Answer
a) $\dot{m}_s=0.0174\text{ kg/s}$
b) $\dot{W}_{\text {in }, \mathrm{R}}=10.9\text{ kW}$
c) $\eta_{I I}=42.0\%$
Work Step by Step
(a) The thermal efficiency of the reversible heat engine is $$
\eta_{\text {th.sev }}=1-\frac{T_0}{T_s}=1-\frac{(25+273.15) \mathrm{K}}{(150+273.15) \mathrm{K}}=0.2954
$$ The COP of the reversible refrigerator is $$
\mathrm{COP}_{\mathrm{R}, \mathrm{ser}}=\frac{T_L}{T_0-T_L}=\frac{(-15+273.15) \mathrm{K}}{(25+273.15)-(-15+273.15) \mathrm{K}}=6.454
$$ The COP of the reversible absorption refrigerator is $$
\mathrm{COP}_{\text {abs sev }}=\eta_{\text {therev }} \mathrm{COP}_{\mathrm{R}, \mathrm{rev}}=(0.2954)(6.454)=1.906
$$ The heat input to the reversible heat engine is $$
\dot{Q}_{\text {in }}=\frac{\dot{Q}_L}{\mathrm{COP}_{\mathrm{abs}, \mathrm{ncv}}}=\frac{70 \mathrm{~kW}}{1.906}=36.72 \mathrm{~kW}
$$ Then, the rate at which the steam condenses becomes $$
\dot{m}_s=\frac{\dot{Q}_{\text {in }}}{h_{f g}}=\frac{36.72 \mathrm{~kJ} / \mathrm{s}}{2113.8 \mathrm{~kJ} / \mathrm{kg}}=\mathbf{0 . 0 1 7 4 k g / s}
$$ (b) The power input to the refrigerator is equal to the power output from the heat engine $$
\dot{W}_{\text {in }, \mathrm{R}}=\dot{W}_{\text {out }+ \text { HE }}=\eta_{\text {th, nev }} \dot{Q}_{\text {in }}=(0.2954)(36.72 \mathrm{~kW})=10.9 \mathrm{~kW}
$$ (c) The second-law efficiency of an actual absorption chiller with a COP of $0.8$ is $$
\eta_{I I}=\frac{\mathrm{COP}_{\text {actual }}}{\mathrm{COP}_{\text {abs Jev }}}=\frac{0.8}{1.906}=0.420=\mathbf{4 2 . 0} \%
$$
