CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A refrigerator whose coefficient of performance K=5, extracts heat from the cooling compartment at the rate of 250J/cycle. What is the work done per cycle to operate the refrigerator? How much heat is discharged per cycle to the room which acts as the high-temperature reservoir?


Open in App
Solution

Step 1: Given data

  1. The coefficient of performance of the refrigerator is η=5.
  2. The cooling rate of the cold reservoir is Qc=250J/cycle.

Step 2: Formula Used

  1. We know that the performance of a refrigerator is the ratio of the amount of heat extracted from the cold body to the work done on it.
  2. The performance or efficiency of a refrigerator is defined by the form, η=QCW=heatextractedfromthecoldbodyworkdoneonthemachine.
  3. Heat discharged by a refrigerator is the sum of total heat taken by the machine and the work done by the machine, i.e, Qd=Qc+W

Step 3: Diagram

Step 4: Calculation

As we know, the work done by a refrigerator is W=QCη

So,

W=2505=50orW=50joule.

Therefore, the work done per cycle to operate the refrigerator is 50joule.

Now heat discharged by the refrigerator is Qd=Qc+W.

So,

Qd=250+50=300orQd=300joules

Therefore, heat discharged per cycle to the room is 300joules.


flag
Suggest Corrections
thumbs-up
0
BNAT
mid-banner-image