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The coefficient of performance (COP) is a crucial concept when discussing heat pump efficiency. It is defined as the ratio of the heat output (\text{Q_{out}}) of the pump to the net power input (\text{W_{net}}). Essentially, the COP tells us how much heat the pump transfers for every unit of electricity it consumes. It is expressed through the following formula: \[ \text{COP} = \frac{Q_{out}}{W_{net}} \] A higher COP means the heat pump is more efficient, as it can produce more heating power with the same amount of electrical input. In our exercise, the COP is given as 2.5, which is relatively efficient.

To solve for the net power required to operate the heat pump, we rearrange the COP equation. We use the given heat transfer rate (\text{Q_{out}} = 20 kW) and the COP to find the net power (\text{W_{net}}). Here’s the rearranged formula: \[ \text{W_{net}} = \frac{Q_{out}}{\text{COP}} \] Plugging in the values: \[ \text{W_{net}} = \frac{20 \text{ kW}}{2.5} = 8 \text{ kW} \] Therefore, the heat pump requires 8 kW of power to operate, as it needs this input to provide the 20 kW heat transfer at the given efficiency.

Calculating the electricity cost involves understanding the total energy consumption and multiplying it by the electricity rate. First, we find the total energy consumption when the heat pump runs for 200 hours a month: \[ \text{Energy} = \text{W_{net}} \times \text{Hours} = 8 \text{ kW} \times 200 \text{ h} = 1600 \text{ kWh} \] Next, we use the given electricity rate (\text{\text{\text{\text{\text{0.08 \text{\text{\text{per kWh}}}}}}}}) to find the monthly cost: \[ \text{Cost} = \text{Energy} \times \text{Price per kWh} = 1600 \text{ kWh} \times 0.08 \text{\text{\text{\text{\text{per kWh}}}}}}\] This results in: \[ \text{Cost} = 128 \text{\text{\text{\text{\text{\text{}}}}}}100\text{{USD}} \] If the heat pump runs consistently for 200 hours a month at this rate, the electricity cost would be $128 per month.