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NEET Easy

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\%.

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