Question

A Carnot engine operates between 900K and 300K.The efficiency of the engine has to be increased to 80% 

1. By how much should the temperature of the source alone be increased? 

2. By how much should the temperature of the sink alone be lowered?

Answer 1

Efficiency of a carnot engine is given by,

where T₂ = Temperature of the sink

T₁ = Temperature of the source

∴ Temperature of the source alone has to be increased by 600K.

2. 0.8 = 1 – \(\frac {T_2}{900}\)

∴ T₁ = 900 (1-0.8) = 180 K

∆T = 300-180 = 120K

Therfore temperature of the sink alone has to be lowered by 120K.