Answer
$
\dot{Q}_{\mathrm{L}, \max }=6.58×10^5 \text{ kJ/h}$
Work Step by Step
The maximum COP that this refrigeration system can have is
$$
\mathrm{COP}_{\mathrm{R}, \max }=\left(1-\frac{T_0}{T_s}\right)\left(\frac{T_L}{T_0-T_L}\right)=\left(1-\frac{298 \mathrm{~K}}{383 \mathrm{~K}}\right)\left(\frac{255}{298-255}\right)=1.316
$$ Thus, $$
\dot{Q}_{\mathrm{L}, \max }=\mathrm{COP}_{\mathrm{R}, \max } \dot{Q}_{\mathrm{gen}}=(1.316)\left(5 \times 10^5 \mathrm{~kJ} / \mathrm{h}\right)=\mathbf{6 . 5 8} \times 10^5 \mathbf{~ k J / h}
$$
