Answer
$2 . 9 1 6\text{ kW}$
Work Step by Step
The power input is $$
\dot{W}_{\mathrm{in}}=\frac{\dot{Q}_L}{\mathrm{COP}}=\frac{(24,000 \mathrm{Btu} / \mathrm{h})\left(\frac{1 \mathrm{~kW}}{3412 \mathrm{Btu} / \mathrm{h}}\right)}{2.05}=\frac{7.034 \mathrm{~kW}}{2.05}=\mathbf{3 . 4 3 1}\ \mathbf{k W}
$$ From an energy balance on the cycle, $$
\dot{Q}_H=\dot{Q}_L+\dot{W}_{\text {in }}=7.034+3.431=10.46 \mathrm{~kW}
$$ The mass flow rate of the water is then determined from $$
\dot{Q}_H=\dot{m} c_{p w} \Delta T_w \longrightarrow \dot{m}=\frac{\dot{Q}_H}{c_{p w} \Delta T_w}=\frac{10.46 \mathrm{~kW}}{\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(12^{\circ} \mathrm{C}\right)}=\mathbf{0 . 2 0 8 6\ k g} / \mathbf{s}
$$ The exergy of the heat transferred from the low-temperature medium is
$$
\dot{E x}_{\dot{Q}_L}=-\dot{Q}_L\left(1-\frac{T_0}{T_L}\right)=-(7.034 \mathrm{~kW})\left(1-\frac{20+273}{0+273}\right)=0.5153 \mathrm{~kW}
$$ The second-law efficiency of the cycle is $$
\eta_{\mathrm{II}}=\frac{\dot{E} x_{\dot{Q}_L}}{\dot{W}_{\text {in }}}=\frac{0.5153}{3.431}=0.1502=\mathbf{1 5 . 0} \%
$$ The exergy destruction is the difference between the exergy supplied and the exergy recovered
$$
\dot{E} x_{\text {dest }}=\dot{W}_{\text {in }}-\dot{E} x_{\dot{Q}_L}=3.431-0.5153=\mathbf{2 . 9 1 6\ k W}
$$
