Answer to Question #303799 in Physical Chemistry for Chemistry

Question

A Carnot engine absorbs heat from a source at a higher temperature of 800 Kelvin

and the efficiency of the Carnot engine is 50%.

To what value the higher temperature should be increased for efficiency to increase

to 80%, if the lower temperature is kept constant.

Answer 1

$\eta$ $=\frac{t_2-t_1}{t_2}\%$ or $(1-\frac{T_2}{T_1})$

Where $t_2$ is the source temperature

$t_1=$ intake temperature

$\frac{50}{100}=\frac{800-t_1}{800}\implies\>t_1=400K$

Where the efficiency is $80\%$

$\frac{80}{100}=\frac{t_2-400}{t_2}$

$t_2=2000K$