- Home
- Standard 11
- Physics
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
Similar Questions
normal
normal