The sink in Carnot's heat engine is at 300k and the engine works at an efficiency 0.4. If the efficiency if the engine is to be increased to 0.5. Find by how many kelvin the temperature if the source should be increased.

Explanation

Given `eta=0.4, n. = 0.5 , T_(2)=300k`
We know that `eta =1-(T_2)/(T_1)` i.e `0.4 = 1-(T_2)/(T_1) or (T_2)/(T_1)=0.6`
or. `T_2=0.6 T_1` or `T_1 = (T_2)/(0.6)=(300)/(0.6)=500K`
Using `eta =1 - (T_2)/(T_1), T_1= 300 + theta " " i.e 05 =1-(06 T_1)/(300+ theta)`
`i. e, 0.5 1-((0.6 xx 500)/(300 + theta)) " " i.e 0.5 = 1/2`
`therefore 2 xx 0.6 xx 500 = 300 + theta, " " 600. 0 = 300 + theta `
` therefore = 600 0-300 = 300`
The source temperature should be increased by 300 K .