1

concept

## Refrigerators

6m

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Hey guys, we've talked a lot about heat engines in the last couple of videos and in this video, I'm going to introduce you to another type of device called a refrigerator and what we're going to see is that there's a lot of similarities between refrigerators and heat engines. So let's go ahead and check this out here. Now. Before we get started, I want to talk about about an idea which we've kind of taken for granted up until now, which is the idea that he always flows from hotter to colder and never the other way around. We talked a lot about this when we studied kalorama tree. The idea here is that if you have a really hot coffee cup and you hold it in your hand, heat will always flow from the hotter coffee to the colder hand, you would never expect that by holding a hot coffee cup that your hand would get colder and the coffee cup would get warmer. That would just be insane. In fact, this is so true. This is actually one of the other statements of the second law of thermodynamics. This is sometimes referred to as the Claudius or the refrigerator statement, which is that in a cycle it's impossible for heat to flow from colder to hotter without an input of work. And that's actually the most important sort of part of that sentence there, without an input of work. So sometimes you might see this written as spontaneously from colder to hotter. The way I kind of like to think about this is like a ball that's falling from higher to lower heights, if you have a table like this, a ball always fall downwards because it's going from higher to lower potential, you would never see the ball spontaneously go upwards against gravity because that's just impossible right? It would have to gain energy to do that, so that's why you have to supply some work. So that leads me to what a refrigerator is, because the refrigerator is basically just a type of machine, kind of like a heat engine. And what it does is it takes in work in order to pump the heat energy from colder to hotter. So it doesn't violate the second law of thermodynamics because you're going to supply work in order to do that. So the way this kind of works here is a heat engine would take the natural flow of heat from hot to cold and you would extract some work out of it. But a refrigerator, like the one in your house, what it does is it takes heat from the food and liquid and air that's inside of this compartment right here, that's the cold reservoir and it takes work from electricity, this work is supplied from the power outlet and then it basically pumps heat out to the outside air and that is Q. H. That's the hot reservoir. So if you'll notice here a heat engine and a refrigerator are basically just backwards. They are backwards, processes. All the arrows are reversed and you're actually now pumping in work in order to extract heat out to the hot reservoir. Alright, that's basically the fundamental difference between a heat engine and a refrigerator. Now, the only really equation that you need to know here is that the uh the change in the internal energy for a heat engine was just equal to zero because it's a cyclic process. It's the same thing for a refrigerator. They're both cyclic processes again, just running in reverse. So what that means here is that this work is still just gonna be Q H minus Q. C. As long as everything is absolute values and positive numbers. This number here, this work is always gonna be the difference between these two. The other one is the efficiency of the heat engine. Now remember that the efficiency of a heat engine was basically just how good that heat engine was at doing work. And there's a similar sort of term for refrigerators, which is called the coefficient of performance. A cooperation of performance is basically how good a refrigerator is. It's given by the letter K. Here and in order to kind of like think about this, What I would like to think about is that the efficiency was always work over Q. H. So the efficiency of a heat engine. This work here is kind of like what you got out of it, right, you got out some useful work from this heat engine, but you don't get that work for free. you have to pay the engine something, which is the heat that's taken from the hot reservoir in order to get it. So this efficiency is like what you got out of it, The work divided by what you paid to get it. And that was this qh here, the co finisher, official performance is kind of the same idea, except that some of the letters are different. What you get out of it, what you really want to get out of the refrigerator. If you want to get all the heat extracted from this cold reservoir, write a really good refrigerator is going to make things really, really cold on the inside and what you pay to get. It is actually just the work that is supplied from electricity, the power outlet. So this is the equation for the coefficient of performance and just how we can get from this equation and rewrite this in terms of this, then we can actually take this equation and we can rewrite this in terms of other variables as well. So, really, these are just the two equations that you need to know for the refrigerators, that's really all there is to it. So, let's go ahead and take a look at the problems here, because there are types of problems that you'll see are gonna be very similar to heat engine problems. So here we have a refrigerator and what it does is it's taking in 600 kg jewels of heat from the food inside. Remember that food inside is gonna be the cold reservoir. So this Q. C. Here is equal to 600 this is going to be killed jules. And what it does is it releases 720 kg jewels to the much warmer room. So that's the hot reservoir. So that's Q. H. This Q. H. here is equal to 720. Notice how this number is bigger than this one and that totally makes sense. So what happens now we want to calculate in part a the work that's required to run the refrigerator basically we're going to calculate what is W. And in order to do that, we're just gonna use these equations over here. Now we don't not told anything about the coefficient of performance yet. So we can actually just use this equation right here which is the work is always the difference between the hot and cold the Q. H. And Q. C. So this w here is just always the absolute value of Q. H minus Q. C. And so that's just gonna be 7 20 kg joules minus 600 kg jewels. And this equals 120. So this w here is just equal to 120 killer jewels. And that's the answer. Alright so let's move on now to part B. Part B. Now asks what is the coefficient of performance for the refrigerator. Now we're just gonna straight up just use this equation right here, we can use um Casey oh sorry, this is gonna be okay. Que is either QC divided by w or its Qc divided by a q h minus q C. So do we have enough information to use any of these equations? We actually have all of the numbers, right? We just filled out this energy flow diagram here. We've got all the numbers so it actually doesn't matter which equation we use, we're gonna get the right answer. So your coefficient of performance here is going to be Kay Qc which is the 600 that's what you got out of it, what you paid to get? It was the 120 kg jewels of work. So it doesn't matter if you actually convert this into jewels because the ratio is gonna be the same, this is gonna be a coefficient of performance of five. Now these coefficients of performances are always just gonna be numbers like this. Usually there'll be numbers from like 3-10 or something like that. Alright, so that's it for this one guys, let me know if you have any questions

