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?
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