Hey, guys. So let's check out our example. We've got some kind of a household mini refrigerator with some coefficient of performance and a power input. We have this sample of water here that we want to freeze down into ice, and we want to calculate ultimately how long it takes for us to freeze this water to ice at this temperature. So, which variable is "how long"? Usually, where we're asked how long it takes for something to happen, that's going to be a delta T. So, where in our equations are we going to get a delta T out of?
Well, if we take a look here, the only thing we have any information about is the coefficient of performance, which is k. So k=3, and k=QcW. This is basically what you get out of it. The W is what you get in, is what you pay to sort of get it. Right? The heat extracted from the cold reservoir. Alright.
So, we don't have the Qc and we also don't have what the work is. But if you remember that this power input is related to work. If you remember that power is related to Wt, then what we can do here is we can say that W=power×Δt. So, I'm actually going to move this down here because basically what we're going to do here is we're going to replace this W here in terms of power. So this QcW becomes Qcpower×time.
In order to get the delta t, I'm going to have to move the delta t over to the other side and then the k down and basically they're going to trade places. What you're going to end up with is this equation here. The delta t, the amount of time it takes, is going to be the heat divided by the power times the coefficient of performance. This is an equation we've seen before. This is not, but, you know, this is just an equation that we've sort of gotten to from this problem here.
We can actually relate to the time that it takes for something to happen to the heat divided by the power and the coefficient of performance. So, if you look through your variables, we actually do know what the power is, and we do know what the coefficient of performance is. So all we really have to do is figure out what's the heat that we have to extract from the cold reservoir. Well, in this case, what happens is the cold reservoir is really just the sample of water that I have.
So what happens is I'm extracting this heat from this cold reservoir, the sample of ice, to freeze it. So how do I figure this out? What's this Qc? Well, if you think about what's going on here, we really have a phase change. We actually have a combination of a temperature and a phase change. So if you look at the sort of temperature, the q versus t graph of what's going on, what happens here is that we have, remember that the water kind of looks like this. So what happens here is that we're starting off sort of in the water region of the graph. Right? So this is like a 100°C. This is 0 degrees Celsius. We're starting over here at 20. This is my initial. And then we want to go down the graph like this. So basically, make all the water at 0 degrees Celsius. And then we want to completely freeze it, which means it's going to go all the way across the phase change, and it's going to end up right over here. So this is my final.
So what happens here is that the heat that I need to extract from the cold reservoir is actually a combination of both the Q=mcΔt and the Q=ml. It's going to be both a temperature and a phase change. So what happens here is this Qc is equal to Qtotal, and this is just equal to let me just scoot this down. And this is equal to mc×Δt+ml. And this is going to be the m the latent heat of fusion. Alright?
So now I'm just going to go ahead and start plugging in some numbers here. So I've got the mass, which is remember, 0.5. So I've got 0.5. Then I've got the c for water, which is 4186. I have that down here just in case you forgot it. And the temperature change. Well, the temperature change in this step right here is going from 20, that's my sort of starting point, down to 0. So in other words, final minus initial would be 0 minus 20. Then we have to add this to the total amount of water that's changing phase. So now we have to do 0.5, right, because all of it is going to freeze into ice, times the latent heat of fusion for water, which is 3.34×105. And I'm just reading that off of my table of constants here. So what happens is, by the way, also you have some negative signs that you have to account for because technically we're freezing the ice. So what happens is you're going to have to insert a negative sign here and a negative sign here.
So w