A Carnot engine has efficiency of 40% when sink temperature is $27^{o} C$ . In order to increase efficiency by 10%, the source temperature will have to be increased by :

{135.7}^{o} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

135.7 K

No worries! We‘ve got your back. Try BYJU‘S free classes today!

100 K

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{35.7}^{o} C

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $100 K$
Efficiency of the carnot engine $(η) = 1 - \frac{T_{c}}{T_{h}}$

$0.4 = 1 - \frac{300}{T}$

$T = \frac{300}{0.6} = 500 K$

For efficiency to be 50%,

$0.5 = 1 - \frac{300}{T}$

$T = \frac{300}{0.5} = 600 K$

So, the temperature has to be increased by $600 K - 500 K = 100 K$

Suggest Corrections

Similar questions

40 %

27^{o} C

10 %

27^{o} C

300 K

40 %

50 %

300

K

40

50

300 K

40 %

60 %

Join BYJU'S Learning Program