A carnot engine operating between temperatures T1 and T2 has efficiency 16 . When T2 is lowered by 62 K, its efficiency increases to 13 . Then T1 and T2 are, respectively:

η1=T1−T2T1=16 η2=T1−(T2−62)T1=13 ⇒T1−T2T1+62T1=13 16+62T1=13 62T1=16 ∴ T1=62×6=372K T1−T2T1=16 1−T2T1=16 T2372=56 ⇒T2=310K

A carnot engine operating between temperatures T1 and T2 has efficiency 16 . When T2 is lowered by 62 K, its efficiency increases to 13 . Then T1 and T2 are, respectively:

η1=T1−T2T1=16 η2=T1−(T2−62)T1=13 ⇒T1−T2T1+62T1=13 16+62T1=13 62T1=16 ∴ T1=62×6=372K T1−T2T1=16 1−T2T1=16 T2372=56 ⇒T2=310K

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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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372 K and 330 K

330 K and 268 K

310 K and 248 K

372 K and 310 K

answer is D.

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Detailed Solution

$η_{1} = \frac{T_{1} - T_{2}}{T_{1}} = \frac{1}{6}$

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

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

$\frac{1}{6} + \frac{62}{T_{1}} = \frac{1}{3}$

$\frac{62}{T_{1}} = \frac{1}{6}$

$∴ T_{1} = 62 \times 6 = 372 K$

$\frac{T_{1} - T_{2}}{T_{1}} = \frac{1}{6}$

$1 - \frac{T_{2}}{T_{1}} = \frac{1}{6}$

$\frac{T_{2}}{372} = \frac{5}{6}$

$\Rightarrow T_{2} = 310 K$

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A carnot engine operating between temperatures T1 and T2 has efficiency 16 . When T2 is lowered by 62 K, its efficiency increases to 13 . Then T1 and T2 are, respectively:

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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: