Answer
$\dot{W}_i=1.71\ kW$
Work Step by Step
$Q_L=mc_v\Delta T$
Given $m=800\ kg,\ c_v=0.72\ kJ/kg.°C,\ \Delta T=(35-20)°C$
$Q_L=8640\ kJ$
Heat removed in 30 min:
$\dot{Q}_L=Q_L/\Delta t$
$\dot{Q}_L=4.8\ kW$
$COP_R=\dot{Q}_L/\dot{W}_i=2.8$
$\dot{W}_i=1.71\ kW$
