Question

The efficiency of Carnot heat engine working between two temperature is 60%. If the temperature of the source alone is decreased by 100 K ,the efficiency becomes 40%. Find the temperature of the source & sink.

Answer 1

Given efficiency of heat engine ${\eta }_1=60$%

Let temperature of source be $T_1$ and temperature of sink be $T_2$

Now as per question when temperature of source in increased by 100 K its efficiency becomes 40%.Thus efficiency of carnot engine now will be,

${\eta }_2=40$%

${\eta }_1=1-\frac{T_2}{T_1}$-----(A) and ${\eta }_2=1-\frac{T_2}{T_1+100}$-----------(B)

Comparing both the efficiency with sink temperature $T_2$ we get

$(1-0.60)T_1=(1-0.40)(T_1-100)$

$T_1=300K$

Putting the above value in A we get,

$T_2=300\left(0.40\right)=120K$