Question

The length of the column of mercury in a thermometer is $4.0$ $cm$ when the thermometer is immersed in ice water and $24.0$ $cm$ when the thermometer is immersed in boiling water.
(a) What should be the length at room temperature, ${22.0^ \circ }$ $C$ $?$
(b) If the mercury column is $25.4$ $cm$ long when the thermometer is immersed in a chemical solution, what is the temperature of the solution$?$

Answer 1

In this case, we'll use the thermal expansion formula, which states that matter has a tendency to change shape, volume, and area in response to a change in temperature, and the material's coefficient of thermal expansion is defined as the degree of expansion divided by the temperature change.

Formula used:
The general formula for linear thermal expansion is:
$\Delta l = {l_0}\alpha \Delta T$
The other formula used is
$l' = {l_0}\left( {1 + \alpha \Delta T} \right)$
Where, $\Delta l,l'$ is change in length, ${l_0}$ is initial length or the original length, $\alpha$ is coefficient of expansion and $\Delta T$ is the change in temperature.

Complete step by step answer:
The above problem revolves around the concept of thermal expansion where a substance or a material tends to undergo expansion in its size and this change in its size needs to be measured. In order to find the change in its length we first need to know the concept of thermal expansion. Thermal expansion is said to be the increase in size of the body or the material at hand when it is heated.

When a liquid is heated it has a tendency to increase in its length and hence the type of expansion considered here is the linear expansion. Linear expansion is defined as the change or the increase in length of the substance considered when heated. Here, the substance that undergoes thermal expansion is given to be mercury inside a thermometer.

The expansion occurs when an object expands due to increase in temperature because heated molecules move faster and take up more space. We know that the molecules and atoms which make up the liquid are bonded to each other by intermolecular forces that keep them bound to each other.

When these molecules are provided heat they tend to vibrate about their positions and gain kinetic energy in order to move more violently with more energy leading to the breakage of the intermolecular bonds between them. Hence due to this, the distance of separation between these molecules will tend to increase resulting in expansion.

Let us now extract the data given in the question. We are given that when the thermometer is immersed in ice water initially then the mercury is at a level of $4\;cm$ and then it is provided some heat for the expansion to occur and the length of the mercury column to increase to a level $24\;cm$. This means that the original length, say ${l_i}$, before heating was $4\;cm$ and the final length, say ${l_f}$, after the thermometer immersed in boiled water is $24\;cm$. Hence we now need to find the change in its length which is given by $\Delta l$. Hence,
$\Delta l = {l_f} - {l_i}$

As discussed above, we now substitute the final and initial lengths of the mercury column.
$\Rightarrow \Delta l = 24 - 4$
$\Rightarrow \Delta l = 20\;cm$
We are given that the heat is provided by immersing the thermometer into boiling water and hence the temperature change will be equivalent to the boiling temperature of water which is $100{}^ \circ C$, since this is the standard boiling point of water. Hence, the change in temperature is $\Delta T$ is $100{}^ \circ C$. In part a) of the problem we are asked to find the change in the length of the mercury column at room temperature. But in order to do this we must first find out the coefficient of linear expansion, $\alpha$, which is constant at all temperatures.

Given, ${l_0} = 4\;cm$, $\Delta l = 20\;cm$ and $\Delta T = 100{}^ \circ C$.
Now we will apply the formula of thermal expansion.
Therefore, $\Delta l = {l_0}\alpha \Delta T$
By rearranging the terms to make $\alpha$ the subject we get:
$\alpha = \dfrac{{\Delta l}}{{{l_0}\Delta T}}$ --------($1$)
Now by substituting all the given values in equation ($1$) we get:
$\alpha = \dfrac{{20}}{{4 \times 100}}$
$\Rightarrow \alpha = \dfrac{{20}}{{400}}$
From here we will get the coefficient value $\alpha = 0.05$

(a) we will use the formula:
$l' = {l_0}\left( {1 + \alpha \Delta T} \right)$ -------($2$)
Now the change in temperature will become $22{}^ \circ C$ since we are asked to find the change in length at room temperature given to be $22{}^ \circ C$. Also we use the value of the coefficient of linear expansion we had determined in the above formula since it is a constant. Given, $\Delta T = {22^ \circ }C$, $\alpha = 0.05$ and ${l_0} = 4\;cm$.
Now substituting all these values in equation ($2$) we get:
$l' = 4\left( {1 + 0.05 \times 22} \right)$
On solving we will get:
$\Rightarrow l' = 4\left( {1 + 1.1} \right)$
$\Rightarrow l' = 4\left( {2.1} \right)$
$\Rightarrow l' = 8.4\;cm$
So the length of the mercury column at room temperature is $8.4$ $cm$ .

(b) when the mercury column is $25.4$ $cm$, we are asked to find the change in temperature. It is given that when the thermometer is immersed in a chemical solution the mercury column increases in length which is given to be $25.4\;cm$. This means that the change in length, that is, $l' = 25.4\;cm$. We already know the value of $\alpha$. Hence, we will use the same formula that is:
$l' = {l_0}\left( {1 + \alpha \Delta T} \right)$
By rearranging the terms to make the change in temperature, $\Delta T$ as the subject since we need to find this temperature we get:
$\Rightarrow \left( {1 + \alpha \Delta T} \right) = \dfrac{{l'}}{{{l_0}}}$
$\Rightarrow \alpha \Delta T = \dfrac{{l'}}{{{l_0}}} - 1$
$\Rightarrow \Delta T = \dfrac{{l' - {l_0}}}{{\alpha {l_0}}}$ ---------($3$)
By substituting the given values in equation ($3$) we get:
$\Rightarrow \Delta T = \dfrac{{25.4 - 4}}{{(0.05)(4)}}$
On solving we will get:
$\Rightarrow \Delta T = \dfrac{{21.4}}{{0.2}}$
$\Rightarrow \Delta T = {107^ \circ }C$
Here $107$ is in degrees Celsius so we will convert it into Kelvin by adding $273$ . Hence,
$\therefore \Delta T = 107 + 273$
So our answer will be $380$ $K$.

Note: There may be a confusion on which formula for linear expansion to use since there are two formulas as mentioned above. One of the formula, that is, $\Delta l = {l_0}\alpha \Delta T$, gives the change in the length which is associated with the initial and final lengths of the substance while the formula $l' = {l_0}\left( {1 + \alpha \Delta T} \right)$ gives equation for the changed or the increased length. Hence both of these formulas are different and used for different situations as seen in the solution. The thermal expansion is of three types : linear expansion, area expansion and volume expansion.