Physics > Thermometry and Thermal Expansion > 1.0 Introduction
Thermometry and Thermal Expansion
1.0 Introduction
2.0 Thermal Expansions and it's types
3.0 Relation between $\alpha ,\,\beta \,$ and $\gamma $
4.0 Variation of Density with Temperature
5.0 The Expansion of Water
6.0 Questions
1.1 Thermometers
Depending on the temperature scale, there are many thermometers like Celsius, Fahrenheit, Kelvin, Reaumur scale. They vary in the upper limit and the lower limit of the scale.So, we can convert from one scale to the other scale as
100 parts in Celsius scale = 180 parts of Fahrenheit scale
1 part of Celsius = 9/5 of Fahrenheit scale$$\begin{equation} \begin{aligned} {T_F} = 32 + \frac{9}{5}{T_C} \\ \Delta {T_F} = \frac{9}{5}{T_C} \\\end{aligned} \end{equation} $$S0,$$\frac{{{T_C} - 0}}{{100}} = \frac{{{T_F} - 32}}{{180}} = \frac{{{T_R} - 0}}{{80}} = \frac{{{T_K} - 273}}{{100}}$$
Question 2. Find the temperature in Celsius, if the temperature in Fahrenheit degrees is ${98.6^O}C$
Solution:
The formula, $$\begin{equation} \begin{aligned} {T_F} = 32 + \frac{9}{5}{T_C} \\ \Rightarrow {T_C} = \frac{5}{9}\left( {{T_F} - 32} \right) \\\end{aligned} \end{equation} $$Putting ${T_F} = 98.6$ , we obtain $$ \Rightarrow {T_C} = 37.0^O$$
Question 3. The steam point and the ice point of a mercury thermometer are marked as ${80^O}$ and ${10^O}$. At what temperature on the centigrade scale the reading of this thermometer will be ${59^O}$?
Solution:
Let the relation between the thermometer reading and centigrade be $y = ax + b$
Given at $x = 100,\;y = 80$ and at $x = 0,\;y = 10$$$\begin{equation} \begin{aligned} \therefore \quad 80 = 100a + b,\quad 10 = b \\ \Rightarrow a = 0.7 \\\end{aligned} \end{equation} $$Now, we have to find $x$ and $y = 59$ $$\begin{equation} \begin{aligned} 59 = 0.7x + b \\ x = 70 \\\end{aligned} \end{equation} $$The answer is ${70^O}C$
Question 4. Two thermometers one marked in Fahrenheit the other in Celsius, are placed in a bath. At what temperature will both thermometers read the same?
Solution:
Let the same temperature be $T$, the same reading in both thermometers,$$\begin{equation} \begin{aligned} {T_F} = 32 + \frac{9}{5}{T_C} \\ T = 32 + \frac{9}{5}T \\ \Rightarrow T = - {40^O} \\\end{aligned} \end{equation} $$
