Question

An ideal Carnot engine whose efficiency is 40% receives heat at 500 K. If its efficiency were 50%, then in take temperature for same exhaust temperature would be

• A. 700 K
• B. 900 K
• C. 800 K
• D. 600 K

Answer 1

The correct option is D 600 K

From $\eta =1-\frac{T_2}{T_1}$
$\frac{T_2}{T_1}=1-\eta =1-\frac{40}{100}=\frac{3}{5}$
Therefore $T_2=\frac{3}{5}{T}_1=\frac{3}{5}\times 500=300K$
Again $\frac{T_2}{T_1^{\prime }}=1-\eta$ or $\frac{300}{T_1^{\prime }}=1-\frac{50}{100}=\frac{1}{2}$
or $T_1=600K$