A carnot engine operating between temperature T1 and T2 has the efficiency of 16- When the temperature T2 is lowered by 62K- it efficiency becomes 13- Then temperature T1 and T2 respectively are

The efficiency of the Carnot engine is defined as, nbsp;η=1 T_2T_1 We have, 16=1 T_2T_1 and on decreasing the temperature by 62K the efficiency (η) becomes, 1 ...

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Standard XI

Chemistry

Carnot Cycle

A carnot engi...

Question

A carnot engine operating between temperature $T_{1}$ and $T_{2}$ has the efficiency of $\frac{1}{6}$ . When the temperature $T_{2}$ is lowered by 62K, it efficiency becomes $\frac{1}{3}$ , Then temperature $T_{1}$ and $T_{2}$ respectively are

T_{1} = 370 K, T_{2} = 330 K

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T_{1} = 330 K, T_{2} = 268 K

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T_{1} = 310 K, T_{2} = 248 K

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T_{1} = 372 K, T_{2} = 310 K

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Solution

The correct option is D $T_{1} = 372 K, T_{2} = 310 K$
The efficiency of the Carnot engine is defined as, $η = 1 - \frac{T_{2}}{T_{1}}$
We have,
$\frac{1}{6} = 1 - \frac{T_{2}}{T_{1}}$
and on decreasing the temperature by $62 K$ the efficiency ( $η$ ) becomes,
$\frac{1}{3} = 1 - \frac{T_{2} - 62}{T_{1}}$
Solving these two equations, we have $T_{1} = 372 K, T_{2} = 310 K$

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Carnot Cycle

Standard XI Chemistry

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A carnot engine operating between temperature T1 and T2 has the efficiency of 16. When the temperature T2 is lowered by 62K, it efficiency becomes 13, Then temperature T1 and T2 respectively are

The correct option is D T1=372K,T2=310KThe efficiency of the Carnot engine is defined as, η=1−T2T1 We have, 16=1−T2T1 and on decreasing the temperature by 62K the efficiency (η) becomes, 13=1−T2−62T1 Solving these two equations, we have T1=372K,T2=310K

A carnot engine operating between temperature $T_{1}$ and $T_{2}$ has the efficiency of $\frac{1}{6}$ . When the temperature $T_{2}$ is lowered by 62K, it efficiency becomes $\frac{1}{3}$ , Then temperature $T_{1}$ and $T_{2}$ respectively are