A nuclear reactor has pressurized water at 300°C acting as the hot reservoir, and steam is eventually condensed to water at 27.0°C acting as the cold reservoir. What is the maximum theoretical (Carnot) efficiency for a heat engine operating between these two temperatures?
A30.0%
B35.0%
C47.6%
D52.4%
Answer & Solution
Correct answer: C. 47.6%
1. Convert both temperatures to kelvins first, since the Carnot formula requires absolute temperature: Th = 300°C + 273 = 573 K, and Tc = 27.0°C + 273 = 300 K.
2. Apply the Carnot efficiency formula: EffC = 1 - Tc/Th.
3. Substitute the values in kelvins: EffC = 1 - (300 K)/(573 K).
4. Dividing gives 300/573 = 0.524, so EffC = 1 - 0.524 = 0.476.
5. Converting to a percentage: 0.476 x 100 = 47.6%.
6. Option D, 52.4%, is the leftover fraction Tc/Th itself rather than one minus that fraction, a common sign-step error.
7. Options A and B, 30.0% and 35.0%, are not obtained from correctly dividing 300 K by 573 K, so neither follows from the given temperatures.
_Source: OpenStax College Physics (CC BY 4.0), Ch 15 "Thermodynamics", section 15.4 Carnot's Perfect Heat Engine: The Second Law of Thermodynamics Restated_
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