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Q.

A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency 1/6. When $T_{2}$ is lowered by 62 K, its efficiency increases to 1/3. Then $T_{1}$ and $T_{2}$ are, respectively

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a

372 K and 310 K

b

330 K and 268 K

c

372 K and 330 K

d

310 K and 248 K

answer is A.

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Detailed Solution

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$η = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{6} \Rightarrow \frac{T_{2}}{T_{1}} = 1 - \frac{1}{6} = \frac{5}{6}$
$\frac{1}{3} = 1 - \frac{(T_{2} - 62)}{T_{1}} \Rightarrow \frac{T_{2} - 62}{T_{1}} = \frac{2}{3}$
$\frac{5 (T_{2} - 62)}{T_{2}} = \frac{2}{3} \Rightarrow 5 T_{2} - 310 = 4 T_{2}$
$T_{2} = 310 K$ and $T_{1} = \frac{6 \times 310}{5} \Rightarrow T_{1} = 372 K$

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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