A certain liquid has a melting point of -50^{ | vedclass

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(A) The principle of thermometry states that the ratio of the difference between a temperature and the Lower Fixed Point $(LFP)$ to the difference bet

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$25^{\circ} L$ and $75^{\circ} L$,respectively

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(A) The principle of thermometry states that the ratio of the difference between a temperature and the Lower Fixed Point $(LFP)$ to the difference between the Upper Fixed Point $(UFP)$ and the $LFP$ is constant for all scales.Let $C$ be the temperature on the Celsius scale and $L$ be the temperature on the new scale.For the Celsius scale: $LFP = 0^{\circ} C$,$UFP = 100^{\circ} C$.For the new liquid scale: $LFP = -50^{\circ} C$ (which corresponds to $0^{\circ} L$),$UFP = 150^{\circ} C$ (which corresponds to $100^{\circ} L$).The conversion formula is: $\frac{L - 0}{100 - 0} = \frac{C - (-50)}{150 - (-50)}$.$\frac{L}{100} = \frac{C + 50}{200}$.$L = \frac{C + 50}{2} \Rightarrow 2L = C + 50 \Rightarrow C = 2L - 50$.For the melting point of water $(C = 0^{\circ} C)$:$0 = 2L - 50 \Rightarrow 2L = 50 \Rightarrow L = 25^{\circ} L$.For the boiling point of water $(C = 100^{\circ} C)$:$100 = 2L - 50 \Rightarrow 2L = 150 \Rightarrow L = 75^{\circ} L$.Thus,the melting and boiling points of water on this scale are $25^{\circ} L$ and $75^{\circ} L$.

$A$ 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

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