Answer
a) $\dot{Q}_L=61,700\text{ Btu/h}$
b) $COP_{}=0.779$
c) $\eta_{\mathrm{II}}=32.8\%$
Work Step by Step
(a) The rate of cooling provided by the system is
$$
\begin{aligned}
\dot{Q}_L & =\dot{m}_R\left(h_1-h_4\right)=(0.04 \mathrm{lbm} / \mathrm{s})(619.2-190.9) \mathrm{Btu} / \mathrm{lbm} \\
& =17.13\ \mathrm{Btu} / \mathrm{s}=\mathbf{6 1}, \mathbf{7 0 0 B t u} / \mathbf{h}
\end{aligned}
$$
(b) The rate of heat input to the generator is $$
\dot{Q}_{\mathrm{geo}}=\dot{m}_{\mathrm{geo}} c_p\left(T_{\text {geo,in }}-T_{\text {geo; out }}\right)=(0.55 \mathrm{lbm} / \mathrm{s})\left(1.0 \mathrm{Btu} / \mathrm{bmm} \cdot{ }^{\circ} \mathrm{F}\right)(240-200)^{\circ} \mathrm{F}=22.0 \mathrm{Btu} / \mathrm{s}
$$ Then the COP becomes $$
\mathrm{COP}=\frac{\dot{Q}_L}{\dot{Q}_{\mathrm{gen}}}=\frac{17.13 \mathrm{Btu} / \mathrm{s}}{22.0 \mathrm{Btu} / \mathrm{s}}=\mathbf{0 . 7 7 9}
$$ (c) The reversible COP of the system is $$
\mathrm{COP}_{\text {abs, rev }}=\left(1-\frac{T_0}{T_s}\right)\left(\frac{T_L}{T_0-T_L}\right)=\left(1-\frac{(70+460)}{(220+460)}\right)\left(\frac{(25+460)}{70-25}\right)=2.38
$$ The temperature of the heat source is taken as the average temperature of the geothermal water: $(240+200) / 2=220^{\circ} \mathrm{F}$. Then the second-law efficiency becomes $$
\eta_{\mathrm{II}}=\frac{\mathrm{COP}}{\mathrm{COP}_{\text {abs,rev }}}=\frac{0.779}{2.38}=0.328=\mathbf{3 2 . 8} \%
$$
