Question
(a) What is the best coefficient of performance for a refrigerator that cools an environment at and has heat transfer to another environment at ? (b) How much work in joules must be done for a heat transfer of 4186 kJ from the cold environment? (c) What is the cost of doing this if the work costs 10.0 cents per (a kilowatt-hour)? (d) How many kJ of heat transfer occurs into the warm environment? (e) Discuss what type of refrigerator might operate between these temperatures.
Final Answer
- 3.24
Solution video
OpenStax College Physics for AP® Courses, Chapter 15, Problem 42 (Problems & Exercises)
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Video Transcript
This is College Physics Answers with Shaun Dychko. An ideal refrigerator is cooling an environment at negative 30.0 degrees Celsius which we convert into Kelvin by adding 273.15 and it's depositing the heat taken out of this cold temperature reservoir into a high temperature reservoir at 45.0 degrees Celsius and question (a) is asking us what is the best coefficient of performance that we could possibly have? Well the coefficient of performance for a refrigerator is the coefficient of performance for a heat pump minus 1. So the coefficient for a heat pump is the heat that's deposited into the high temperature reservoir divided by the work that it takes to do it. and efficiency is the reciprocal of that— it's w over Q h— and the Carnot efficiency is 1 minus the cold temperature divided by the high temperature. So that means this heat pump coefficient of performance can be rewritten as efficiency to the negative 1 because this is the exponent for flipping a fraction and taking its reciprocal in other words and then we can write this then as 1 minus T c over T h to the negative 1. So we replace the coefficient of performance for a heat pump with 1 minus T c over T h to the negative 1 all minus 1 and then plug in numbers. So we have 1 minus 243.15 Kelvin divided by 318.15 Kelvin all to the negative 1 minus 1 and that's 3.24 is the best coefficient of performance you can have for this refrigerator. So how much work in joules must be done to transfer 4186 kilojoules of energy from the cold environment? So another way to express the coefficient of performance for a refrigerator is to say it's the amount of heat that it draws out of the cold environment divided by the amount of work energy it takes to do that. So we can solve for w by multiplying both sides by w and dividing both sides by the coefficient of performance. So w equals Q c over coefficient of performance so that's 4186 kilojoules divided by 3.242 and that's 1290 kilojoules of work energy needed to operate the refrigerator and the cost will be that expressed in joules by multiplying by 10 to the 3 since that's what the prefix 'kilo' means and then multiply that by 10.0 cents for every 3.60 times 10 to the 6 joules and this works out to 3.59 cents. And part (d) asks how many kilojoules of heat transfer occurs into the warm environment? Well that's gonna be the amount of work energy used to make the transfer plus the amount of energy or heat taken out of the cold environment. So that's 1291.18 kilojoules that we calculated up here plus 4186 kilojoules of heat taken from the cold environment for a total of 5480 kilojoules.