2

Problem

A refrigerator has a coefficient of performance of 2.4. Each cycle, it takes in 3×10^{4} J of heat from the cold reservoir. How much is expelled to the hot reservoir?

A

2.5×10

^{4}JB

4.25×10

^{4}JC

2.1×10

^{4}JD

1.25×10

^{4}J3

example

## How long to freeze water?

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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

4

concept

## Heat Pumps

6m

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Hey guys, so now that we understand how refrigerators work in this video, I'm gonna show you another type of advice that you might run across, which is called a heat pump, and what I'm gonna show you is that they work very similarly to refrigerators. We'll even see a similar equation for the coefficient of performance. The trickiest thing here is understanding conceptually the differences between a heat engine, a refrigerator and a heat pump. So that's what I want to show you in this video. Let's get started. A heat pump, like I said, is just like a refrigerator, It pumps heat from colder to hotter. So let's kind of recap everything we've learned so far, a heat engine, remember, takes the natural flow of heat from hot to cold, it takes some energy from the whole hot reservoir and it extracts some usable work and it produces some usable work here and then it takes whatever works whatever energy it didn't convert to work and expels as waste heat to the cold reservoir. A real life example of this would be like an engine, like a real engine in your car, or like a generator that you might have in your house. A generator, which which uses gasoline, basically can power your home just in case you lose power or something like that. Alright, now a refrigerator is like a heat engine, but it works in reverse. What it does is it extracts heat from the cold reservoir and it requires an input of work like the electrical outlet that your fridges plugged into, and then it expels heat out to the already hot reservoir. Alright, so the key thing here is that a refrigerator doesn't produce work. It requires some work to be done. Alright, a real life example would be like the fridge in your house or even like an air conditioner. Right? You want the inside of your home to be cold. So you have to take that heat and you have to pump it to the outside. Now, let's talk about a heat pump, right? It still pumps heat from colder to hotter. However, what happens is that the reservoirs are switched and that's the sort of main idea of a heat pump. So, if you live somewhere really, really cold, what happens is the cold reservoir is the outside air, right? If you live somewhere north, basically what a heat pump does, is it takes heat from the already cold air outside into sort of like a generator or something like something like that, and then it requires some work. And what it does is it heats that air and then pumps that into your house. So, here's the difference in a fridge, the cold reservoir was inside the hot reservoir was outside in a heat pump. It's sort of inverted the colon reservoir is outside, but now the hot reservoir is inside and that's the main difference. Now, a heat pump still just like a refrigerator requires some work to run. So that's why it's not really like a heat engine, it's more like a fridge, but it's sort of running inside out. All right, so, a real life example is, you know, if you live somewhere cold, you might have something like a space heater or something like that. That's a perfect example. Alright, So what does that mean for the equations? Well, for a refrigerator, the coefficient of performance was QC over w right. The heat extracted from the cold reservoir, that's what you get out of it, divided by the work. Which is what you pay to get it for a heat pump. It's a little bit different because what you're really getting out of it is actually