\begin{equation} \frac{\Delta L}{L}=\alpha_{L}\Delta T \tag{2} \end{equation} or, \begin{equation} \Delta L=\alpha_{L}L\Delta T \tag{3} \end{equation} Constant $\alpha_L$ characterizes the thermal expansion properties of a particular material and it is known as

Similarly, we can define

$K^{-1}$ is the unit of these coefficients of expansions. These three coefficient are not strictly constant for a substance and there value is depends on temperature range in which they are being measured. As an example, figure below shows that coefficient of volume expansion increase with temperature and takes a constant value above $500K$

- Water exhibits an anomalous behaviour.
- The volume of the water decreases as temperature increases from 0 °C and 4 °C. The volume of a given amount of water increases as the temperature increases further of 4 °C
- So, water has a maximum density at 4 °C.
- This anomalous behaviour has the favourable effect for animals living in water. Since the density of the water is maximum at 4°C, water at the bottom of the lakes remains at 4 °C in winter even if the surface freezes. This allow marine life to survive. If water did not have this property, lakes and ponds would freeze from the bottom up, which would destroy much of their animal and plant life.

A blacksmith fixes iron ring on the rim of the wooden wheel of a bullock cart. The diameter of the rim and the iron ring are 5.243 m and 5.231 m respectively at 27 °C. To what temperature should the ring be heated so as to fit the rim of the wheel? ( $\alpha =1.20 \times 10^{-5} C^{-1} $)

Given,

$T_1 = 27 ^0$ C

$L_1 = 5.231$ m

$L_2 = 5.243$ m

So,

$L_2 =L_1 [1+ \alpha (T_2 - T_1)]$

$5.243 m = 5.231 m [1 + 1.20 \times 10^{-5} (T_2 - 27)]$

or $T_2 = 218 ^0$C.

An iron ring has to fit over a cylindrical rod. The diameter of the rod and iron ring are 6.445 cm and 6.420 cm at 20 ° C.The ring can slip over the iron rod only if the diameter of the ring is about .008 cm larger than iron rod.To what temperature should the ring be heated so as to fit the cylindrical rod? ( $\alpha =1.20 \times 10^{-5} C^{-1} $)

Given,

$T_1 = 20 ^0$ C

$L_1 = 6.420$ m

$L_2 = 6.445 + .008=6.453$ m

So,

$L_2 =L_1 [1+ \alpha (T_2 - T_1)]$

$6.453 m = 6.420 m [1 + 1.20 \times 10^{-5} (T_2 - 20)]$

or $T_2 = 450 ^0$C.

when a circular disc having hole is heated. if the hole size increase or decreases?

The hole size increases

When the lid of the glass jar is tight, we often put it a hot water and then it is easy to open. Why?

The metal expands on heating. You might say , glass also expands ,but metal expands more than glass for same temperature difference. So it becomes easy to open

**Notes****Assignments****Revision Notes**