Question

The efficiency of a Carnot engine is $60\%$. If the temperature of source is $127^{\circ }C$, the sink must be maintained at

• A. $113K$
• B. $+113^{\circ }C$
• C. $-113^{\circ }C$
• D. $-113K$

Answer 1

Answer: (C)

Efficiency of a Carnot engine, $\eta =1-\frac{T_2}{T_1}$
where $T_1$ and $T_2$ are the temperature of the source and sink respectively.
Here, $\eta =60\%=\frac{60}{100}=0.6$
$T_1=127^{\circ }C=\left(127+273\right)K=400K,T_2=?$
$\therefore 0.6=1-\frac{T_2}{400}$
$\frac{T_2}{400}=1-0.6=0.4$
$T_2=160K={\left(160-273\right)}^{\circ }C$
$=-113^{\circ }C$