Question

The efficiency of a heat engine is $\frac{1}{6}$. Its efficiency double when the temperature of sink decreases by ${62}^{o}C$, its efficiency doubles. Then what is the temperature of source?

• A. $372K$
• B. $470K$
• C. $542K$
• D. $1042K$

Answer 1

The correct option is A $372K$

$\eta =1-\frac{T_2}{T_1}$
$\frac{1}{6}=1-\frac{T_2}{T_1}$ so $\frac{T_2}{T_1}=\frac{5}{6}$ so $T_2=\frac{5T_1}{6}$
Now ${\eta }^1=1-\frac{T_2^1}{T_1}⟹2\times \frac{1}{6}=1-\frac{(T_2-62)}{T_1}⟹\frac{T_2-62}{T_1}=\frac{2}{3}$
$3T_2-186=2T_1⟹3\left(\frac{5T_1}{6}\right)-186=2T_1⟹\frac{5}{2}{T}_1-2T_1=186$
$\frac{T_1}{2}=186⟹T_1=372K$