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Hi there.
00:01
So for this problem, we have an air conditioner that takes air from the room with a lowest temperature of 70 fahrenheit degrees and transferring to the outdoors, which is at the hottest temperature of 96 fahrenheit degrees, and transferring to the outdoors, which is at the hottest temperature of 96 fahrenheit celsius.
00:30
Use and fahrenheit degrees, sorry.
00:34
So what we need to calculate is for each jule of electrical energy required to run the refrigerator, how many joules of heat are transferred from the room.
00:55
So we are given the value for the work done, and that is equal to one jewel.
01:05
And we need to find the value for the heat.
01:12
Now, we start with the definition of the coefficient of performance, which is the lowest temperature over the hottest temperature minus the lowest temperature.
01:28
And we also know that the word done is equal to the heat over, the coefficient of performance.
01:40
So substituting there the expression for the coefficient of performance, we know that the word done is equal to the heat times the ratio between the hottest temperature over the lowest temperature minus one.
01:59
And then solving for the heat, we will obtain that that is equal to the work done over the hottest temperature over the lowest temperature minus 1...