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

Question

Dr. Abubacker · Accepted Answer

(d) From $\eta = 1 - \frac{T_{2}}{T_{1}}\  \Rightarrow \ \frac{T_{2}}{T_{1}} = 1 - \eta = 1 - \frac{400}{100} = \frac{3}{5}$
$	herefore\ \ T_{2} = \frac{3}{5}T_{1} = \frac{3}{5} 	imes 500 = 300\ K$
Again $\frac{T_{2}}{T_{1}^{'}} = 1 - \eta$ or $\frac{300}{T_{1}^{'}} = 1 - \frac{50}{100} = \frac{1}{2}$
or $T_{1} = 600\ K$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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