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The freezer uses 730 kW \( \cdot \) h of energy per year. There are 365 days in a year, so first calculate the daily energy usage:\[\text{Daily Energy Usage} = \frac{730 \text{ kW} \cdot \text{h}}{365 \text{ days}} = 2 \text{ kW} \cdot \text{h per day}\]

The freezer operates for 5 hours every day. To determine the power it requires while operating, divide the daily energy usage by the number of operating hours:\[\text{Power} = \frac{2 \text{ kW} \cdot \text{h per day}}{5 \text{ hours}} = 0.4 \text{ kW}\]Thus, the power required while operating is 0.4 kW.

The Coefficient of Performance (COP) for a refrigerator operating between two temperatures \( T_C = -5.0^\circ C \) and \( T_H = 20.0^\circ C \) is given by:\[COP = \frac{T_C + 273.15}{T_H - T_C}\]Convert temperatures to Kelvin and substitute:\[T_C = -5.0^\circ C + 273.15 = 268.15 \text{ K}\]\[T_H = 20.0^\circ C + 273.15 = 293.15 \text{ K}\]\[COP = \frac{268.15}{293.15 - 268.15} = \frac{268.15}{25.0} \approx 10.73\]Thus, the theoretical maximum COP is approximately 10.73.

To find the maximum amount of ice that can be produced, first determine the energy required to freeze water from 20.0^\circ C to ice at -5.0^\circ C. This process requires three steps: cooling water from 20.0^\circ C to 0^\circ C, freezing it, and then cooling ice from 0^\circ C to -5.0^\circ C. * Cooling water: \( Q_1 = mc_w\Delta T_1 \)* Freezing water: \( Q_2 = mL_f \)* Cooling ice: \( Q_3 = mc_i\Delta T_2 \)Where:- \( m \) is the mass of water in kg - \( c_w = 4.18 \text{ kJ/kg}\cdot\text{K} \) (specific heat of water)- \( c_i = 2.09 \text{ kJ/kg}\cdot\text{K} \) (specific heat of ice)- \( L_f = 334 \text{ kJ/kg} \) (latent heat of fusion)- \( \Delta T_1 = 20^\circ C \), \( \Delta T_2 = -5^\circ C \)The total energy \( Q_{total} = Q_1 + Q_2 + Q_3 \) needed:\(\begin{align*}Q_{total} & = m \cdot (4.18 \times 20 + 334 + 2.09 \times 5) \& = m \cdot (83.6 + 334 + 10.45) \& = m \cdot 428.05 \text{ kJ}\end{align*}\)

The maximum energy available to the freezer per hour is given by its power and the COP:\[\text{Energy} = 0.4 \text{ kW} \times 10.73 \times 1 \text{ hour} \times 3600 \text{ seconds/hour} = 15434.4 \text{ kJ}\]To find the mass \( m \) that can be converted to ice, use the total energy required per kilogram from Step 4:\[15434.4 \text{ kJ} = m \cdot 428.05 \text{ kJ} \implies m \approx \frac{15434.4}{428.05} \approx 36.07 \text{ kg}\]So, the theoretical maximum amount of ice that can be made in an hour is approximately 36.07 kg.