A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the surroundings at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2023)

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Q: A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the surroundings at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2023)

Solution: COP = Q_in / W = 200 J/s / (250 J/s - 200 J/s) = 200 / 50 = 4.

Steps: 7

Step 1: Identify the heat removed from the inside of the refrigerator, which is given as 200 J/s. This is Q_in.
Step 2: Identify the heat expelled to the surroundings, which is given as 250 J/s. This is Q_out.
Step 3: Calculate the work done by the refrigerator (W) using the formula: W = Q_out - Q_in.
Step 4: Substitute the values: W = 250 J/s - 200 J/s = 50 J/s.
Step 5: Use the formula for the coefficient of performance (COP): COP = Q_in / W.
Step 6: Substitute the values into the COP formula: COP = 200 J/s / 50 J/s.
Step 7: Calculate the COP: COP = 4.

A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the surroundings at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2023)

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