Question

The temperature of sink of Carnot engine is ${27}^{\circ }C$. Efficiency of engine is $25\%$. Then temperature of source is

• A. ${227}^{\circ }C$
• B. ${327}^{\circ }C$
• C. ${127}^{\circ }C$
• D. ${27}^{\circ }C$

Answer 1

Answer: (C)

${127}^{\circ }C$

Given,

$T_2={27}^{0}C=27+273=300K$

$\eta =25$%$=\frac{25}{100}=0.25$

$T_1=?$

The efficiency of Carnot engine,

$\eta =1-\frac{T_2}{T_1}$

$0.25=1-\frac{300}{T_1}$

$0.75-\frac{300}{T_1}=0$

$T_1=\frac{300}{0.75}=400K$

$T_2=400-273$

$T_1={127}^{0}C$

The temperature of the source is ${127}^{0}C$.

The correct option is C.