Question

A refrigerator, whose coefficient of performance K=5, extracts heat from the cooling compartment at the rate of 250 J/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?
(A) 50 J
(B) 200 J
(C) 300 J
(D) none

Answer 1

Hint: These machines either heat engine or refrigerator works on the principle of the second law of thermodynamics which states that there cannot be a flow of heat from a body at a lower temperature to a body at a higher temperature without work.



Complete step by step answer:
The refrigerator transfer heat from a cold body to a hot body at the expense of mechanical energy. Coefficient of performance is the ratio of heat extracted from colder body to the work needed to transfer it. Now given,
Work done=?
\[K=\dfrac{{{Q}_{2}}}{W}\]
Given \[{{Q}_{2}}\]= 250 J
Putting back into the above-mentioned formula we get,
\[W=\dfrac{250}{5}=50\]J
Thus, 50 J of work is needed to be done per cycle to achieve such coefficient of performance. Now we need to find heat discharged into the room.
H= 250+50
=300 J
Thus 300 J of heat is being discharged into the room. Hence the correct option is (C)

Additional Information: Conditions such as temperature and humidity also needed to be considered while designing and working on the heat pump. Efficiency decreases if temperature difference increases or when freezing takes place.



Note:Since, coefficient of performance gives us the efficiency of the machine, higher it is more efficient is the machine is. Its value can be greater than one unlike for heat engine whose efficiency is always less than one.