Question

A mercury thermometer reads 80${}^o$C when the mercury is at $5.2cm$ mark and ${60}^o$C when the mercury is at $3.9cm$ mark. The temperature (in ${}^{o}C$) when the mercury level is at $2.6cm$ mark is $4\times {10}^x$. Find the value of $x$.

Answer 1

$\frac{l_1-l_2}{l_1-l_3}=\frac{\alpha l_0(T_1-T_2)}{\alpha l_0(T_1-T_2)}=\frac{T_1-T_2}{T_1-T_3}$
or $\frac{5.2-3.9}{5.2-2.6}=\frac{80-60}{80-T_3}$
$1.3(80-T_3)=2.6\left(20\right)$ or $T_3={40}^{o}C$