Question

What is the temperature of source in carnot cycle of 10% efficiency when heat exhausts at 270K

Answer 1

Explanation:

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

The temperature of source in carnot cycle of 10% efficiency when heat exhausts at 270K is 300 K.

η = 1 − $\frac{T_{1} }{T_{2} }$

Therefore,  $\frac{T_{1} }{T_{2} }=1-$η$=1-\frac{10}{100}$=$\frac{90}{100}$

or

$T_{1}$ = $\frac{100 T_{2} }{90}$

$=\frac{100}{90}*270$

= 300 K