Question

The temperature of sink of a Carnot engine is 27?C. If the efficiency of engine be 25%, then the temperature of source must be

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

Answer 1

Answer: (B)

$127{}^\circ C$

Answer 2

Efficiency of Car not engine is given by, $\eta =1-\frac{{{T}_{2}}}{{{T}_{1}}}$ where ${{\text{T}}_{\text{2}}}$ is temperature of sink in kelvin and ${{\text{T}}_{1}}$ is temperature of source in kelvin. Given, $\eta =25$ ${{T}_{2}}=27{{\,}^{o}}C=27+273=300\,K$ $\therefore$ $0.25=1-\frac{300}{{{T}_{1}}}$ $\Rightarrow$ $\frac{300}{{{T}_{1}}}=1-0.25$ $\Rightarrow$ $\frac{300}{{{T}_{1}}}=0.75$ $\Rightarrow$ ${{T}_{1}}=\frac{300}{0.75}=400\,K=127\,{{\,}^{o}}C$