Thermodynamics Question 21
Question 21 - 01 February - Shift 2
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:
(1) $16.5 K$
(2) $33 K$
(3) $66 K$
(4) $62 K$
Show Answer
Answer: (4)
Solution:
Formula: Efficiency of Carnot Engine
$T_H=99^{\circ} C=99+273$ $=372 K$.
$1-\frac{T_C}{T_H}=\frac{1}{3}$
$\frac{T_C}{T_H}=\frac{2}{3} \quad(1) \Rightarrow T_C=\frac{2}{3} \times 372$
$ =2 \times 124=248 K $
$1-\frac{T_C+X}{T_H}=\frac{1}{6}$
$\frac{5}{6}=\frac{T_C+X}{T_H}$
$\frac{5}{6}=\frac{248+X}{372}$
$248+X=5 \times 62$
$X=310-248=62 K$
