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=2 C $
or $C=\frac{150}{2}$
$=75^{\circ}L$