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 :

• A. $900 K$
• B. $300 K$
• C. $1000 K$
• D. $360 K$

Answer 1

Answer: (C)

$1000 K$

Answer 2

But η=1−T1​T2​​
∴21​=1−600T2​​
⇒600T2​​=21​⇒T2​=300K
Now efficiency is increased to 70% and T2​=300
K, Let temp of source T1​=T
⇒107​=1−T300​
⇒T300​=1−107​
⇒T300​=103​
∴T=1000K