Answer
$\dot{Q}_{\text {heat gain }}=44,310\text{ kJ/h}$
Work Step by Step
Noting that $1$ ton of refrigeration is equivalent to a cooling rate of $211 \mathrm{~kJ} / \mathrm{min}$, the rate of heat gain of the house in steady operation is simply equal to the cooling rate of the air-conditioning system, $$
\dot{Q}_{\text {heat gain }}=\dot{Q}_{\text {cooling }}=(3.5 \mathrm{ton})(211 \mathrm{~kJ} / \mathrm{min})=738.5 \mathrm{~kJ} / \mathrm{min}=\mathbf{4 4 , 3 1 0 ~ k J} / \mathbf{h}
$$
