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?

Hey, guys. So let's get some practice here. So we're told that this refrigerator has a coefficient of performance of 2.4. So remember that is k. We're told that each cycle, it's gonna take in this amount of heat from the cold reservoir to make it colder, and it has to expel some heat to the hot reservoir. And that's when we wanna calculate, how much heat gets expelled to the hot reservoir. So let me just quickly draw the energy flow diagram because it's a good practice to do it for refrigerators. So this is my hot reservoir and it's connected to my engine. Remember some work is required to go into this engine in order to expel heat from the hot to the hot reservoir, and it has to take in heat from the cold reservoir. Remember, think of this as like the refrigerator in your home. You wanna extract heat to make the stuff in your refrigerator cold. So what happens here is that we're extracting this amount of heat from the cold reservoir, that's q c. And what we're told here is that's 3 times 10 to the 4th. So that's 3 times 10 to the 4th. Then we're gonna expel some heat out to the hot reservoir, and that's what we wanna find here. Okay? Now, unfortunately, we can't use this equation, this w equals qh minus qc, because if you look we don't have the work and we also don't have the qh. So we're gonna have to use another equation. Something else that we know about this problem is that we have the coefficient of performance. We have k. So we have k is equal to 2.4. And if you look we have equations for K. So we have that K is equal to qc over w. Do we wanna use this? Well, again, our target variable here is q h. So we don't wanna use this format here because it's not it doesn't have our target variable in it. Instead, we're gonna use this equation. So the sort of the expanded version of it. So we're gonna have to use qc divided by qh0minusqc. So now if you look at this, we really do only have one variable that we're missing. This is our target variable. We have qc and we also have the coefficient of performance. So really all the physics has done here and we're really just doing algebra to solve for this qh. Okay? So this is gonna be 2.4 equals the QC is 3 times 10 to the 4th divided by QH minus 3 times 10 to the 4th. Alright? So how do I get this sort of q h? Well, it's kind of tricky because this q h here is sort of, you know, it's subtracted from this denominator here. So I can't just pull this out and move this to the top. I actually have to pull the whole entire denominator and move it up to the top like this. So I have to multiply this to the other side. What you end up getting here is 2.4 times qh minus 3 times 10 to the 4th. So that's equal to 3 times 10 to the 4th. Okay? So now, we do we have a couple of things that you can do. You could distribute this 2.4 into both of these terms, but that might take actually a little bit longer. So instead, what you can do is you could just immediately divide by the 2 point 4. What you're gonna get here is qh minus 3 times 10 to the 4th is equal to and when you work this out here, when you divide by 2.4 to get rid of this, what you're gonna get is 1.25 times 10 to the 4th. Now, all you have to do is just add 3 times 10 to the 4th to both sides. So you cross that out and your QH is gonna be 4.25 times 10 to the 4th joules. That's the amount of heat that gets expelled out to the hot reservoir. 4.25 times 10 to the 4th. Alright? So if you want to, you can plug these numbers, you know, the QH and your QC, back into your coefficient of performance equation, and you should get back to 2.4. That's just a good way to double check. Alright? So your answer choice is gonna be answer choice b. That's it for this one, guys.

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23. The Second Law of Thermodynamics

Refrigerators

Physics

23. The Second Law of Thermodynamics

Refrigerators

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Multiple Choice

A refrigerator has a coefficient of performance of 2.4. Each cycle, it takes in 3×10⁴ J of heat from the cold reservoir. How much is expelled to the hot reservoir?

2.5×10⁴ J

4.25×10⁴ J

2.1×10⁴ J

1.25×10⁴ J

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Verified step by step guidance

Understand the concept of the coefficient of performance (COP) for a refrigerator, which is defined as the ratio of the heat removed from the cold reservoir (Qc) to the work input (W): COP = Qc / W.

Rearrange the formula to solve for the work input (W): W = Qc / COP.

Substitute the given values into the equation: Qc = 3×10^4 J and COP = 2.4, to find W.

Use the first law of thermodynamics, which states that the total energy input (heat removed from the cold reservoir plus work done) equals the energy expelled to the hot reservoir (Qh): Qh = Qc + W.

Substitute the values of Qc and W into the equation Qh = Qc + W to find the amount of heat expelled to the hot reservoir.