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A Carnot engine, whose efficiency is 20% receives heat at 450 K. If its efficiency is increased to 30%, then the intake temperature for the same exhaust temperature is
- a)514 K
- b)534 K
- c)554 K
- d)574 K
Correct answer is option 'A'. Can you explain this answer?
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Most Upvoted Answer
A Carnot engine, whose efficiency is 20% receives heat at 450 K. If it...
Given:
- Efficiency of Carnot engine = 20%
- Initial intake temperature = 450 K
To find:
- Intake temperature when efficiency is increased to 30%
Formula:
The efficiency of a Carnot engine is given by the equation:
Efficiency = 1 - (Tc/Th)
Where:
- Efficiency is the ratio of useful work output to the heat input
- Tc is the absolute temperature of the cold reservoir
- Th is the absolute temperature of the hot reservoir
Calculation:
1. Given that the efficiency of the Carnot engine is 20%, we can write the equation as:
0.20 = 1 - (Tc/450)
2. Rearranging the equation, we get:
Tc/450 = 1 - 0.20
3. Solving for Tc, we find:
Tc/450 = 0.80
Tc = 0.80 * 450
Tc = 360 K
4. Now, we need to find the intake temperature when the efficiency is increased to 30%. Let's assume the new intake temperature is Th.
5. Using the efficiency equation again, we can write:
0.30 = 1 - (360/Th)
6. Rearranging the equation, we get:
360/Th = 1 - 0.30
7. Solving for Th, we find:
360/Th = 0.70
Th = 360/0.70
Th ≈ 514.29 K
Thus, the intake temperature when the efficiency is increased to 30% is approximately 514 K. Therefore, option 'A' is the correct answer.
- Efficiency of Carnot engine = 20%
- Initial intake temperature = 450 K
To find:
- Intake temperature when efficiency is increased to 30%
Formula:
The efficiency of a Carnot engine is given by the equation:
Efficiency = 1 - (Tc/Th)
Where:
- Efficiency is the ratio of useful work output to the heat input
- Tc is the absolute temperature of the cold reservoir
- Th is the absolute temperature of the hot reservoir
Calculation:
1. Given that the efficiency of the Carnot engine is 20%, we can write the equation as:
0.20 = 1 - (Tc/450)
2. Rearranging the equation, we get:
Tc/450 = 1 - 0.20
3. Solving for Tc, we find:
Tc/450 = 0.80
Tc = 0.80 * 450
Tc = 360 K
4. Now, we need to find the intake temperature when the efficiency is increased to 30%. Let's assume the new intake temperature is Th.
5. Using the efficiency equation again, we can write:
0.30 = 1 - (360/Th)
6. Rearranging the equation, we get:
360/Th = 1 - 0.30
7. Solving for Th, we find:
360/Th = 0.70
Th = 360/0.70
Th ≈ 514.29 K
Thus, the intake temperature when the efficiency is increased to 30% is approximately 514 K. Therefore, option 'A' is the correct answer.
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A Carnot engine, whose efficiency is 20% receives heat at 450 K. If its efficiency is increased to 30%, then the intake temperature for the same exhaust temperature isa)514 Kb)534 Kc)554 Kd)574 KCorrect answer is option 'A'. Can you explain this answer?
Question Description
A Carnot engine, whose efficiency is 20% receives heat at 450 K. If its efficiency is increased to 30%, then the intake temperature for the same exhaust temperature isa)514 Kb)534 Kc)554 Kd)574 KCorrect 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, whose efficiency is 20% receives heat at 450 K. If its efficiency is increased to 30%, then the intake temperature for the same exhaust temperature isa)514 Kb)534 Kc)554 Kd)574 KCorrect 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, whose efficiency is 20% receives heat at 450 K. If its efficiency is increased to 30%, then the intake temperature for the same exhaust temperature isa)514 Kb)534 Kc)554 Kd)574 KCorrect answer is option 'A'. Can you explain this answer?.
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