Thermodynamics Question 4
Question 4 - 25 January - Shift 1
A Carnot engine with efficiency $50 %$ takes heat from a source at $600 K$. In order to increase the efficiency to $70 %$, keeping the temperature of sink same, the new temperature of the source will be :
(1) $360 K$
(2) $1000 K$
(3) $900 K$
(4) $300 K$
Show Answer
Answer: (2)
Solution:
Formula: Efficiency of Carnot Engine
Source
Initially $\eta=\frac{1}{2}$
But $\quad \eta=1-\frac{T_2}{T_1}$
$\therefore \quad \frac{1}{2}=1-\frac{T_2}{600}$
$\Rightarrow \frac{T_2}{600}=\frac{1}{2} \quad$ matho $\Rightarrow T_2=300 K$
Now efficiency is increased to $70 %$ and $T_2=300$
$K$, Let temp of source $T_1=T$
$\Rightarrow \frac{7}{10}=1-\frac{300}{T}$
$\Rightarrow \frac{300}{T}=1-\frac{7}{10}$
$\Rightarrow \frac{300}{T}=\frac{3}{10} \quad \therefore T=1000 K$
