Question

A certain liquid has a melting point of $-50^{\circ }C$ and a boiling point of $150^{\circ }C$. A thermometer is designed with this liquid and its melting and boiling points are designated at $0^{\circ }L$ and $100^{\circ }L$ The melting and boiling points of water on this scale are

• A. $25^{\circ }L$ and $75^{\circ }L$, respectively
• B. $0^{\circ }L$ and $100^{\circ }L$, respectively
• C. $20^{\circ }L$ and $70^{\circ }L$, respectively
• D. $30^{\circ }L$ and $80^{\circ }L$, respectively

Answer 1

Answer: (A)

From principle of thermometry,

$\frac{T-T_{LFP}}{T_{UFP}-T_{LFP}}$ =a constant for every

thermometric scale

Now, for any temperature L on a thermometer designed with given liquid and equivalent temperature C on centigrade scale, we have

$\left(\frac{L-T_{LFP}}{T_{UFP}-T_{LFP}}\right)$ Liquid based scale $=\left(\frac{C-T_{LFP}}{T_{UFP}-T_{LFP}}\right)$ Centigrade scale

$\Rightarrow \frac{L-\left(-50\right)}{150-\left(-50\right)}=\frac{C-0}{100-0}$

$\frac{L+50}{150+150}=\frac{C}{100}$

$\Rightarrow L+50=2C$

Now at $0^{\circ }L$, centigrade scale reading will be

$0+50-2C$ or $C-\frac{50}{2}-25^{\circ }L$

and at $100^{\circ }L$, centigrade scale reading will be

$10+50=2C$

or $C=\frac{150}{2}$

$=75^{\circ }L$