Question

You place your lunch leftovers in the refrigerator. Suppose the refrigerator needs to remove 1.4690E+4 J of thermal energy from your lunch to cool it to the temperature of the inside of the refrigerator. In the meantime, this means the refrigerator produces 1.8957E+4 J of thermal energy that it expels into the kitchen as a result. What is the total work done by the compressor motor in the refrigerator? (Ignore any thermal loses due to friction in the motor.)

What is the Coefficient of Performance for the refrigerator?

Your refrigerator actually acts like a heater in your kitchen. Suppose you have a small electric space heater that has a power output of 1.2kW. How long would this heater have to run to produce the same amount of heat as the refrigerator produced while cooling your leftovers?

It takes electricity to run the motor on the refrigerator. If your cost of electricity is 6 cents per kilowatt*hour, how much does it cost to cool your lunch down?

Answer #1

a )

given

Q_{H} = 1.8957 X 10^{4} J

Q_{C} = 1.4690 X 10^{4} J

the work done = thermal energy expelled - thermal energy from the launch

W = Q_{H} - Q_{C }

W = 1.8957 X 10^{4} - 1.4690 X 10^{4}

W = 4267 J

**the total work done by the compressor motor in the
refrigerator is W = 4267 J**

b )

The Coefficient of Performance for the refrigerator is

COP = Q_{H} / W

COP = 1.8957 X 10^{4} / 4267

**COP = 4.4426**

c )

the given power P = 1.2 kW

P = 1200 W

the power P = W / t

then t = W / P

t = 4267 / 1200

**t = 3.555 sec**

d )

here we got W = 4267 and 1 kWH = 36 X 10^{5} J

one kWH = 1.18 X 10^{-3} kWH

given

6 cents per kilowatt*hour

6 X 1.18 X 10^{-3}**= 7.08 X 10 ^{-3}
cents**

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