Question

5. A certain heat engine operates between 1000 K and 500 K a) What is the maximum efficiency of the engine? (ANS: 0.5) b) Calculate the maximum work that can be done per kJ of heat supplied by the hot source. (ANS: 0.5 kJ) c) How much heat is discharged into the cold sink in a reversible process for each kJ supplied by the hot source? (ANS: 0.25)

Answer 1

Solution : Given that: - Heat Engine - love to 2500k = 1000k = 500€. -Temperature heart Source CTU -temperature froeit link. (T) a) porcentage (%) Maximum efficiency of the Engine = T, XTz 11-12 X100 = 1000-500 loc 1000 5 500 10 = 504 . Maximum . Efficiency Maximum efficency of the engine = 50% of the engine = Too =0.5 (b) treat Supplied by the hot Source Qi) => 1.005. Marimum. Werk done w) max) Heat Supplied by the hor Source (QI)

mak (maus 0.5 x lio 0.5 w hat - Maximum work done by the Surce Wimax) = 0.5 kJ teat Supplied by the Source = looky - Q 05 = 10-02 loo 20:25 A @ 5.0.5 X015 10:25 13) & Heal is discharged into the cold sink (Q) 0.85 kJ.