A refrigerator removes heat from the inside at a rate of 200 J/s and expels it t
Practice Questions
Q1
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)
0.8
1.25
1.5
2.0
Questions & Step-by-Step Solutions
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)
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.
Coefficient of Performance (COP) – The COP of a refrigerator is a measure of its efficiency, defined as the ratio of heat removed from the cold reservoir (Q_in) to the work input (W).
Heat Transfer – Understanding the rates of heat removal and expulsion is crucial for calculating the COP.
Work Calculation – The work done by the refrigerator is the difference between the heat expelled and the heat absorbed.
