Answer
$\mathbf{2 . 9 7} \mathbf{~ k W}$
Work Step by Step
In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser
pressure. From the refrigerant tables (Tables A-12 and A-13), $$
\begin{aligned}
& P_1=320\ \mathrm{kPa} \mid h_1=h_{g @ 320 \mathrm{kPa}}=251.93 \mathrm{~kJ} / \mathrm{kg} \\
& \text { sat. vapor }\} s_1=s_{g @ 320 \mathrm{kPa}}=0.93026 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K} \\
& \left.\begin{array}{l}
P_2=1.4\ \mathrm{MPa} \\
s_2=s_1
\end{array}\right\} h_2=282.60 \mathrm{~kJ} / \mathrm{kg} \\
& \left.\begin{array}{l}
\begin{array}{l}
P_3=1.4 \mathrm{MPa} \\
\text { sat. liquid }
\end{array}
\end{array}\right\} h_3=h_{f @ 1.4 \mathrm{MPa}}=127.25 \mathrm{~kJ} / \mathrm{kg} \\
& h_4 \cong h_3=127.25 \mathrm{~kJ} / \mathrm{kg} \text { (throttling) }
\end{aligned}
$$ The heating load of this heat pump is determined from $$
\begin{aligned}
\dot{Q}_H & =\left[\dot{m} c\left(T_2-T_1\right)\right]_{\text {water }} \\
& =(0.12 \mathrm{~kg} / \mathrm{s})\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)(45-15)^{\circ} \mathrm{C} \\
& =15.05 \mathrm{~kW}
\end{aligned}
$$ and $$
\dot{m}_R=\frac{\dot{Q}_H}{q_H}=\frac{\dot{Q}_H}{h_2-h_3}=\frac{15.05 \mathrm{~kJ} / \mathrm{s}}{(282.60-127.25) \mathrm{kJ} / \mathrm{kg}}=0.09686 \mathrm{~kg} / \mathrm{s}
$$ Then, $$
\dot{W}_{\text {in }}=\dot{m}_R\left(h_2-h_1\right)=(0.09686 \mathrm{~kg} / \mathrm{s})(282.60-251.93) \mathrm{kJ} / \mathrm{kg}=\mathbf{2 . 9 7} \mathbf{~ k W}
$$
