A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the outside at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2022)
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A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the outside at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2022)
Q: A refrigerator removes heat from the inside at a rate of 200 J/s and expels it to the outside at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2022)
Step 1: Identify the heat removed from the inside of the refrigerator, which is given as 200 J/s. This is denoted as Q_c.
Step 2: Identify the heat expelled to the outside, which is given as 250 J/s. This is denoted as Q_h.
Step 3: Calculate the work done (W) by the refrigerator. This is the difference between the heat expelled and the heat removed: W = Q_h - Q_c = 250 J/s - 200 J/s.
Step 4: Perform the calculation for W: W = 250 J/s - 200 J/s = 50 J/s.
Step 5: Use the formula for the coefficient of performance (COP) of the refrigerator: COP = Q_c / W.
Step 6: Substitute the values into the formula: COP = 200 J/s / 50 J/s.
