Question-40 An engine operating between temperatures T_1 and T_2 has efficiency \frac{1}{5}. When T_2 is lowered by 45\ K, its efficiency becomes \frac{1}{2}. Temperatures T_1 and T_2 are respectively
(A) 100\ K,\ 70\ K
(B) 160\ K,\ 120\ K
(C) 140\ K,\ 110\ K
(D) 150\ K,\ 120\ K
Answer : (D)
Explanation:
Given,
η₁ = 1/5 = 1 − \frac{T_2}{T_1} ⇒ 4/5 = \frac{T_2}{T_1} ⇒ T₂ = \frac{4}{5} T₁ ….(i)
Also,
η₂ = 1/2 = 1 − \frac{(T_2 − 45)}{T_1}
Substituting (i),
\frac{\left(\frac{4}{5}T_1 - 45\right)}{T_1} = 1 - \frac{1}{2} \frac{4}{5}T_1 - 45 = \frac{1}{2}T_1∴ T₁ = 150 K
∴ T₂ = \frac{4}{5} \times 150 = 120 K ….[From(i)]