Gujarati
Hindi
11.Thermodynamics
normal

A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac {1}{6}$ . When $T_2$ is lowered by $60\,K$ ; its efficiency increases to $\frac {1}{3}$. Then $T_1$ and $T_2$ are respectively

A

$360\,K$ and $300\,K$

B

$372\,K$ and $330\,K$

C

$330\,K$ and $268\,K$

D

$310\,K$ and $248\,K$

Solution

$\eta=1-\frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{1}{6}$

$\therefore \frac{\mathrm{T}_{2}}{\mathrm{T}_{1}}=\frac{5}{6}$

$\frac{1}{3}=1-\frac{\mathrm{T}_{2}-62}{\mathrm{T}_{1}}$

$\therefore \frac{\mathrm{T}_{2}-60}{\mathrm{T}_{1}}=\frac{2}{3}$

$\therefore \quad \mathrm{T}_{1}=360 \mathrm{K}$ and $\mathrm{T}_{2}=300 \mathrm{K}$

Standard 11
Physics

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