00:01
That the coefficient of performance ideal would be equal to the lower operating temperature divided by the higher operating temperature minus the lower operating temperature.
00:09
So we can say that the coefficient of performance effective would be equal to 20 % of this.
00:16
So now we can solve.
00:18
The effective coefficient of performance would be equal to 0 .20 and then times 273 plus 24 kelvin divided by 4 .4 .5.
00:32
Kelvin and this is equaling 4 .24.
00:40
Now, for part b, they want us to find the work per cent, so we can say that the coefficient of performance would be equal to the lower, the heat exhaust, rather, divided by the work done.
00:53
We know that the work done would then be equal to the heat exhaust divided by the effective coefficient of performance.
00:59
And so we can say that the work done per unit time would be equal to the heat exhaust per unit time divided by the coefficient of performance effective.
01:08
So we can say, and at this point we can just solve...