Question 285: Thermodynamics & KTG

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 :

Accepted Answer

Given $\eta=\frac{1}{3}$
When $\mathrm{T}_2 $ raised by x ($\mathrm{T}_2 $ = $\left(\mathrm{T}_2+\mathrm{x}ight)$), its efficiency decreases to $$\frac{1}{6}$$
$\eta^{\prime}=\frac{1}{6}$
Temperature of hot reservior $\left(\mathrm{T}_1ight)=99^{\circ} \mathrm{C}$
$$
=99+273=372^{\circ} \mathrm{K}
$$
As we know, efficiency of carnot engine
$$
\begin{aligned}
& \eta=1-\frac{T_2}{T_1}=\frac{1}{3} .....(1) \\
& \eta^{\prime}=1-\frac{\left(T_2+xight)}{T_1}=\frac{1}{6} .....(2) \\
& \eta^{\prime}=\frac{T_1-\left(T_2+xight)}{T_1}=\frac{1}{6}
\end{aligned}
$$
From equation (1)
$$
\frac{1}{3}=1-\frac{\mathrm{T}_2}{372}
$$
$$
\begin{aligned}
& \frac{1}{3}=\frac{372-\mathrm{T}_2}{372} \\
&\Rightarrow 372-\frac{372}{3}=\mathrm{T}_2 \\
&\Rightarrow \mathrm{~T}_2=248 \mathrm{~K}
\end{aligned}
$$
By putting the value of $\mathrm{T}_2$ in equation (2)
$$
\begin{aligned}
& \frac{T_1-\left(T_2+xight)}{T_1}=\frac{1}{6} \\
&\Rightarrow \frac{372-(248+x)}{372}=\frac{1}{6} \\
&\Rightarrow 372-24-x=\frac{372}{6} \\
&\Rightarrow 124-x=62 \\
&\Rightarrow 124-62=x \\
&\Rightarrow x=62 \mathrm{~K}
\end{aligned}
$$

Answer

66 K

Answer

62 K

Answer

16.5 K

Answer

33 K

Thermodynamics & KTG

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 :

Choose the correct answer: