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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?
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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.
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A Carnot engine having an efficiency of 1/10 is being used as a refrig...
⇒ 100 - 10 =
= 90 JAttention JEE Students!
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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?.
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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
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