Question 305: Thermodynamics & KTG

Question

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 :

Accepted Answer

$\eta=1-\frac{T_{	ext {sink }}}{T_{	ext {source }}}$
$50 \%$ efficiency $\Rightarrow \frac{1}{2}=1-\frac{T_{	ext {sink }}}{T_{	ext {source }}}$
$\frac{1}{2}=1-\frac{T_{	ext {sink }}}{600} \Rightarrow T_{	ext {sink }}=300$
Now, $70 \%$ efficiency $\Rightarrow \frac{7}{10}=1-\frac{T_{	ext {sink }}}{T_{	ext {source }}}$
$\frac{300}{T_{	ext {source }}}=\frac{3}{10}$
$T_{	ext {source }}=1000 \mathrm{~K}$

Answer

300 K

Answer

1000 K

Answer

900 K

Answer

360 K

Thermodynamics & KTG

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 :

Choose the correct answer: