Calculating COP of Refrigerator: Solve with W=P(delta)t

[jdog6]

A vessel containing 8.43 kg of water at 21.1C is put into a refrigerator. The 0.1655 hp motor runs for 9 min to cool the liquid to the refrigerator's low temperature 4C.

What is the COP of the refrigerator?

W=P(delta)t
= 123.463Wx540s
= 66670.02 J

COP = Qc/W

Please Help.

Thank you.

 

[Andrew Mason]

A vessel containing 8.43 kg of water at 21.1C is put into a refrigerator. The 0.1655 hp motor runs for 9 min to cool the liquid to the refrigerator's low temperature 4C.
What is the COP of the refrigerator?
W=P(delta)t
= 123.463Wx540s
= 66670.02 J
COP = Qc/W
Please Help.
Thank you.

The Coefficient of Performance for a refrigerator is defined as the ratio of the heat removed from the cold reservoir to the work added to the system. What is the work or energy input? (I think you have correctly worked this out based on one hp = 746 J/sec). What is the heat removed (in J.) ?

AM

 

[quicknote]

I would first like to clarify that COP stands for Coefficient of Performance, which is a measure of the efficiency of a refrigerator in transferring heat. To calculate the COP of a refrigerator, we need to consider the amount of heat removed (Qc) and the amount of work done (W). In this scenario, the work done by the refrigerator's motor is 66670.02 J, as calculated above.

To determine the amount of heat removed, we can use the formula Q=mcΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. Plugging in the values given, we get:

Q = (8.43 kg)(4186 J/kg°C)(21.1°C - 4°C)
= 116,492.74 J

Now, we can calculate the COP of the refrigerator using the formula COP = Qc/W:

COP = 116,492.74 J/66670.02 J
= 1.75

This means that for every joule of work done by the refrigerator's motor, 1.75 joules of heat are removed. In other words, the refrigerator is 1.75 times more efficient in removing heat than the work it takes to run the motor.

I hope this helps clarify the calculation of COP and its importance in evaluating the efficiency of refrigerators. Let me know if you have any further questions or need any additional assistance.