00:01
Okay, so this problem is analyzing the power used in an air conditioner.
00:09
And so just an overview, the equations are pretty simple.
00:14
The math is pretty simple.
00:16
I think the main challenge in this problem is sort of translating the language of air conditioners, power, whatever, to what you would actually need for the formulas given to you in the chapter.
00:29
So i'm just going to go ahead and write out our few.
00:31
Give -ins so it's an 800 watt air conditioner and then your coefficient of performance is 2 .8 and your cold reservoir is 294 that's converting the 21 degrees c to to kelvin and then your your hot temperature is 308 kelvin again i always just i generally convert the temperature right away so you don't forget.
01:06
Okay, so the first question asks, heat removal from the unit.
01:13
So that's, oh, actually, before i even answer that, i want to talk about this power is 800 watts.
01:21
So the fact that it means that it says that it operates on 800 watts means like that's overall the work that's being done by the machine, or the rate of work.
01:33
And so basically in broad strokes, you can kind of divide everything by time and say that this p is this worked about it by time.
01:41
And then the rate of coolings are always going to be the q divided by time.
01:44
And i'll show you exactly what i mean by that.
01:48
Great.
01:49
So for a, we basically want to get the q coal divided by time because, again, that asks for the rate of heat removal.
01:59
So the heat is flowing from the cool reservoir.
02:05
If you look at like a diagram on refrigerators at the end of the chapter, i thought that was kind of helpful for solving this problem.
02:15
It's just looking at that end of the chapter thing.
02:17
Great.
02:18
So you want to get this.
02:20
So you start out with qc equals this coefficient.
02:25
And then these are all magnitude.
02:26
So if i forget to do magnitudes, sorry.
02:32
I'm just trying to remember they often belong there.
02:36
You can take this equation, divide it both by time, and then boom, you got what you're looking for.
02:43
And then we know this work divided by time.
02:45
It's just the power.
02:46
So that's just going to be your coefficient, 2 .8 times 800 watts.
02:56
And then when i calculated that, i got 2 .24 10 to 3 watts.
03:09
Okay.
03:10
So for part b, you want to get, let's see, what is the wording that they used, the rate of which heat is discharged to the outside air.
03:19
Immediately, if you, again, look at the diagram on the refrigerator, that's going to be your q hot.
03:26
And if you look at the diagram, there's equations.
03:33
And you see from the equations that work is q hot minus q cold with magnitude.
03:39
So you can say then that q hot is q cold plus the work with just the magnitude.
03:49
So we don't have to worry about the signs.
03:51
Sort of an energy conservation equation.
03:54
And then if you divide kind of both sides by time, then you get q hot is equal to the rate of q cold to point...