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 outside at a rate of 250 J/s. What is the coefficient of performance (COP) of the refrigerator? (2022)
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 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.
Step 7: Perform the division: COP = 200 / 50 = 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_c) 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 removed.
