C}
\end{array}\right\} \begin{array}{l}
h_1=248.82 \mathrm{~kJ} / \mathrm{kg} \\
s_1=0.95414 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}
\end{array} \\
& \left.\begin{array}{l}
P_2=1.2\ \mathrm{MPa} \\
T_2=70^{\circ} \mathrm{C}
\end{array}\right\} h_2=300.63 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_{2 s}=1.2\ \mathrm{MPa} \\
s_{2 s}=s_1
\end{array}\right\} h_{2 s}=287.23 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
P_3=1.15\ \mathrm{MPa} \\
T_3=44^{\circ} \mathrm{C}
\end{array}\right\} h_3=h_{f @ 44^{\circ} \mathrm{C}}=114.30 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \cong h_3=114.30 \mathrm{~kJ} / \mathrm{kg}(\text { throttling })
\end{aligned}
$$ Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from $$
\dot{Q}_L=\dot{m}\left(h_1-h_4\right)=(0.07 \mathrm{~kg} / \mathrm{s})(248.82-114.30) \mathrm{kJ} / \mathrm{kg}=9.416\ \mathbf{k W}
$$ and $$
\dot{W}_{\text {in }}=\dot{m}\left(h_2-h_1\right)=(0.07 \mathrm{~kg} / \mathrm{s})(300.63-248.82) \mathrm{kJ} / \mathrm{kg}=\mathbf{3 . 6 2 7}\ \mathbf{k W}
$$ (b) The isentropic efficiency of the compressor is determined from $$
\eta_C=\frac{h_{2 s}-h_1}{h_2-h_1}=\frac{287.23-248.82}{300.63-248.82}=0.7414=\mathbf{7 4 . 1} \%
$$ (c) The COP of the refrigerator is determined from its definition, $$
\mathrm{COP}_{\mathrm{R}}=\frac{\dot{Q}_L}{\dot{W}_{\text {in }}}=\frac{9.416 \mathrm{~kW}}{3.627 \mathrm{~kW}}=\mathbf{2 . 6 0}
$$
