Answer
a) $\dot{W}_{rev}=0.20\ hp$
b) $\dot{I}=0.50\ hp$
c) $\eta=28.9\%$
Work Step by Step
a) From the coefficient of performance:
$COP_{R,rev}=\dfrac{1}{T_H/T_L-1}$
Given $T_H=535\ R,\ T_L=480\ R$
$COP_{R,rev}=8.73$
Since
$COP_{R,rev}=\dfrac{\dot{Q}_L}{\dot{W}_{rev}},\quad \dot{Q}_L=75\ Btu/min=1.77\ hp$
$\dot{W}_{rev}=0.20\ hp$
b) Given $\dot{I}=\dot{W}_{u}-\dot{W}_{rev},\quad \dot{W}_{u}=0.70\ hp$
$\dot{I}=0.50\ hp$
c) Second law efficiency:
$\eta=\dfrac{\dot{W}_{rev}}{\dot{W}_{u}}=28.9\%$
