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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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

Similar Questions

The water on the surface of a lake is just about to freeze. The most likely temperature at the bottom of the lake is (in $^{\circ} C$)

Standardisation of thermometers is obtained with

What is absolute zero temperature?

If the temperature of a body increases by $30^{\circ}C$ on the Celsius scale,then the increase in temperature on the Fahrenheit scale is .... $^oF$.

In a mercury thermometer, the ice point (lower fixed point) is marked as $10^{\circ}$ and the steam point (upper fixed point) is marked as $130^{\circ}$. At $40^{\circ} C$ temperature, what will this thermometer read (in $^{\circ}$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module