A Carnot engine operating between temperatures T1 and T2 has efficiency 1-6- When T2 is lowered by 62 K- it-s efficiency increases to 1-3- Then T1 and T2 are- respectively -

The correct option is D 372 K and 310 K η_1 =1 T_2T_1= 16 ........(1) η_2 =1 T_2 62T_1= 13 ....(2)By solving these equations , ...

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Byju's Answer

Standard XII

Chemistry

Carnot Cycle

A Carnot engi...

Question

A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency $\frac{1}{6}$ . When $T_{2}$ is lowered by $62$ $K$ , it's efficiency increases to $\frac{1}{3}$ . Then $T_{1}$ and $T_{2}$ are, respectively :

372

K

and

310

K

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372

K

and

330

K

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330

K

and

268

K

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310

K

and

248

K

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Solution

The correct option is D $372$ $K$ and $310$ $K$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{6}$ ........(1)

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

By solving these equations , we get
$T_{1} = 372 K$

$\frac{T_{1} - T_{2}}{T_{1}} = \frac{1}{6}$
$T_{2} = 310 K$

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

The correct option is D 372 K and 310 Kη1=1−T2T1=16........(1)η2=1−T2−62T1=13....(2)By solving these equations , we getT1=372KT1−T2T1=16T2=310K

A Carnot engine operating between temperatures $T_{1}$ and $T_{2}$ has efficiency $\frac{1}{6}$ . When $T_{2}$ is lowered by $62$ $K$ , it's efficiency increases to $\frac{1}{3}$ . Then $T_{1}$ and $T_{2}$ are, respectively :

The correct option is D $372$ $K$ and $310$ $K$
$η_{1} = 1 - \frac{T_{2}}{T_{1}} = \frac{1}{6}$ ........(1)

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

By solving these equations , we get
$T_{1} = 372 K$

$\frac{T_{1} - T_{2}}{T_{1}} = \frac{1}{6}$
$T_{2} = 310 K$