Question

A Carnot cycle operates between source and sink temperatures of 250°C and – 15°C. If the system receives 90 kJ from the source, find : 

(i) Efficiency of the system ; 

(ii) The net work transfer ; 

(iii) Heat rejected to sink

Answer 1

Temperature of source, T₁ = 250 + 273 = 523 K 

Temperature of sink, T₂ = – 15 + 273 = 258 K 

Heat received by the system, Q₁ = 90 kJ

η_carnot = 1 – \(\cfrac{T_2}{T_1}\) = 1 - \(\cfrac{258}{523}\)

= 0.506 = 50.6%.

(ii) The net work transfer, W = η_carnot × Q₁

= 0.506 × 90 = 45.54 kJ.

(iii) Heat rejected to the sink, Q₂ = Q₁ – W 

[\(\because\) W = Q₁ – Q₂] 

= 90 – 45.54 = 44.46 kJ.