Answer
c) $178.5\text{ kJ/kg}$
Work Step by Step
(a) In this normal vapor-compression refrigeration cycle, the refrigerant enters the compressor as a saturated vapor
at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure.
(b) $$
\begin{aligned}
& h_4=h_3=h_{f @45{ }^{\circ}C}=101 \mathrm{~kJ} / \mathrm{kg} \\
& h_1=h_{g@-5^{\circ}\mathrm{C}}=248.1 \mathrm{~kJ} / \mathrm{kg} .
\end{aligned}
$$ The heat absorbed by the refrigerant in the evaporator is $$
q_L=h_1-h_4=248.1-101=147.1 \mathrm{~kJ} / \mathrm{kg}
$$ (c) The COP of the air conditioner is $$
\operatorname{COP}_{\mathrm{R}}=\operatorname{SEER}\left(\frac{1 \mathrm{~W}}{3.412 \mathrm{Btu} / \mathrm{h}}\right)=\left(16 \frac{\mathrm{Btu} / \mathrm{h}}{\mathrm{W}}\right)\left(\frac{1 \mathrm{~W}}{3.412 \mathrm{Btw} / \mathrm{h}}\right)=4.689
$$ The work input to the compressor is $$
\mathrm{COP}_{\mathrm{R}}=\frac{q_L}{w_{\text {in }}} \longrightarrow w_{\text {in }}=\frac{q_L}{\mathrm{COP}_{\mathrm{R}}}=\frac{147.1 \mathrm{~kJ} / \mathrm{kg}}{4.689}=31.4 \mathrm{~kJ} / \mathrm{kg}
$$ The enthalpy at the compressor exit is $$
w_{\text {in }}=h_2-h_1 \longrightarrow h_2=h_1+w_{\text {in }}=248.1 \mathrm{~kJ} / \mathrm{kg}+31.4 \mathrm{~kJ} / \mathrm{kg}=279.5 \mathrm{~kJ} / \mathrm{kg}
$$ The heat rejected from the refrigerant in the condenser is then
$$
q_H=h_2-h_3=279.5-101=178.5 \mathrm{~kJ} / \mathrm{kg}
$$
