The efficiency of a Carnot engine operating between steam point and ice point is:
Show Hint
100%
50%
77%
27%
11%
The Correct Option is D
Solution and Explanation
The efficiency (\( \eta \)) of a Carnot engine is given by: \[ \eta = 1 - \frac{T_{{cold}}}{T_{{hot}}} \] where \( T_{{cold}} = 273 \, {K} \) (ice point) and \( T_{{hot}} = 373 \, {K} \) (steam point). Calculating: \[ \eta = 1 - \frac{273}{373} \approx 0.268 = 26.8\% \] This rounds to approximately 27%.
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