Question

For 100% efficiency of a Carnot engine the temperature of the sink should be

• A. 273${}^o$ C
• B. 0${}^o$ C
• C. -273${}^o$ C
• D. none of these

Answer 1

The correct option is C -273${}^o$ C

The efficiency of the cycle is given as $\eta =1-\frac{T_2}{T_1}$.

Here $T_1$ is the temperature of the source and $T_2$ is the temperature of the sink.

For 100% efficiency we get $1=1-\frac{T_2}{T_1}$
$\frac{T_2}{T_1}=0$

$T_2=0K=-{273}^{o}C$