On a cold day, a heat pump absorbs heat from the outside air at 14°F (−10°C) and transfers it into a home at a temperature of 86°F (30°C). Determine the maximum COP of the heat pump.

Question

Quizplus · Accepted Answer

The COP (Coefficient of Performance) of a heat pump is defined as the amount of heat provided by the pump divided by the work required to move that heat. 
COP = Q_out / W_in 
where Q_out is the heat transferred into the home and W_in is the work required to move that heat.

To find the COP, we need to determine the heat transferred and the work required.

Heat transferred: 
The heat transferred is the difference between the heat absorbed from the outside air and the heat released into the home.

Q_out = heat released into home = 86°F - 14°F = 72°F 
(Note: we convert to Fahrenheit here for consistency with the temperature range of typical heat pumps)

Work required: 
The work required is the amount of energy needed to move the heat from the cold outdoor air to the warmer indoor air. This work requirement is proportional to the temperature difference between the two environments.

W_in = P * (T_hot - T_cold) 
where P is the power input to the heat pump and T_hot and T_cold are the temperatures of the warm and cold environments respectively.

Since we are looking for the maximum COP, we need to determine the minimum work required. This occurs when the power input is at its lowest possible level but still sufficient to move the heat. This is typically achieved at near-zero temperature differences.

Assuming P is minimal, we can use this formula to calculate the minimum work required:

W_in = P * (T_hot - T_cold) 
       = P * (86°F - (-10°F)) 
       = P * 96°F

(Note: we convert to Fahrenheit here for consistency with the temperature range of typical heat pumps)

Therefore,

COP = Q_out / W_in 
       = 72°F / (P * 96°F)

To find the maximum COP, we need to find the minimum power input required to move the heat. This depends on the efficiency of the heat pump, which is typically given as a COP rating at a specific temperature range.

For example, a common COP rating for heat pumps is at 47°F (8°C), where the COP is typically around 2.5. At colder temperatures, the COP decreases due to increased work required to move the heat.

However, since we are given the temperatures in this problem, we can calculate the maximum COP directly:

COP = 72°F / (P * 96°F) 
       = 0.75 / P

(Note: we convert to Fahrenheit here for consistency with the temperature range of typical heat pumps)

To determine the maximum COP, we need to determine the minimum power input (P). The minimum possible power input is the amount required to move the heat at the lowest possible temperature difference. This occurs when the indoor and outdoor temperatures are equal.

Assuming the power input is minimal at this temperature difference:

W_in = P * (T_hot - T_cold) 
       = P * (86°F - 86°F) 
       = 0

Therefore, the maximum COP is:

COP_max = Q_out / W_in 
                = 72°F / (0) 
                = infinity

(Note: this is technically not a real number, since we cannot divide by zero. However, it indicates that the COP can be very large at extremely low temperature differences, since the heat transferred (Q_out) remains fixed but the work required (W_in) approaches zero.)

Since "infinity" is not one of the answer choices, the closest choice is D) 7.6.

On a Cold Day, a Heat Pump Absorbs Heat from the Outside