Question

A vertical glass tube of length $\mathbf{\text{L}}\mathbf{=}\mathbf{1}\mathbf{.280}\mathbf{}\mathbf{000}\mathbf{\text{\hspace{0.17em}m}}$is half filled with a liquid at$\mathbf{20.000}\mathbf{}\mathbf{000}\mathbf{°}\mathbf{\text{C}}$. How much will the height of the liquid column change when the tube and liquid are heated to$\mathbf{30.000}\mathbf{}\mathbf{000}\mathbf{°}\mathbf{\text{C}}$? Use coefficients${\mathbf{\alpha }}_{\mathrm{glass}}\mathbf{=}\mathbf{1.000}\mathbf{}\mathbf{000}\mathbf{}\mathbf{\times }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{5}}}{\mathbf{\text{K}}}\mathbf{\text{and}}{\mathbf{\beta }}_{\mathrm{liquid}}\mathbf{=}\mathbf{}\mathbf{4.000}\mathbf{}\mathbf{000}\mathbf{\times }\mathbf{}{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{/}\mathbf{\text{K}}\mathbf{.}$

Short answer

Change in the height of the liquid column of the glass tube is$1.3\times {10}^{-4}\text{m}$

Step-by-step solution

The given data

1. Height$\text{h}=1.28\text{\hspace{0.17em}m}$
2. Initial height${\text{h}}_0=\text{h}/2=0.64\text{\hspace{0.17em}m}$
3. Temperature$\mathrm{\Delta }\text{T}={\text{T}}_2-{\text{T}}_1=30°-20°=10°\text{C}$
4. Coefficient of thermal expansion${\mathrm{\alpha }}_{\mathrm{glass}}=1\times {10}^{-5}/\text{K}$
5. Coefficient of liquid${\mathrm{\beta }}_{\mathrm{liquid}}=4\times {10}^{-5}/\text{K}$

Understanding the concept of thermal expansion

Consider a vertical glass tube filled with liquid at a particular temperature. If we heat the glass tube, the liquid inside the glass tube will change its height. In this case, the cross-section area expands according to how the glass expands on heating. This can be analyzed by considering the area. Also, the height of the glass tube can be calculated by taking the ratio of the volume of the liquid to the total area of the glass.

Formula:

Base area of the tube,$\text{A}=\mathrm{\pi }{\text{r}}^2$ ,,,(i)

The height of the tube,$\text{h}=\frac{\text{V}}{\text{A}}$ …(ii)

The linear expansion of a body,$\mathrm{\Delta }\text{L}=\text{L}\mathrm{\alpha }\mathrm{\Delta }\text{T}$ …(iii)

Where$\mathrm{\alpha }$ is the coefficient of linear expansion of body.

The volume change in expansion of a body,$\mathrm{\Delta }\text{V}={\text{V}}_0\mathrm{\beta \Delta }\text{T}$ …(iv)

Where, $\mathrm{\beta }$is the volume expansion coefficient of body.

Calculation of change in the height of the liquid column

To calculate the height change in the glass tube, let the initial volume of the liquidbe${\text{V}}_0$ , the initial height be ${\text{h}}_0$, and the initial cross section area be ${\text{A}}_0$.

On differentiating the area equation (i), we get

$\begin{array}{c}\mathrm{dA}=2\mathrm{\pi r}\mathrm{dr}\\ =2\mathrm{\pi r}(\mathrm{r\alpha }\mathrm{dT})(\because \mathrm{dr}=\mathrm{r\alpha }\mathrm{dT},\text{fromequation}(\mathrm{iii}\left)\right)\\ =2\mathrm{\alpha }\left({\mathrm{\pi r}}^2\right)\mathrm{dT}\\ =2\mathrm{\alpha A}\mathrm{dT}\text{(fromequation(i))}\end{array}$

Therefore, the height from equations (ii), (iii) and (iv) will be given as:

$\begin{array}{c}\mathrm{h}=\frac{{\mathrm{V}}_0+{\mathrm{V}}_0{\mathrm{\beta }}_{\mathrm{liquid}}\mathrm{\Delta T}}{{\mathrm{A}}_0+{\mathrm{A}}_{0}2{\mathrm{\alpha }}_{\mathrm{glass}}\mathrm{\Delta T}}(\mathrm{V}={\mathrm{V}}_0+{\mathrm{V}}_0{\mathrm{\beta }}_{\mathrm{liquid}}\mathrm{\Delta T}\text{and}\mathrm{A}={\mathrm{A}}_0+{\mathrm{A}}_{0}2{\mathrm{\alpha }}_{\mathrm{glass}}\mathrm{\Delta T})\\ =\frac{{\mathrm{V}}_0(1+{\mathrm{\beta }}_{\mathrm{liquid}}\mathrm{\Delta T})}{{\mathrm{A}}_0(1+2{\mathrm{\alpha }}_{\mathrm{glass}}\mathrm{\Delta T})}\\ ={\mathrm{h}}_0\left(\frac{1+{\mathrm{\beta }}_{\mathrm{liquid}}\mathrm{\Delta T}}{1+2{\mathrm{\alpha }}_{\mathrm{glass}}\mathrm{\Delta T}}-1\right)(\because {\mathrm{V}}_0/{\mathrm{A}}_0={\mathrm{h}}_0)\\ =\left(0.64\text{\hspace{0.17em}m}\right)\left(\frac{1+(4\times {10}^{-5}/\text{K})(10°\text{C})}{1+2(1\times {10}^{-5}/\text{K})(10°\text{C})}-1\right)\left(\text{consideringthegivenquantities}\right)\\ =1.3\times {10}^{-4}\text{m}\end{array}$

Hence, the value of the height of the liquid column is$1.3\times {10}^{-4}\text{m}$