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A heat pump with a COP of 2. 6 is used to supply energy to a house at a rate of 7500 kJ/h. Determine the rate of energy absorption from the outdoor air. 0. 48 kW 2. 88 kW 1. 28 kW 0. 80 kW
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A heat pump with a COP of 2. 6 is used to supply energy to a house at a rate of 7500 kJ/h. Determine the rate of energy absorption from the outdoor air.

0. 48 kW

2. 88 kW

1. 28 kW

0. 80 kW

User Alberto Martinez
by
8.2k points

1 Answer

5 votes

Answer:

0.80 kW

Step-by-step explanation:

power output of the motor= 7500 kJ/hr = 12500/6 Watt.

Coefficient of performance COP = 2.6

power input to the motor= 12500/(6×2.6) = 801.3 Watts

User Vidhee
by
8.1k points
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