An air-conditioner with an average COP of 2.5 consumes 15 kWh of electricity during a certain day. The amount of heat removed by this air-conditioner that day isQuestion 4Select one:a.23.5 MJb.54 MJc.21.6 MJd.135 MJe.6 MJ
Question
An air-conditioner with an average COP of 2.5 consumes 15 kWh of electricity during a certain day. The amount of heat removed by this air-conditioner that day isQuestion 4Select one:a.23.5 MJb.54 MJc.21.6 MJd.135 MJe.6 MJ
Solution
The Coefficient of Performance (COP) of an air conditioner is the ratio of the cooling or heating provided by the unit to the energy consumed by the unit. In this case, the COP is given as 2.5.
The energy consumed by the air conditioner is given as 15 kWh.
To find the amount of heat removed by the air conditioner, we multiply the energy consumed by the air conditioner by the COP.
So, Heat removed = Energy consumed x COP = 15 kWh x 2.5 = 37.5 kWh
Now, we need to convert this energy from kWh to MJ (Mega Joules).
1 kWh = 3.6 MJ
So, 37.5 kWh = 37.5 x 3.6 MJ = 135 MJ
So, the amount of heat removed by the air conditioner that day is 135 MJ.
Therefore, the correct answer is d. 135 MJ.
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