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 $

$1-\frac{T_C+X}{T_H}=\frac{1}{6}$

$\frac{5}{6}=\frac{T_C+X}{T_H}$

$\frac{5}{6}=\frac{248+X}{372}$

$248+X=5 \times 62$

$X=310-248=62 K$