this you're actually getting this heat that gets pumped into your house or the hot reservoir. So, here what happens is that we replace the QC with a Q. H. That's all there is to it. Now, it's just a Q H. In the numerator. Alright, so that's all there is to it guys, that's sort of a conceptual difference. Let's go ahead and take a look at our example. Alright, so here we have a heat pump. It has a coefficient performance of 3.6. Now, one thing I forgot to mention here is that the coefficient performance for heat pumps. We did note as K hp is just sort of like, you know, so you don't get confused between them. So, we have here is that K H p is equal to 3.6? We also have as a power supply of seven times 10 to the third watts. So remember that power is p this is equal to seven times 10 to the third watts. Now be careful here because this means that it's seven times 10 to the third jewels per second. That's what I want is what we want to do in this problem is calculate the heat energy that's delivered into a home over four hours of use. So really want to heat energy. Remember that's Q. Except we want it delivered into a home. So remember, according to this diagram, we're using this diagram, that's qh from, that's what you get out of the equation. So we're really looking for Q. H delivered over four hours. Alright, so how do I calculate that? But we only have one equation for heat pumps. It's just this one right over here. So we've got here is that K hp is equal to the heat delivered to the hot reservoir divided by the work. Now, remember this isn't qh over four hours. This is just Q. H. Over W. This is just the equation for one cycle, but we can actually modify it so that it works for any amount of time. So, remember that the relationship between work and power is that power is W over delta T. So what happens is I can rewrite this and say w is equal to power times time. So what happens is we can basically take this cage P. Equation in these terms, we can rewrite them. This is gonna be K. Qh and this is the work done, this is gonna be power times the delta T. Over four hours. Right? So I'm writing this over four hours of use, I just have the power and I just need to multiply by delta T. For four hours. Now what I do that basically what I get here is that this qh isn't just for one cycle, This QH becomes the heat energy delivered over four hours. So this is how I get to my target variable over here. It just comes straight from this equation here. Alright. So basically if I move this to the other side and start plugging in numbers right times P times delta T. For four hours. So my K. H. P. Is equal to 3.6. Actually gonna go ahead and start that another line. So we have 3.6 times the power which is seven times 10 to the third. And then we have the delta T. For four hours. We want it in seconds. Because remember this is jules per second. So we have here is I have four hours times 60 minutes per hour times 60 seconds per minutes. You'll see that the hours and minutes cancels, leave you only with seconds. And if you go ahead and work this out, what you're gonna get is you're gonna get 3.63 times 10 to the eighth. And that's in jewels. So thats how much heat energy gets pumped into your home over four hours. Alright, so that's it for this one guys, let me know if you have any questions.

Additional resources for Refrigerators

PRACTICE PROBLEMS AND ACTIVITIES (4)

- A certain brand of freezer is advertised to use 730 kW•h of energy per year. (c) What is the theoretical maxim...
- A certain brand of freezer is advertised to use 730 kW•h of energy per year. (a) Assuming the freezer operates...
- A refrigerator has a coefficient of performance of 2.25, runs on an input of 135 W of electrical power, and ke...
- The coefficient of performance K = H/P is a dimensionless quantity. Its value is independent of the units used...