How long to freeze water?

by Patrick Ford
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Hey guys, so let's check out our example. So we've got some kind of a household mini refrigerator with some coefficient of performance and a power input and 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? Well, usually when we asked how long it takes for something to happen, that's going to be a delta T. So where in our equations are we gonna get a delta T out of? Well, if we take a look here, really, the only thing we have any information about is the coefficient of performance, which is K. So K equals three and K is equal to the the QC over W. Right? 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 what happens here is that 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 W over time, then what we can do here is we can say that w is equal to power times delta T. So I'm actually gonna move this down here because basically we're gonna do here is we're gonna re replace this W here in terms of power. So this QC over W becomes Qc divided by power times time. Okay, so what happens here is that in order to get the delta T. I'm gonna have to move the delta T over to the other side and then the K down. And basically they're gonna trade places. What you're gonna end up with this is this equation here, the delta T. The amount of time it takes is gonna 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, 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 uh the Q versus T. Graph of what's going on, what happens here is that we have to 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 of the graph? Right, So this is like 100 this is 100 C. This is zero 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°C 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 gonna 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 equals M. C, delta T. And the Q equals M. L. It's going to be both a temperature and a phase change. So what happens here is this Q. C. Is equal to Q. Total and this is just equal to just scoot this down and this is equal to M. C. Times delta T. Plus M. L. And this is gonna be the M the latent heat of fusion. Alright, So now I'm just gonna 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 to the sea for water, which is 41-86, that 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 zero. So in other words, final minus initial would be 0 -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 gonna is gonna freeze into ice times. The the latent heat of fusion for water, which is 3.34 Times 10 to the 5th. And I'm just reading that off of my table of constants here. So what happens is by the way, also you have some some negative signs that you have to account for because technically we're freezing the ice. So what happens is you're gonna have to insert a negative sign here and a negative sign here. So what I want to mention here is that we actually want to end up with a positive number in this equation because otherwise your delta T. Is gonna be negative. So really all you have to do is just say, well the total amount of heat that you actually need to extract is gonna be this absolute sign, the absolute value sign of all these numbers plugged in. So when you work this out, what you're gonna get here is that you have to extract 2.1 times 10 to the fifth jewels. So that's the total amount of heat that you have to extract from the ice. So now what we have to do is now that we have this Q. C. We're really just gonna plug this into our delta T. Equation. So I'm gonna move this down here and my delta T. Is going to be Q. C. Which is 2.1 times 10 to the fifth, divided by the power, the power is 85 watts into the coefficient of performance is going to be three. So if you work this out, what you're gonna get is you're gonna get about 823 seconds. Alright, so that's how long it's gonna take for you to freeze this amount of water. So if you think about this this is kind of like 15 minutes, which kind of should make some sense, right? So if you stick a, you know, a little bit of water in the in the fridge, it doesn't just definitely doesn't freeze in like a minute or two, it's gonna take some time, maybe like 15 minutes, something like that. So that's pretty reasonable. Alright, so that's it for this one guys, let me know if you have any questions