A household fridge removes heat, determine power the fridge draws

[joejoe121]

Homework Statement

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

Relevant Equations

W_in = Q_L / COP

I'm confused about the wording "runs one-fourth of the time". My professor said we multiply the Q_L by four, but I don't get it conceptually.

 

[haruspex]

Homework Statement: A household refrigerator runs one-fourth of the time and removes heat
from the food compartment at an average rate of 800 kJ/h. If the COP of the
refrigerator is 2.2, determine the power the refrigerator draws when running.
Relevant Equations: W_in = Q_L / COP

I'm confused about the wording "runs one-fourth of the time". My professor said we multiply the Q_L by four, but I don't get it conceptually.

If it only runs a quarter of the time but overall averages using power P, what power must it average while running?

 

[joejoe121]

If it only runs a quarter of the time but overall averages using power P, what power must it average while running?

is to four times the power while running to compensate for the time not running?

 

[Steve4Physics]

@joejoe121, you are employed to move boxes from A to B.
On average you move 80 boxes per hour.
However, you are sneaking-off to read Physics Forums posts for 45minutes each hour.
At what rate are you moving boxes when you are are actually working?

Edited for clarity.

 

[haruspex]

is to four times the power while running to compensate for the time not running?

Yes.

 

[Lnewqban]

I'm confused about the wording "runs one-fourth of the time".

We say that a refrigerator "runs" when the refrigeration cycle is active.
That means that the compressor has been energized, making the refrigerant fluid circulate through the tubes, and extracting thermal energy from inside the refrigerator body and releasing it into the surrounding air.

Most refrigerators run for a few minutes until reaching the desired internal temperature, then, stop and remain not running for about three times that period of time (time during which that internal temperature slowly increases).

Example of 20-minute cycle: it runs for 5 minutes and remains stopped for 15 minutes.

 

[joejoe121]

@joejoe121, you are employed to move boxes from A to B.
On average you move 80 boxes per hour.
However, you are sneaking-off to read Physics Forums posts for 45minutes each hour.
At what rate are you moving boxes when you are actually working?

Edited for clarity.

If im working only for 15 minutes at a time or 0.25 hours moving 80 boxes is 320 boxes per hour

 

[Steve4Physics]

If im working only for 15 minutes at a time or 0.25 hours moving 80 boxes is 320 boxes per hour

Yes. You can apply the same logic to the refrigerator.

In the original question the refrigerator removes (average) 800kJ heat per hour*. But it is only running for 15mins in each hour. So:
- what is the rate of heat removal (in kJ/h) when the refrigerator is running?
- what input power (in kJ/h) is required by the refgerator when it is running?

We usually give power in units of watts, W (or kilowatts, kW). So you may need to convert your final answer from kJ/h to W or kW.

*Note: 800kJ/h is rather big for a 'household refrigerator'. So check you have the correct value.

 

[joejoe121]

Yes. You can apply the same logic to the refrigerator.

In the original question the refrigerator removes (average) 800kJ heat per hour*. But it is only running for 15mins in each hour. So:
- what is the rate of heat removal (in kJ/h) when the refrigerator is running?
- what input power (in kJ/h) is required by the refrigerator when it is running?

We usually give power in units of watts, W (or kilowatts, kW). So, you may need to convert your final answer from kJ/h to W or kW.

*Note: 800kJ/h is rather big for a 'household refrigerator'. So, check you have the correct value.

3200 kj/h is the heat removal rate when running.
and is input power work? If so, then I believe its 1454.54 kw/h or 0.404 kw/s.