Answer
$\dot{Q}_{\text {cooling }}=51,840 \text{ kJ/h}$
Work Step by Step
The COP of the heat pump is given to be 3.4. Then the COP of the air-conditioning system becomes $$
\mathrm{COP}_{\text {air-cond }}=\mathrm{COP}_{\text {heat pump }}-1=3.4-1=2.4
$$ Then the rate of cooling (heat absorption from the air) becomes $$
\dot{Q}_{\text {cooling }}=\mathrm{COP}_{\text {air-cond }} \dot{W}_{\text {in }}=(3.4)(6 \mathrm{~kW})=20.4 \mathrm{~kW}=51,840 \mathrm{~kJ} / \mathrm{h}
$$ Since $1 \mathrm{~kW}=3600 \mathrm{~kJ} / \mathrm{h}$. We conclude that this heat pump can meet the cooling needs of the room since its cooling rate is greater than the rate of heat gain of the room.
