Question

A household refrigerator runs one-fourth of the time and removes heat from the food compartment at an average rate of \(800 \mathrm{kJ} / \mathrm{h}\). If the COP of the refrigerator is \(2.2,\) determine the power the refrigerator draws when running.

Short answer

Question: Determine the power the refrigerator draws when running, given that it runs one-fourth of the time, removes heat at an average rate of 800 kJ/h, and has a COP of 2.2. Answer: The refrigerator draws 25.25 watts of power when running.

Step-by-step solution

Understand the relationship between COP, power, and heat transfer

The Coefficient of Performance (COP) of the refrigerator is given by the formula: COP = (Heat removed from the compartment (Q) / Work input (W)) We are given COP = 2.2, and heat removal rate (Q') = 800kJ/h. Our goal is to find the power draw (W'), which is the work input per unit time.

Convert heat removal rate from kJ/h to kW

First, we need to convert the heat removal rate from kJ/h to kW. To do this, we will make use of the conversion factor: 1 kJ/h = (1/3600) kW Q' = 800 kJ/h = 800 * (1/3600) kW = 0.2222 kW

Calculate the work input per unit time (power draw)

Now, we can rearrange the COP formula and calculate the power draw (W') using the given COP and the heat removal rate (Q') we have just found: W' = Q' / COP W' = 0.2222 kW / 2.2 = 0.101 kW

Consider runtime fraction of the refrigerator

We are informed that the refrigerator runs one-fourth of the time. To find the average power draw during its runtime, we multiply the power draw by the runtime fraction: Average Power Draw = 0.101 kW * (1/4) = 0.02525 kW

Convert average power draw to watts

Finally, we will convert the obtained average power draw in kW to watts: Power Draw (W) = 0.02525 kW * 1000 = 25.25 W Therefore, the refrigerator draws 25.25 watts of power when running.