Answer
$2 7 . 1 \%$
Work Step by Step
The power input is $$
\dot{W}_{\mathrm{in}}=\dot{Q}_H-\dot{Q}_L=5500-3500=2000 \mathrm{~kJ} / \mathrm{h}=(2000 \mathrm{~kJ} / \mathrm{h})\left(\frac{1 \mathrm{~kW}}{3600 \mathrm{~kJ} / \mathrm{h}}\right)=\mathbf{0 . 5 5 5 6\ k W}
$$ The COP is $$
\mathrm{COP}_{\mathrm{R}}=\frac{\dot{Q}_L}{\dot{W}_{\text {in }}}=\frac{(3500 / 3600) \mathrm{kW}}{0.5556 \mathrm{~kW}}=\mathbf{1 . 7 5}
$$ The COP of the Carnot cycle operating between the space and the ambient is $$
\mathrm{COP}_{\text {Camot }}=\frac{T_L}{T_H-T_L}=\frac{258 \mathrm{~K}}{(298-258) \mathrm{K}}=6.45
$$ The second-law efficiency is then $$
\eta_{\Pi}=\frac{\mathrm{COP}_{\mathrm{R}}}{\mathrm{COP}_{\text {Camot }}}=\frac{1.75}{6.45}=0.271=\mathbf{2 7 . 1} \%
$$
