1. 7. A Carnot engine has an efficiency of 0.45%. If the operating temperature in Celsius is such
that Ti-0.30TH, what is the operating temperatures of this engine?
Solution
given
e = 0.45%
= 0.0045
let T1 and T2 cold and hotreservir temperatures respectively
we know, e = 1 - (T1 + 273)/(T2 + 273) =
0.0045 = 1 - (0.30*T2 + 273)/(T2 + 273)
==> T2 = 1.77 C (TH) <<<<<-----------Answer
T1 = 0.03*T2
= 0.3*1.77
= 0.531 C (TL) <<<<<-----------Answer