A Carnot engine operating between temperatures T₁ and T₂ has efficiency 1/6. When T₂ is lowered by 62 K its efficiency increases to 1/3. Then T₁ and T₂ are, respectively: (a) 372 K and 330 K (b) 330 K and 268 K (c) 310 K and 248 K (d) 372 K and 310 K

Answer: (d). T₂ = 5/6T₁ and T₂ - 62 = 2/3T₁. Subtract: T₁/6 = 62 ⇒ T₁ = 372 K, T₂ = 310 K. So the answer is (d) 372 K and 310 K.

A Carnot engine operating between temperatures T₁ and T₂ has… | JEE Main 2011 (AIEEE)

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Heat Engines, Refrigerators & Second Law

A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{6}$ . When $T_2$ is lowered by $62\ K$ its efficiency increases to $\frac{1}{3}$ . Then $T_1$ and $T_2$ are, respectively:

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Answer: (d)

$T_2 = \frac{5}{6}T_1$ and $T_2 - 62 = \frac{2}{3}T_1$ .
Subtract: $\frac{T_1}{6} = 62 \Rightarrow T_1 = 372\text{ K}$ , $T_2 = 310\text{ K}$ .
So the answer is (d) $372\ K$ and $310\ K$ .

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