NEET Exam  >  NEET Questions  >  A Carnot engine whose sink is at 300 K has an... Start Learning for Free
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?
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
View all questions of this test
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.
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

Top Courses for NEET

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?
Question Description
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?.
Solutions 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? 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 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? defined & explained in the simplest way possible. Besides giving the explanation of 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 detailed solution 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? has been provided alongside types of 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? theory, EduRev gives you an ample number of questions to practice 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? tests, examples and also practice NEET tests.
Explore Courses for NEET exam

Top Courses for NEET

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev