Question

The efficiency of the Carnot engine is 25% when the temperature of sink is 27C. Then temperature of source is:

Answer 1

\(\eta = 1 - \frac{T_2}{T_1}\)

\(\frac{25}{100} = 1 - \frac{300}{T_1}\)

\(\frac14 = 1- \frac{300}{T_1}\)

\(\frac{300}{T_1} = \frac34\)

\(T_2 = \frac{300\times4}{3}\)

\(T_1 = 400 K° \)

\(T_1 = 127°C\)