Question

A vertical glass tube of length L = 1.4m is half filled with a liquid at 20.5 ˚C. How much will the height of the liquid column change when the tube is heated to 34.3 ˚C? Take αglass = 1.20× 10-5/K and βliquid = 3.99× 10-5/K.

Answer 1

Given is:-

Length of the glass tube = 1.4m

$\alpha _{glass} = 1.20 \times 10^{-5 }/K$ and $\beta _{liquid} = 3.99 \times 10^{-5 }/K$

Here, we need to consider the cross-sectional area expansion of the glass and the volume expansion of the liquid:

$\Delta A = A_o (2\alpha)\Delta T$

$\Delta V = V_o \beta \Delta T$

$h = \frac{V}{A} = \frac{V_o + \Delta V}{A_o + \Delta A} = \frac{V_o(1+ \beta \Delta T)}{A_o (1+2 \alpha \Delta T)} = h_o \frac{(1 + \beta \Delta T)}{(1+2 \alpha \Delta T)}$

$\Delta h = h - h_o = h_o[\frac{( 1 + \beta \Delta T)}{(1+2 \alpha \Delta T)} -1]$ eq-1

here $h_o = \frac{1.4}{2} = 0.7m$ and $\Delta T = 34.3-20.5 = 13.8 ^\circ C$

by putting all the values in eq-1, we get

$\Delta h = h - h_o = 0.7[\frac{( 1 + 3.99 \times 10^{-5} \times 13.8)}{(1+2 \times 1.20 \times 10^{-5} \times 13.8)} -1]$

$\Delta h = 1.535 \times 10^{-4}m$ this is the change in the height of the liquid column.