Question

The steam point and the ice point of a mercury thermometer are marked as ${80}^{\circ }$ and ${20}^{\circ }$. What will be the temperature in centigrade mercury scale when this thermometer reads ${32}^{\circ }$ ?

Answer 1

Ice point = ${20}^{\circ }\left(T_0\right)$, $T_1={32}^{\circ }C$

steam point $={80}^{\circ }\left(T_{100}\right)$

$T=\frac{T_1-T_0}{T_{100}-T_0}\times 100$

$=\frac{32-20}{80-20}\times 100$

$=\frac{12}{60}\times 100=\frac{120}{6}={20}^{\circ }C$