Question

a carnot engine has an efficiency of 50% when its sink temperature is 27°c what must be the change in its source temperature for increasing its efficiency to 60%?

Answer 1

Taking the Sink Temperature(T2) = 27 degree celsius = 300 kelvin

Efficiency(eta) be 25% = .25

Source Temperature(T1) = ?

Efficiency= (T1 - T2) / T1

Efficiency= 1 -T2/T1

T2/T1 = 1 - Efficiency

T1 = T2/(1 - Efficiency)

T1= 300/(1-.25)

T1 = 400 kelvin = 127 degree celsius

Answer 2

Given efficiency=50%

We know,

Efficiency =1-T2/T1

50%=1-300/T1

50/100=1-300/T1

1/2=(T1-300)/T1

T1=2T1-600

T1=600k

The answer will be 600k

..........thx.......