### Video Transcript

A refrigerator has a coefficient of performance of 3.0. And it requires an input of 2.0 times 10 to the power of two joules of work per cycle. How much heat per cycle is removed from the cold reservoir? How much heat per cycle is discarded to the hot reservoir?

Okay, so in this question, we’ve got a refrigerator which has a coefficient of performance of 3.0. We’re told that it requires an input of 2.0 times 10 to the power of two joules of work per cycle. We’re asked to work out the amount of heat per cycle removed from the cold reservoir and the heat per cycle discarded to the hot reservoir. Now, let’s start by discussing first what a refrigerator actually does.

A refrigerator is in thermal contact with a hot reservoir and a cold reservoir. What it does is to pump heat from the cold reservoir into the hot one with the help of the input work. So let’s say the heat pumped from the cold reservoir is cold 𝑄 one, the heat pumped into the hot reservoir is called 𝑄 two, and the input work is called 𝑊. Now, we’ve labeled all the important bits of information on our diagram. So let’s now start by looking at the coefficient of performance.

The coefficient of performance of a heater or a refrigerator is defined as the useful heat, 𝑄 sub useful, divided by the input work. Now, the definition of useful heat varies depending on what kind of appliance we have. In this case, we have a refrigerator. The purpose of a refrigerator is to pump heat away from the cold reservoir. And so, in our case, the useful heat is the heat pumped away from the reservoir, which is 𝑄 sub one. This is known as the useful heat because what we’re trying to do with a refrigerator is to pump heat away from the cold reservoir in order to keep it cool. So we can say that our coefficient of performance here is equal to 𝑄 sub one divided by 𝑊.

But we’ve been told in the question that this is equal to 3.0. And we’ve been told in the question that the input work, 𝑊, is 2.0 times 10 to the power of two joules. And this is the same as 200 joules. So we can rearrange to find out what 𝑄 sub one is. 𝑄 sub one is 3.0 times 200 joules. And this happens to be 600 joules. Now, if we look at the very first sub question, we’re asked to find out the amount of heat removed per cycle from the cold reservoir. This value is precisely 𝑄 one. Because, remember, the work that we use, the input work, was the work per cycle. Therefore, 𝑄 one is the amount of heat removed from the cold reservoir per cycle. So our final answer to this part of the question is that the amount of heat removed per cycle from the cold reservoir is 600 joules.

Now, let’s work towards finding the amount of heat per cycle discarded to the hot reservoir. In other words, we’re trying to find out the value of 𝑄 sub two. In order to do this, we need to use the law of conservation of energy. Now, the law of conservation of energy technically states that no energy is created nor destroyed. It’s simply transferred from one form to another.

But in this situation, we need to word it slightly differently. Basically, we can express the law of conservation of energy here as the energy going into the refrigerator is equal to the energy coming out of the refrigerator because all of the energy that goes into the refrigerator must come out again. The refrigerator cannot hold on to some energy or like destroy it or do something else with it. So anything that goes in, that’s the work and 𝑄 one, must all come out again as 𝑄 two. In other words, 𝑊 plus 𝑄 one is equal to 𝑄 two.

We were told in the question that 𝑊 is equal to 2.0 times 10 to the power of two joules which is 200 joules. And we’ve just calculated that 𝑄 one is equal to 600 joules. So 𝑄 two is equal to 200 joules plus 600 joules. And this ends up being 800 joules. And so our final answer to this part of the question is that the heat per cycle discarded to the hot reservoir is 800 joules.