Question

On a thermometer, the freezing point of water is marked as ${20}^{\circ }C$ and the boiling point of water is marked as ${150}^{\circ }C$ . A temperature of ${60}^{\circ }C$ will be read on this thermometer as ________ ${}^{\circ }C$.

Answer 1

We know that $\frac{T_t-T_0}{T_{100}-T_0}=\frac{T_c-0}{100-0}$

Given: $T_{100}={150}^{0}C,T_0={20}^{0}C$
$T_C={60}^{\circ }C$

Let $T_t$ be temperature on this new thermometer corresponding to ${60}^{\circ }C$ on Celsius scale

Using the relation

$\frac{T_t-T_0}{T_{100}-T_0}=\frac{T_c-0}{100-0}$

$\frac{T_t-20}{150-20}=\frac{60}{100}$

$T_t=0.6\times 130+20$

= ${98}^{\circ }C$