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A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]
- a)325 K
- b)250 K
- c)380 K
- d)275 K
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By ho...
We know that efficiency of Carnot Engine
where, T1 is temp. of source & T2 is temp. of sink
Now efficiency to be increased by 50%
Increase in temp = 750 – 500 = 250 K
Most Upvoted Answer
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By ho...
To find the increase in temperature of the source required to increase the efficiency of the Carnot engine by 50% of its original efficiency, we can use the formula for the efficiency of a Carnot engine:
Efficiency = 1 - (T_sink / T_source)
Given:
Efficiency = 40%
T_sink = 300 K
Let's calculate the original temperature of the source (T_source) using the given efficiency:
Efficiency = 1 - (T_sink / T_source)
0.4 = 1 - (300 / T_source)
0.4 = (T_source - 300) / T_source
Cross-multiplying, we get:
0.4T_source = T_source - 300
0.4T_source - T_source = -300
-0.6T_source = -300
T_source = -300 / -0.6
T_source = 500 K
Therefore, the original temperature of the source is 500 K.
Now, let's calculate the new efficiency of the Carnot engine after increasing it by 50% of the original efficiency:
New efficiency = 40% + (50% of 40%)
New efficiency = 40% + (0.5 * 0.4)
New efficiency = 40% + 0.2
New efficiency = 60%
Let's find the new temperature of the source (T_source_new) using the new efficiency:
New efficiency = 1 - (T_sink / T_source_new)
0.6 = 1 - (300 / T_source_new)
0.6 = (T_source_new - 300) / T_source_new
Cross-multiplying, we get:
0.6T_source_new = T_source_new - 300
0.6T_source_new - T_source_new = -300
-0.4T_source_new = -300
T_source_new = -300 / -0.4
T_source_new = 750 K
Therefore, the new temperature of the source is 750 K.
To find the increase in temperature of the source required, we subtract the original temperature of the source from the new temperature of the source:
Increase in temperature = T_source_new - T_source
Increase in temperature = 750 K - 500 K
Increase in temperature = 250 K
Hence, the increase in temperature of the source required to increase the efficiency by 50% of the original efficiency is 250 K. Therefore, the correct answer is option B.
Efficiency = 1 - (T_sink / T_source)
Given:
Efficiency = 40%
T_sink = 300 K
Let's calculate the original temperature of the source (T_source) using the given efficiency:
Efficiency = 1 - (T_sink / T_source)
0.4 = 1 - (300 / T_source)
0.4 = (T_source - 300) / T_source
Cross-multiplying, we get:
0.4T_source = T_source - 300
0.4T_source - T_source = -300
-0.6T_source = -300
T_source = -300 / -0.6
T_source = 500 K
Therefore, the original temperature of the source is 500 K.
Now, let's calculate the new efficiency of the Carnot engine after increasing it by 50% of the original efficiency:
New efficiency = 40% + (50% of 40%)
New efficiency = 40% + (0.5 * 0.4)
New efficiency = 40% + 0.2
New efficiency = 60%
Let's find the new temperature of the source (T_source_new) using the new efficiency:
New efficiency = 1 - (T_sink / T_source_new)
0.6 = 1 - (300 / T_source_new)
0.6 = (T_source_new - 300) / T_source_new
Cross-multiplying, we get:
0.6T_source_new = T_source_new - 300
0.6T_source_new - T_source_new = -300
-0.4T_source_new = -300
T_source_new = -300 / -0.4
T_source_new = 750 K
Therefore, the new temperature of the source is 750 K.
To find the increase in temperature of the source required, we subtract the original temperature of the source from the new temperature of the source:
Increase in temperature = T_source_new - T_source
Increase in temperature = 750 K - 500 K
Increase in temperature = 250 K
Hence, the increase in temperature of the source required to increase the efficiency by 50% of the original efficiency is 250 K. Therefore, the correct answer is option B.
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A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]a)325 Kb)250 Kc)380 Kd)275 KCorrect answer is option 'B'. Can you explain this answer?
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A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]a)325 Kb)250 Kc)380 Kd)275 KCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]a)325 Kb)250 Kc)380 Kd)275 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]a)325 Kb)250 Kc)380 Kd)275 KCorrect answer is option 'B'. Can you explain this answer?.
A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]a)325 Kb)250 Kc)380 Kd)275 KCorrect answer is option 'B'. Can you explain this answer? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]a)325 Kb)250 Kc)380 Kd)275 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A Carnot engine whose sink is at 300 K has an efficiency of 40%. By how much should the temperature of source be increased so as to increase, its efficiency by 50% of original efficiency ? [2006]a)325 Kb)250 Kc)380 Kd)275 KCorrect answer is option 'B'. Can you explain this answer?.
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