Question

A Carnot engine operates at an efficiency of $40\%$ with the sink at ${27}^{\circ }C$. Find the amount by which the temperature of the source must be increased in order to increase the efficiency by $10\%$.

• A. $600\text{K}$
• B. $100\text{K}$
• C. $500\text{K}$
• D. $300\text{K}$

Answer 1

The correct option is B $100\text{K}$

Let $T_1$ be the initial temperature of the source hence efficiency of heat engine is given by,
$\eta =40\%=\frac{40}{100}$
$\Rightarrow \eta =1-\frac{T_2}{T_1}...\left(i\right)$
$\Rightarrow \frac{40}{100}=1-\frac{(273+27)}{T_1}$
$\Rightarrow T_1=500\text{K}$
For the efficiency to be increased by $10\%$ i.e. new efficiency is ${\eta }^{\prime }=50\%$
let $T_1^{\prime }$ be the new temperature of the source
again from eq.$\left(i\right)$,
$\Rightarrow \frac{50}{100}=1-\frac{(273+27)}{T_1^{\prime }}$
or $T_1^{\prime }=600\text{K}$
The required increase in the temperature of the source,
$\Delta T=T_1^{\prime }-T_1=600\text{K}-500\text{K}$
$\therefore \Delta T=100\text{K}$