The efficiency of an ideal Carnot engine operating between two temperatures T1 (source) and T2 (sink) is given by:
A. (T1 - T2) / T2
B. T2 / (T1 - T2)
C. 1 - (T1/T2)
D. 1 - (T2/T1)
Answer: Option D
Solution (By JKSSB Mock Tests)
The thermal efficiency (η) of a reversible Carnot engine depends exclusively on the absolute temperatures of the hot source (T1) and the cold sink (T2). The formula is η = 1 - (T2/T1), or (T1 - T2) / T1. Temperatures must be in Kelvin. It shows no engine can be 100% efficient unless the sink is at absolute zero.
Explanation:
η = 1 - T₂/T₁: depends only on absolute temperatures, independent of working fluid or design. Universal maximum efficiency limit. Memory tip: 'Carnot: only T_hot and T_cold matter; real engines less efficient'. Thermodynamics concept frequently tested in competitive exams.
Explanation:
Acceleration phase: s₁ = ½a t² = ½×2×100 = 100 m, v = a t = 20 m/s. Constant velocity phase: s₂ = v×t = 20×10 = 200 m. Total = 300 m. Motion in two parts.
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