Question

A Carnot engine works between the source and the sink with efficiency 40%. How much temperature of the sink be lowered keeping the source temperature constant so that it's efficiency increases by 10%

Answer 1

1/6 of sink temperature should be lowered keeping the source temperature constant to increase the efficiency by 10 %.

• Let tsink = temperature of sink

and, tsource = temperature of source

• According to the question,

.40 = 1 – (tsink / tsource)

(tsink / tsource) = .60         ………(i)

• Let decrease in temperature be x. Then,

.50 = 1 – [(tsink - x) / tsource]

(tsink - x) / tsource = .50    ……..(ii)

• On dividing (i) by (ii) and solving,

x = (1/6) tsink