JEE Exam > JEE Questions > A Carnot engine having an efficiency of 1/10 ...
Start Learning for Free
A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:
- a)90 J
- b)1 J
- c)99 J
- d)100 J
Correct answer is option 'A'. Can you explain this answer?
| FREE This question is part of | Download PDF Attempt this Test |
Most Upvoted Answer
A Carnot engine having an efficiency of 1/10 is being used as a refrig...
Given:
Efficiency of the Carnot engine = 1/10
Work done on the refrigerator = 10 J
To find:
The amount of heat absorbed from the reservoir at the lower temperature
Explanation:
A Carnot engine operates between two reservoirs at different temperatures: a high-temperature reservoir (TH) and a low-temperature reservoir (TL). It works on the Carnot cycle, which consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
The efficiency of a Carnot engine is given by the equation:
Efficiency = 1 - (TL/TH)
We are given that the efficiency is 1/10. So, we can write:
1/10 = 1 - (TL/TH)
Simplifying the equation, we get:
TL/TH = 9/10
We know that the efficiency of a refrigerator is the ratio of the heat absorbed from the cold reservoir to the work done on the refrigerator. Mathematically, it is given by:
Efficiency of refrigerator = QL/W = TL/(TH - TL)
We are given that the work done on the refrigerator is 10 J. So, we can write:
1/10 = TL/(TH - TL)
Substituting the value of TL/TH from the efficiency equation, we get:
1/10 = (9/10)/(TH - (9/10)TH)
1/10 = (9/10)/(1/10)TH
TH = 9
Now, we can find the value of TL using the equation TL/TH = 9/10:
TL/9 = 9/10
TL = 81/10
The amount of heat absorbed from the reservoir at the lower temperature is given by:
QL = TL * (W/(TH - TL))
QL = (81/10) * (10/(9 - 81/10))
QL = (81/10) * (10/(9 - 729/10))
QL = (81/10) * (10/(81/10))
QL = 81
Therefore, the correct answer is option A) 90 J.
Efficiency of the Carnot engine = 1/10
Work done on the refrigerator = 10 J
To find:
The amount of heat absorbed from the reservoir at the lower temperature
Explanation:
A Carnot engine operates between two reservoirs at different temperatures: a high-temperature reservoir (TH) and a low-temperature reservoir (TL). It works on the Carnot cycle, which consists of four reversible processes: isothermal expansion, adiabatic expansion, isothermal compression, and adiabatic compression.
The efficiency of a Carnot engine is given by the equation:
Efficiency = 1 - (TL/TH)
We are given that the efficiency is 1/10. So, we can write:
1/10 = 1 - (TL/TH)
Simplifying the equation, we get:
TL/TH = 9/10
We know that the efficiency of a refrigerator is the ratio of the heat absorbed from the cold reservoir to the work done on the refrigerator. Mathematically, it is given by:
Efficiency of refrigerator = QL/W = TL/(TH - TL)
We are given that the work done on the refrigerator is 10 J. So, we can write:
1/10 = TL/(TH - TL)
Substituting the value of TL/TH from the efficiency equation, we get:
1/10 = (9/10)/(TH - (9/10)TH)
1/10 = (9/10)/(1/10)TH
TH = 9
Now, we can find the value of TL using the equation TL/TH = 9/10:
TL/9 = 9/10
TL = 81/10
The amount of heat absorbed from the reservoir at the lower temperature is given by:
QL = TL * (W/(TH - TL))
QL = (81/10) * (10/(9 - 81/10))
QL = (81/10) * (10/(9 - 729/10))
QL = (81/10) * (10/(81/10))
QL = 81
Therefore, the correct answer is option A) 90 J.
Free Test
FREE
| Start Free Test |
Community Answer
A Carnot engine having an efficiency of 1/10 is being used as a refrig...
⇒ 100 - 10 =
= 90 JAttention JEE Students!
To make sure you are not studying endlessly, EduRev has designed JEE study material, with Structured Courses, Videos, & Test Series. Plus get personalized analysis, doubt solving and improvement plans to achieve a great score in JEE.
|
Explore Courses for JEE exam
|
|
Similar JEE Doubts
Top Courses for JEEView all
A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer?
Question Description
A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer?.
A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer?.
Solutions for A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A Carnot engine having an efficiency of 1/10 is being used as a refrigerator. If the work done on the refrigerator is 10 J, the amount of heat absorbed from the reservoir at lower temperature is:a)90 Jb)1 Jc)99 Jd)100 JCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice JEE tests.
|
|
Explore Courses for JEE 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
|
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup