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 $\quad \eta=1-\frac{T_2}{T_1}$

$\therefore \quad \frac{1}{2}=1-\frac{T_2}{600}$

$\Rightarrow \frac{T_2}{600}=\frac{1}{2} \quad$ $\Rightarrow T_2=300 K$

Now efficiency is increased to $70 \%$ and $T_2=300$

$K$, Let temp of source $T_1=T$

$\Rightarrow \frac{7}{10}=1-\frac{300}{T}$

$\Rightarrow \frac{300}{T}=1-\frac{7}{10}$

$\Rightarrow \frac{300}{T}=\frac{3}{10} \quad \therefore T=1000 K$