Carnot CycleHardQuestion 31054
Question
Two Carnot engines A (between T₁ and T) and B (between T and T₂) have same efficiency. Temperature T is:
- A(T₁+T₂)/2
- B√(T₁T₂)Correct
- CT₁T₂/(T₁+T₂)
- DT₁/T₂
Correct answer
√(T₁T₂)
Explanation
η_A = 1−T/T₁ = η_B = 1−T₂/T. So T/T₁ = T₂/T → T² = T₁T₂ → T = √(T₁T₂) (geometric mean).