A refrigerator removes heat from the interior at a rate of 200 W. If the coefficient of performance (COP) is 4, what is the power consumed by the refrigerator?
Practice Questions
1 question
Q1
A refrigerator removes heat from the interior at a rate of 200 W. If the coefficient of performance (COP) is 4, what is the power consumed by the refrigerator?
50 W
100 W
200 W
800 W
COP = Qc/W, thus W = Qc/COP = 200 W / 4 = 50 W.
Questions & Step-by-step Solutions
1 item
Q
Q: A refrigerator removes heat from the interior at a rate of 200 W. If the coefficient of performance (COP) is 4, what is the power consumed by the refrigerator?
Solution: COP = Qc/W, thus W = Qc/COP = 200 W / 4 = 50 W.
Steps: 7
Step 1: Understand the problem. We have a refrigerator that removes heat from inside it at a rate of 200 Watts (W).
Step 2: Know what the Coefficient of Performance (COP) means. The COP is a measure of the efficiency of the refrigerator, given as 4 in this case.
Step 3: Use the formula for COP. The formula is COP = Qc / W, where Qc is the heat removed (200 W) and W is the power consumed by the refrigerator.
Step 4: Rearrange the formula to find W. We can rearrange it to W = Qc / COP.
Step 5: Substitute the values into the formula. We have Qc = 200 W and COP = 4, so W = 200 W / 4.
Step 6: Calculate the power consumed. Dividing 200 W by 4 gives us 50 W.
Step 7: Conclude that the power consumed by the refrigerator is 50 W.
