Question

An ideal engine has an efficiency of $\frac{1}{12}$. When the temperature of sink is reduced by ${70}^{\circ }\text{C}$, its efficiency gets doubled. The temperature of the source is

• A. ${567}^{\circ }\text{C}$
• B. ${467}^{\circ }\text{C}$
• C. ${767}^{\circ }\text{C}$
• D. ${367}^{\circ }\text{C}$

Answer 1

The correct option is A ${567}^{\circ }\text{C}$

Given, initial efficiency , $\eta =\frac{1}{12}$
Reduction in the temperature of sink $={70}^{\circ }\text{C}$.
Final efficiency, ${\eta }^{\prime }=2\times \frac{1}{12}=\frac{1}{6}$
The efficiency of an ideal heat engine(carnot's engine) is,
$\eta =1-\frac{T_2}{T_1}....\left(i\right)$
where $T_1$ and $T_2$ are temperature of source and sink respectively.

From Eq.$\left(i\right)$ for initial case,
$\frac{1}{12}=1-\frac{T_2}{T_1}$
$\Rightarrow T_2=\frac{11T_1}{12}...\left(1\right)$
For final case when temperature of sink is reduced by ${70}^{\circ }\text{C}$.
${\eta }^{\prime }=1-\frac{T_2^{\prime }}{T_1}$
$\Rightarrow \frac{1}{6}=1-\frac{T_2-70}{T_1}$
$\because$Value of difference in temperature on celcius and kelvin scale is same.
$\Rightarrow \frac{T_2-70}{T_1}=\frac{5}{6}$
$\Rightarrow T_2=\frac{5T_1}{6}+70...\left(2\right)$
From Eq.$\left(1\right)$ and $\left(2\right)$
$\frac{11T_1}{12}=\frac{5T_1}{6}+70$
$\Rightarrow \frac{T_1}{12}=70$
$\Rightarrow T_1=840\text{K}$
$\therefore T_1=840-273={567}^{\circ }\text{C}$
The temperature of the source is ${567}^{\circ }\text{C}$