Question

A reversible power cycle whose thermal efficiency is 40% receives 50 kJ by heat transfer from a hot reservoir at 600 K and rejects energy by heat transfer to a cold reservoir at temperature TC. Determine the energy rejected, in kJ, and TC, in K.

Answer 1

Step-by-step explanation:

thermal efficiency = 40%

efficiency = (T₂ -T₁ ) / T₂ where T₂ is temperature of hot reservoir and T₁ is temperature of cold reservoir.

(600 -T₁ ) / 600 = .4

600 -T₁ = 240

T₁ = 360K

Energy converted into work = 50 x .4 = 20 kJ

heat rejected = 50 - 20 = 30 kJ