Question

A Carnot engine operating between two reservoirs has efficiency $\frac{1}{3}$. When the temperature of cold reservoir raised by $x$, its efficiency decreases to $\frac{1}{6}$. The value of $x$, if the temperature of hot reservoir is $99^{\circ }C$, will be :

• A. $62K$
• B. $16.5K$
• C. $33K$
• D. $66K$

Answer 1

Answer: (A)

$T_H=99^{\circ }C=99+273$
$=372K$
$1-\frac{T_C}{T_H}=\frac{1}{3}$
$\frac{T_C}{T_H}=\frac{2}{3}(1)\Rightarrow T_C=\frac{2}{3}\times 372$
$=2\times 124=248K$
$1-\frac{T_C+X}{T_H}=\frac{1}{6}$
$\frac{5}{6}=\frac{T_C+X}{T_H}$
$\frac{5}{6}=\frac{248+X}{372}$
$248+X=5\times 62$
$X=310-248=62K$