A Carnot engine operating between temperatures T1 and T2 has efficiency . When T2 is lowered by 62 K ; its efficiency increases to . Then T1 and T2 are respectively :

Engineering

Physics

Thermodynamics

Question

A Carnot engine operating between temperatures T₁ and T₂ has efficiency $\frac{1}{6}$ . When T₂ is lowered by 62 K ; its efficiency increases to $\frac{1}{3}$ . Then T₁ and T₂ are respectively :

310 K and 248 K

330 K and 268 K

372 K and 330 K

372 K and 310 K

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$η = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{6} \Rightarrow \frac{T_{2}}{T_{1}} = 1 - \frac{1}{6} = \frac{5}{6}$

$\frac{1}{3} = 1 - \frac{(T_{2} - 62)}{T_{1}} \Rightarrow \frac{T_{2} - 62}{T_{1}} = \frac{2}{3}$

$\frac{5 (T_{2} - 62)}{T_{2}} = \frac{2}{3}$

5T₂ – 310 = 4T₂

T₂ = 310 and $T_{1} = \frac{6 \times 310}{5}$

T₁ = 372 K