NEET Exam > NEET Questions > A refrigerator works b/w -3 degree cel. and 2...
Start Learning for Free
A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J?
Most Upvoted Answer
A refrigerator works b/w -3 degree cel. and 27 degree cel. The work do...
Refrigerator Efficiency Calculation
In order to determine the work done by the compressor for each joule of heat absorbed from the freezer, we can use the formula for the coefficient of performance (COP) of a refrigerator. The COP is defined as the ratio of the heat extracted from the cold reservoir to the work input to the system.
Given Data:
- Temperature of the freezer (Tc) = -3 degree Celsius = 270 K
- Maximum temperature of the refrigerator (Th) = 27 degree Celsius = 300 K
Calculating COP:
COP = Th / (Th - Tc)
COP = 300 / (300 - 270)
COP = 300 / 30
COP = 10
Understanding Work Done:
The work done by the compressor for each joule of heat absorbed from the freezer can be calculated using the formula:
Work done = Heat absorbed / COP
Since we are interested in the work done for each joule of heat absorbed, we can consider the heat absorbed to be 1 Joule.
Calculating Work Done:
Work done = 1 / 10
Work done = 0.1 Joule
Therefore, the correct answer is (iv) 1/10 J. The compressor does 0.1 Joule of work for each Joule of heat absorbed from the freezer.
Community Answer
A refrigerator works b/w -3 degree cel. and 27 degree cel. The work do...
C
Attention NEET Students!
To make sure you are not studying endlessly, EduRev has designed NEET study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in NEET.
|
Explore Courses for NEET exam
|
|
Similar NEET Doubts
Top Courses for NEETView all
A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J?
Question Description
A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J?.
A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J?.
Solutions for A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? defined & explained in the simplest way possible. Besides giving the explanation of
A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J?, a detailed solution for A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? has been provided alongside types of A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? theory, EduRev gives you an
ample number of questions to practice A refrigerator works b/w -3 degree cel. and 27 degree cel. The work done by the compressor for each joule of heat absorbed from the freezer is (i)9Joule ii)10 J iii) 1/9 J (iv)1/10 J? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Suggested Free Tests
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup