Question
A reversible engine with an efficiency of 32.0% has TC = 304.0 K.
(a) What is TH?
(b) How much heat is exhausted for every 0.138 kJ of work done?
(a) What is TH?
(b) How much heat is exhausted for every 0.138 kJ of work done?
Answers
Elena
η =1 –Tc/Th => Th/Th =1-0.32 = 0.68,
Th = Tc/0.68 = 304/0.68 = 447 K.
η = W/Qh,
Qh = W/ η = 138/0.32 =431.3 J
Th = Tc/0.68 = 304/0.68 = 447 K.
η = W/Qh,
Qh = W/ η = 138/0.32 =431.3 J