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NEET generalGeneralEasy

Question

The efficiency of an ideal heat engine working between the freezing point and boiling point of water, is

A

6.25%

B

20%

C

26.8%

D

12.5%

Step-by-Step Solution

Efficiency η=(1−T2T1)×100\eta = (1 - \frac{T_2}{T_1}) \times 100. Here T2=273 KT_2 = 273 \text{ K} (freezing point) and T1=373 KT_1 = 373 \text{ K} (boiling point). η=(1−273373)×100=100373×100≈26.8%\eta = (1 - \frac{273}{373}) \times 100 = \frac{100}{373} \times 100 \approx 26.8\%.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generalefficiencyengineworkingbetweenfreezing

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