250+ TOP MCQs on Coefficient of Performance of Refrigeration – 2 and Answers

Refrigeration Questions and Answers for Freshers on “Coefficient of Performance of Refrigeration – 2”.

1. Efficiency of the Refrigerator is _________ to the C.O.P of refrigerator.
a) inversely proportional
b) equal
c) independent
d) directly proportional
Answer: c
Clarification: Efficiency is the ratio of work done to heat supplied, whereas C.O.P is the ratio of Refrigeration effect to work done. Hence it is totally independent quantity.

2. What is the C.O.P. of a refrigeration system if the work input is 40 KJ/kg and work output is 80 KJ/kg and refrigeration effect produced is 130 KJ/kg of refrigerant flowing.
a) 3.00
b) 3.25
c) 2.25
d) 3.75
Answer: b
Clarification: C.O.P. = (frac{Refrigeration ,effect}{Work ,Done ,(Output-Input)})
= (frac{130}{80-40})
= 3.25 (unit less).

3. Find the Relative C.O.P. of a refrigeration system if the work input is 100 KJ/kg and refrigeration effect produced is 250 KJ/kg of refrigerant flowing. Also Theoretical C.O.P. is 3.
a) 0.65
b) 0.80
c) 0.83
d) 0.91
Answer: c
Clarification: Actual C.O.P. = (frac{Refrigeration ,effect}{Work ,Done})
= (frac{250}{100})
= 2.5 (unit less)
Relative C.O.P. = (frac{Actual ,C.O.P.}{Theoretical ,C.O.P.})
= (frac{2.5}{3}) = 0.833 i.e. = 83.3 %.

4. If a condenser and evaporator temperatures are 225 K and 100 K respectively, then reverse Carnot C.O.P is _________
a) 0.5
b) 1.5
c) 1.25
d) 1.75
Answer: c
Clarification: Reverse Carnot C.O.P. = (frac{t2-t1}{t1})
= (frac{225-100}{100})
= 1.25.

5. If a condenser and evaporator temperatures are 312 K and X K respectively, and C.O.P. is given as 5 then find the value of X.
a) 52
b) 65
c) 78
d) 82
Answer: a
Clarification: Reverse Carnot C.O.P. = (frac{t2-t1}{t1})
5 = (frac{312-X}{X})
6X = 312
X = 52 K.

6. The C.O.P for reverse Carnot refrigerator is 6. The ratio of lowest temperature to highest temperature will be _____
a) twice
b) three times
c) four times
d) seven times
Answer: d
Clarification: Reverse Carnot C.O.P. = (frac{t2-t1}{t1})
6 = (frac{X-Y}{Y})
6 = (frac{X}{Y}) – 1
Thus, (frac{X}{Y}) = 7 i.e. X=7Y i.e. Higher temperature = 7 times Lower temperature.

7. In general the ratio of lowest to highest temperature with respect to C.O.P. can be denoted by _________
a) C.O.P + 1 = Ratio of temperature
b) C.O.P/2 = Ratio of temperature
c) C.O.P + 4 = Ratio of temperature
d) C.O.P + 2 = Ratio of temperature
Answer: a
Clarification: In general, the ratio of lowest to highest temperature, say (frac{X-Y}{Y}) = C.O.P. Hence, (frac{X}{Y}) – 1 = C.O.P. i.e. (frac{X}{Y}) = C.O.P + 1.

8. The C.O.P for reverse Carnot refrigerator is 2. The ratio of highest temperature to lowest temperature will be _____
a) 4 times
b) 3 times
c) 1/2 times
d) 1/3 times
Answer: d
Clarification: Reverse Carnot C.O.P. = (frac{t2-t1}{t1})
2 = (frac{X-Y}{Y})
2 = (frac{X}{Y}) – 1
Thus, (frac{X}{Y}) = 3 i.e. X=3Y i.e. Lowest temperature = (frac{1}{3}) times Highest temperature.

9. The C.O.P of a reverse Carnot cycle doesn’t depend on which of the following?
a) Moisture
b) Evaporator temperature
c) Condenser temperature
d) Work done
Answer: a
Clarification: C.O.P for a reverse Carnot depends directly on the difference of Evaporator and Condenser temperature, and inversely on the Work done.

10. If a condenser and evaporator temperatures are ‘X’ K and 100 K respectively, and reverse Carnot C.O.P is 2.5 then find out the ‘X’.
a) 100 K
b) 150 K
c) 350 K
d) 200 K
Answer: c
Clarification: Reverse Carnot C.O.P. = (frac{t2-t1}{t1})
2.5 = (frac{X-100}{100})
X = 350 K.