Question

A carnot engine operating between temprature T₁ and T₂ has efficiency 1/6 when T₂ is lowered by 62k , its efficiency increase to 1/3 . Then , T₁ and T₂ are respectively . [AIEEE -2011] ​

Answer 1

Explanation:- As efficiency , η₁ = 1-T₂/T₁

⟹ given :- η₁ = 1/6

and η₂ = 1/3

∴ η₁ = 1-T₂/T₁

⟹ 1/6 = 1- T₂/T₁

⟹ T₂/T₁ = 5/6 _________(i)

∵ given :- T₂ is lowered by 62K , Its efficiency increases to 1/3 .

• ∴ η₂ = 1- T₂ - 62 /T₁

• 1/3 = 1- T₂/T₁ - 62/T₁

• T₂/T₁ + 62/T₁ = 1-1/3

• 5/6 + 62/T₁ = 2/3 [∵ T₂/T₁ = 5/6 from (1) ]

• 62/T₁ = 2/3 - 5/6

• 62/T₁ = 1/6

T₁ = 372K

and , T₂ = 310K