Thermal Expansion

When matter is heated without change in state, it usually expands. According to atomic theory of matter, asymmetry in potential energy curve is responsible for thermal expansion as with rise in temperature say from \(T_{1}\) to \(T_{2}\) the amplitude of vibration and hence energy of atoms increases from \(E_{1}\) to \(E_{2}\) and hence the average distance between them from \(\mathrm{r}_{1}\) to \(\mathrm{r}_{2}\).

Due to this increase in distance between atoms, the matter as a whole expands. Had the potential energy curve been symmetrical, no thermal expansion would have taken place in spite of heating.

Normal Solids

To varying extents, most materials expand when heated and contract when cooled. The increase in any one dimension of a solid is called linear expansion, linear in the sense that the expansion occurs along a line. A rod whose length is \(\mathrm{L}_{0}\) when the temperature is \(\mathrm{T}_{0}\) when the temperature increases to \(\mathrm{T}_{0}+\Delta \mathrm{T}\), the length becomes \(\mathrm{L}_{0}+\Delta \mathrm{L}\), where \(\Delta \mathrm{T}\) and \(\Delta \mathrm{L}\) are the magnitudes of the changes in temperature and length, respectively.

Conversely, when the temperature decreases to \(\mathrm{T}_{0}-\Delta \mathrm{T}\), the length decreases to \(\mathrm{L}_{0}-\Delta \mathrm{L}\). For small temperature changes, experiments show that the change in length is directly proportional to the change in temperature \((\Delta \mathrm{L} \propto \Delta \mathrm{T})\). In addition, the change in length is proportional to the initial length of the rod,

THERMAL EXPANSION, CALORIMETRY  & HEAT TRANSFER

Equation \(\Delta \mathrm{L}=\alpha \mathrm{L}_{0} \Delta \mathrm{T}\) expresses the fact that \(\Delta \mathrm{L}\) is proportional to both \(\mathrm{L}_{0}\) and \(\Delta \mathrm{T}\left(\Delta \mathrm{L} \propto \mathrm{L}_{0} \Delta \mathrm{T}\right)\) by using a proportionality constant \(\alpha\), which is called the coefficient of linear expansion.

Common unit for the coefficient of linear expansion : \(\frac{1}{\mathrm{C}^{\circ}}=\left(\mathrm{C}^{\circ}\right)^{-1}\)

Illustration :

A brass scale correctly calibrated at \(15^{\circ} \mathrm{C}\) is employed to measure a length at a temperature of \(35^{\circ} \mathrm{C}\). If the scale gives a reading of \(75 \mathrm{~cm}\), find the true length. (Linear expansively of brass \(\left.\left.=2.0 \times 10^{-5} C^{\circ}\right)^{-1}\right)\)

Sol. Let the distance between two fixed divisions on the scale at \(15^{\circ} \mathrm{C}\) be \(L_{1}\) and that at \(35^{\circ} \mathrm{C}\) be \(L_{2}\). Clearly, \(\quad\left(L_{2}-L_{1}\right)=\alpha L_{1}(35-15)\)

or \(L_{2}=L_{1}\left(1+20 \times 2.0 \times 10^{-5}\right)\)

\(=L_{1}(1.0004)\)

i.e., at \(35^{\circ} \mathrm{C}\), an actual length of \(L_{2}\) will be read as \(L_{1}\) (note), due to the increased separation of the divisions of the scale. In other words, the observed length will be less than the actual length.

Given : \(\quad L_{1}=75 \mathrm{~cm}\)

\(\therefore \quad L_{2}=75(1.0004) \mathrm{cm}\)

\(=75.03 \mathrm{~cm}\)

Illustration :

Estimate the time lost or gained by a pendulum clock at the end of a week when the atmospheric temperature rises to \(40^{\circ} \mathrm{C}\). The clock is known to give correct time at \(15^{\circ} \mathrm{C}\) and the pendulum is of steel. (Linear expansively of steel is \(12 \times 10^{-6} /{ }^{\circ} \mathrm{C}\) ).

Sol. Rise in temperature \(=40-15=25^{\circ} \mathrm{C}\)

\(\mathrm{T}=2 \pi \sqrt{\frac{\ell}{\mathrm{y}}} \therefore \frac{\Delta \mathrm{T}}{\mathrm{T}}=\frac{1}{2} \frac{\Delta \ell}{\ell}=\frac{1}{2} \alpha \Delta \theta\)

Time lost per second \(=\frac{1}{2} \alpha \Delta \theta\)

\(=\frac{1}{2} \times\left(12 \times 10^{-6} /{ }^{\circ} \mathrm{C}\right) \times\left(25^{\circ} \mathrm{C}\right)\)

\(=150 \times 10^{-6} \mathrm{~s} / \mathrm{s}\)

Therefore, time lost per week (i.e., \(7 \times 86400 \mathrm{~s}\) )

\(=150 \times 10^{-6} \mathrm{~s} / \mathrm{s} \times 7 \times 86400 \mathrm{~s}\)

\(=90.72 \mathrm{~s}\)

Illustration :

A glass rod when measured with a zinc scale, both being at \(30^{\circ} \mathrm{C}\), appears to be of length \(100 \mathrm{~cm}\). If the scale shows correct reading at \(0^{\circ} \mathrm{C}\), determine the true length of the glass rod at (a) \(30^{\circ} \mathrm{C}\) and (b) \(0^{\circ} \mathrm{C}\). ( ' \(\alpha\) ' for glass \(=8 \times 10^{-6} /{ }^{\circ} \mathrm{C}\) and for zinc \(26 \times 10^{-6} /{ }^{\circ} \mathrm{C}\) )

Sol. At \(30^{\circ} \mathrm{C}\), although the reading shown by the zinc scale corresponding to the length of the glass rod is \(100 \mathrm{~cm}\), but the actual length would be more than \(100 \mathrm{~cm}\), the reason being the increased separation between the markings, owing to a rise in temperature (from \(0^{\circ} \mathrm{C}\) to \(30^{\circ} \mathrm{C}\) ).

Now, an actual (original at \(0^{\circ} \mathrm{C}\) ) length of \(100 \mathrm{~cm}\) on the zinc scale (or more precisely, two markings or divisions on the scale, separated by a distance of \(100 \mathrm{~cm}\) ) would, at a temperature of \(30^{\circ} \mathrm{C}\), correspond to a length given by 

\( l=100\left(1+26 \times 10^{-6} \times 30\right) \mathrm{cm} \)

\( =100.078 \mathrm{~cm}\)

\(\therefore\) The true length of the glass rod at \(30^{\circ} \mathrm{C}\) is \(100.078 \mathrm{~cm}\).

Now, at \(0^{\circ} \mathrm{C}\), the length of glass rod would be lesser than that at \(30^{\circ} \mathrm{C}\),

\(\therefore \quad\) Using \(l_{t}=l_{0}(1+\alpha t), l_{0}=\frac{l_{\mathrm{t}}}{1+\alpha \mathrm{t}}\)

\(\therefore \quad\) The length of the rod at \(0^{\circ} \mathrm{C}\), will be

\(l_{0}=\frac{100.078 \mathrm{~cm}}{\left(1+8 \times 10^{-6} \times 30\right)}=100.054 \mathrm{~cm}\)