Hey, guys. So 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 are 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 an idea which we've kind of taken for granted up until now, which is the idea that heat always flows from hotter to colder and never the other way around. We talked a lot about this when we studied calorimetry. The idea here is 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 Clausius 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 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 will 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 a 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 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 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 backward 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 change in the internal energy for a heat engine was just equal to 0 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 going to be \(q_h - q_c\) as long as everything is in absolute values and positive numbers. This number here, this work is always going to be the difference between these 2. The other one is the efficiency, versus 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 term for refrigerators, which is called the coefficient of performance. A coefficient 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 always 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 had 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 coefficient of 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 a refrigerator is you want to get all the heat extracted from this cold Right? 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 2 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 the types of problems that you'll see are going to be very similar to heat engine problems. So here we have a refrigerator, and what it does is it's taking in 600 kilojoules of heat from the food inside. So remember, that food inside is going to be the cold reservoir. So this \(q_c\) here is equal to 600. This is going to be kilojoules. And what it does is it releases 720 kilojoules to the much warmer room. So that's the hot reservoir. So that's \(QH\). This \(QH\) 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 going to use these equations over here. Now we're 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 \(QH\) and \(QC\). So this \(w\) here is just always the absolute value of \(QH\) - \(QC\), and so that's just going to be 720 kilojoules - 600 kilojoules, and this equals 120. So this \(w\) here is just equal to 120 kilojoules, 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 going to straight up just use this equation right here. We can use, \(k\), or sorry, this is going to be \(k\). \(k\) is either \(q_c\) divided by \(w\) or it's \(q_c\) divided by \(q_h\) - \(q_c\). 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 going to get the right answer. So your coefficient of performance here is going to be \(k\), \(q_c\), which is the 600. That's what you got out of it. What you paid to get it was the 120 kilojoules of work. So it doesn't matter if you actually convert this into joules because the ratio is going to be the same. This is going to be a coefficient of performance of 5. Now these coefficients of performances are always just going to be numbers like this. Usually they'll be numbers from like 3 to 10 or something like that. Alright. So that's it for this one, guys. Let me know if you have any questions.

# Refrigerators - Online Tutor, Practice Problems & Exam Prep

### Refrigerators

#### Video transcript

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?

^{4}J

^{4}J

^{4}J

^{4}J

### How long to freeze water?

#### Video transcript

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 Q_{total}, 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

### Heat Pumps

#### Video transcript

Hey, guys. So now that we understand how refrigerators work, in this video, I'm going to show you another type of device that you might run across, which is called a heat pump. And what I'm going to 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 hot reservoir, and it extracts some usable work and it produces some usable work here. And then it takes whatever work 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 you would use as 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 fridge is 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 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 aren't switched. That's the sort of main idea of a heat pump. So if you lived 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 cold 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. Alright? So, a real-life example is, you know, if you live somewhere, you know, 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 Q_{c}W, right? It's the heat extracted from the cold reservoir, that's what you get out of it, divided by the work, which is what you paid 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 Qh. 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 of performance of 3.6. Now one thing I forgot to mention here is that the coefficient of performance for heat pumps, we denote as KHp, just sort of like you so you don't get confused between them. So what we have here is that KHp is equal to 3.6. We also have is a power supply of 7×103 watts. So remember that power is P this is equal to 7×103 watts. Now be careful here because this means that it's 7×103 joules per second, that's what a watt is. What we want to do in this problem is calculate the heat energy that's delivered into a home over 4 hours of use. So really want a 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 Qh delivered over 4 hours. Alright. So how do I calculate that? Well, we only have one equation for heat pumps. It's just this one right over here. So we've got here is that KHp is equal to the heat delivered to the hot reservoir divided by the work. Now remember, this isn't Qh over 4 hours, it's just Qh over W. This is just the equation for one cycle. 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 KHp equation in these terms, we can rewrite them. This is going to be K, Qh and this is the work done, this is going to be power times the delta t over 4 hours. Right? So running this over 4 hours of use, I just have the power and I just need to multiply it by delta t over 4 hours. Now when I do that, basically what I get here is that this Qh doesn't isn't just for one cycle, this Qh becomes the heat energy delivered over 4 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, I can start plugging in numbers. Right? It's times P times delta t for 4 hours. So my KHp is equal to 3.6. Actually, let me go ahead and start that another line. So we have 3.6 times, the power, which is 7×103, and then we have the delta t for 4 hours. We want it in seconds because, remember, this is joules per second. So we have here is, I have 4 hours times 60 minutes per hour times 60 seconds per minutes. You'll see that the hours and minutes cancels, leaving you only with seconds. And if you go ahead and work this out, what you're gonna get is you're gonna get 3.6×108 and that's in joules. So that's how much heat energy gets pumped into your home over 4 hours. Alright. So that's it for this one, guys. Let me know if you have any questions.

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