Question

A Carnot engine whose efficiency is 40%, takes in heat from a source maintained at a temperature of 500 K. It is desired to have an engine of efficiency 60%. Then, the intake temperature for the same exhaust (sink) temperature must be

• A. 1200 K
• B. 750 K
• C. 600 K
• D. 800 K

Answer 1

Answer: (B)

750 K

Answer 2

Efficiency of Carnot engine

$\eta = 1 - \frac{T_{2}}{T_{1}}$

Where $T_{1}$is the temperature of the source and $T_{2}$ is the temperature of the sink

For the 1^st case

$\eta = 40\%,T_{1} = 500K$

$\therefore\frac{40}{100} = 1 - \frac{T_{2}}{500}$

$\frac{T_{2}}{500} = 1 - \frac{40}{100} = \frac{3}{5}$

$T_{2} = \frac{3}{5} \times 500 = 300K$

For the 2^nd case

$\eta = 60\%,T_{2} = 300K$

$\therefore\frac{60}{100} = 1 - \frac{300}{T_{1}}$

$\frac{300}{T_{1}} = 1 - \frac{60}{100} = \frac{2}{5}$

$T_{1} = \frac{5}{2} \times 300 = 750K$