Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • A refrigerator operates on a Carnot cycle between two reservoirs at temperatures <img alt="\mathrm{T_{\text {hot }}=300 \mathrm{~K} \, \, and\, \, T_{\text {cold }}=200 \mathrm{~K}}" src="https://e

A refrigerator operates on a Carnot cycle between two reservoirs at temperatures \mathrm{T_{\text {hot }}=300 \mathrm{~K} \, \, and\, \, T_{\text {cold }}=200 \mathrm{~K}}. Calculate the coefficient of performance (COP) of the refrigerator.
 

Option: 1

16.10 \mathrm{k}


Option: 2

450.56 \mathrm{k}


Option: 3

20.98 \mathrm{k}


Option: 4

5.20 \mathrm{k}


Answers (1)

best_answer

The coefficient of performance (COP) of a refrigerator is given by the formula:
\mathrm{COP}=\frac{Q_{\text {cold }}}{W}
where \mathrm{Q_{\text {cold }}} is the heat removed from the cold reservoir and W is the work done by the refrigerator.
In a Carnot cycle, the efficiency \mathrm{(\eta)} is given by:

\mathrm{\eta=1-\frac{T_{\text {cold }}}{T_{\text {hot }}}}
Since the COP is the inverse of the efficiency, we have:
\mathrm{\mathrm{COP}=\frac{1}{\eta}=\frac{T_{\text {hot }}}{T_{\text {hot }}-T_{\text {cold }}}}

Substitute the given values:

\mathrm{\mathrm{COP}=\frac{300 \mathrm{~K}}{300 \mathrm{~K}-200 \mathrm{~K}}=\frac{300 \mathrm{~K}}{100 \mathrm{~K}}=3}

Therefore, the coefficient of performance of the refrigerator is COP =3.

Posted by

Devendra Khairwa

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main Maths 2025: Five tips to score high marks in ‘toughest’ subject; most important chapters
anam-khan

JEE Main 2025 exam city intimation slip for session 1 soon at jeemain.nta.nic.in
anam-khan

Focus on modern physics, formulas to solve tricky problems; 5 tips to tackle JEE Main physics 2025

Latest Question