Question

The efficiency of a Carnot engine is 40%. If the temperature of the source is 193.6°C, then determine the temperature of the sink.

Answer 1

Given, \(\eta\) = 40%, = \(\frac{40}{100} = \frac{2}{5}\)

and T₁ = 193°C = 193.6 + 273K = 466.6K; T₂ = ?