Question

A Carnot engine operating between temperatures $T_1$ and $T_2$ has efficiency $\frac{1}{6}$. When $T_2$ is lowered by $62K$, its efficiency increases to $\frac{1}{3}$. Then $T_1$ and $T_2$ are, respectively :

• A. $372K$ and $330K$
• B. $330K$ and $268K$
• C. $310K$ and $248K$
• D. $372K$ and $310K$

Answer 1

The correct option is D $372K$ and $310K$

$\text{Given:-}$ $\eta =\frac{1}{6}$

When $T_2$ is lowered by 62 K,then it's efficiency becomes $\frac{1}{3}$

$\text{Solution:-}$

The efficiency of Carnot engine,

$\eta =(1-\frac{T_2}{T_1})$

$\therefore \frac{1}{6}=(1-\frac{T_2}{T_1})$

$(Given,\eta =\frac{1}{6})$

$\frac{T_2}{T_1}=\frac{5}{6}\Rightarrow T_1=\frac{6T_2}{5}..........\left(i\right)$

As per question , when $T_2$ is lowered by 62 K,then it's efficiency

becomes $\frac{1}{3}$

$\therefore \frac{1}{3}=(1-\frac{T_2-62}{T_1})$

$\frac{T_2-62}{T_1}=1-\frac{1}{3}$

$\frac{T_2-62}{\frac{6}{5}{T}_2}=\frac{2}{3}$

$5T_2-310=4T_2\Rightarrow T_2=310K$

From equation (i)

$T_1=\frac{6\times 310}{5}=372K$

$\text{Hence the correct option is D}$