Question

The temperature of the sink of a Carnot engine is ${27}^{\circ }C.$ If the efficiency of the engine is $40\%,$ find the temperature of the source.

• A. ${227}^{0}C$
• B. ${327}^{0}C$
• C. ${127}^{0}C$
• D. ${27}^{0}C$

Answer 1

Answer: (A)

${227}^{0}C$

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

using the Efficiency of carnot engine, putting the value in it.

Given: $T_2={27}^{o}c=300K$ and efficiency $\eta =40$

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

$\frac{40}{100}=1-\frac{300}{T_1}$

$\therefore T_1=500K={227}^{o}C$