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A Carnot engine whose heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?a)Increase by 18°Cb)Increase by 200°Cc)Increase by 120°Cd)Increase by 73°CCorrect answer is option 'B'. 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 heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?a)Increase by 18°Cb)Increase by 200°Cc)Increase by 120°Cd)Increase by 73°CCorrect answer is option 'B'. 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 heat sinks at 27°C, has an efficiency of 25%. By how many degrees should the temperature of the source be changed to increase the efficiency by 100% of the original efficiency?a)Increase by 18°Cb)Increase by 200°Cc)Increase by 120°Cd)Increase by 73°CCorrect answer is option 'B'. Can you explain this answer?.

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