Question

The efficiency of carnot heat engine is 0.2 when the temperature of sink alone is reduced by 100K the efficiency of heat engine gets doubled.what is the temperature of source and sink

Answer 1

In first case, Efficiency of engine N1=0.2

if T1 and T2 are the temperature of Source and Sink then

$0.2=1-\frac{T_{2} }{T_{1} }$     ............equation (1)

Second case when efficiency is doubled by reducing the temp. of sink by 100k

$0.4=1-\frac{T_{2}-100}{T_{1} }$

$0.4=1-\frac{T_{2} }{T_1}+\frac{100}{T_1}$ ...........equation(2)

From equation 1&2

$0.2=\frac{100}{T_1}$

$T_1=500K$

putting this value in equation 1 we get $T_2=400K$