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

Assuming a linear temperature scale, the relation between the two temperature scales will be:

$\frac{T-20}{80-20}=\frac{T_C-0}{100-0}$

$T_C=\frac{5}{3}(T-20)$

Thus, at given temperature,

$T_C=\frac{5}{3}(32-20)={20}^{o}C$

Answer is ${20}^{o}C$