Question

A Carnot engine receives 250 kJ/s of heat from a heat source reservoir at 525 o C and rejects heat to a heat-sink reservoir at 50 o C. What are the power developed and the heat rejected

Answer 1

Answer:

Explanation:

efficiency of carnot engine = (Q₁ - Q₂) / Q₁ = (T₁ - T₂) / T₁

Q₁ = heat absorbed , Q₂ = heat rejected

T₁ = Temperature of source = 525 + 273 = 798 K

T₂ = Temperature of sink = 50 + 273 = 323 K

efficiency = (798 - 323) / 798

= 475 / 798 = .595

(Q₁ - Q₂) / Q₁ = .595

(250 - Q₂) / 250 = .595

250 - Q₂ = 148.75

Q₂ = 101.25 kJ

Heat rejected = 101.25 kJ .

output work = Q₁ - Q₂ = 250 - 101.25 = 148.75 kJ / s

power = 148.75 kJ /s