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Ch 20: The Second Law of Thermodynamics

Chapter 20, Problem 20

A Carnot refrigerator is operated between two heat reservoirs at temperatures of 320 K and 270 K. (c) What is the coefficient of performance of the refrigerator?

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Hey everyone welcome back in this problem. We have a cardinal refrigerator that is used to maintain water at 5°C. Okay. By rejecting heat into the atmosphere at 30°C In each cycle 700 joules of heat is delivered to the high temperature reservoir. Okay. And we are asked to find the coefficient of performance K. of this refrigerator. Now let's recall the coefficient of performance. K. Can be written as the absolute value of Q. C. Okay the quantity of heat um being removed from the cold reservoir divided by Q. H. Absolute value. Okay the amount of heat being delivered to the high temperature reservoir minus the absolute value of Q. C. Okay. All right so in order to find K. What we need is our Q. C. And R. Q. H. Okay let's go ahead and write down what we're given. Okay we're told that the cold water in the refrigerator is five degrees Celsius. Okay so T. C. The temperature in the cold reservoir it's gonna be five degrees Celsius, converting this into Calvin. We're gonna have 278.15 Calvin. Okay we add 273.15. We're told that the heat gets rejected to the atmosphere and so the atmosphere is our high temperature reservoir at 30 degrees Celsius. So T. H. Is going to be 30 degrees Celsius. Okay converting to Calvin again at 2 73.15 we get 303.15 Calvin. We're told that 700 jewels of heat is delivered to the high temperature reservoir in each cycle. Okay this is going to be the quantity Q. H. And so the absolute value of Q. H. is going to be 700 jewels. We don't know Qc yet. How can we find Q. C. Let's recall the relationship between the quantity of heat and the temperatures. We can write that the absolute value of Q. C divided by the absolute value of Q. H. Is equal to the absolute value of T. C. Divided by the absolute value of th. Okay so the ratio of the quantity of heat is equal to the ratio of the temperatures. Okay this is gonna give us Q. C. The absolute value is equal to T. C divided by th so to 78. Calvin divided by 303. Calvin all times Q H. 700 jewels. The unit of Calvin will cancel. We're just gonna be left with the unit of jewels and this is going to be 642 . - eight jewels. All right, so now we have our Q. C. And R. Q. H. We can go ahead and find K. The coefficient of performance that we're looking for. Okay, is going to be equal to Q C. 642. 2 8 jewels divided by QH. 700 jewels minus Q C. 642.27 to 8 jewels K. Unit of jewel on the top unit of jewel on the bottom. When we divide we're going to be left with no units. Okay? And we're gonna get K. Is equal to 11.1 And that is it? Okay. So the coefficient of performance is going to be c. I hope this video helped. Thanks everyone for watching See you in the next one.