Thermal Expansion: Class 11 Physics Notes, Definition, Working Principle, Formula & Real-Life Applications

Case ( i ): When object is expanded only ${\mathcal{l}}_2={\mathcal{l}}_1\left\{1+{\alpha }_0\left({\theta }_2-{\theta }_1\right)\right.$

${\mathcal{l}}_1=$ actual length of object at ${\theta }_1{}^{\circ }\mathrm{C}=$ measure length of object at ${\theta }_1{}^{\circ }\mathrm{C}$ .

${\mathcal{l}}_2=$ actual length of object at ${\theta }_2{}^{\circ }\mathrm{C}=$ measure length of object at ${\theta }_2{}^{\circ }\mathrm{C}$ .

${\alpha }_0=$ linear expansion coefficient of object.

Case (ii): When only measurement instrument is expanded actual length of object will not change but measured value (MV) decreases.

    $\begin{array}{rr}& \mathrm{M}\mathrm{V}={\mathcal{l}}_1\left\{1-{\alpha }_{\mathrm{S}}\left({\theta }_2-{\theta }_1\right)\right\}\\ & {\alpha }_{\mathrm{S}}=\text{ linear expansion coe }\\ & \text{ at }{\theta }_1{}^{\circ }\mathrm{C}\mathrm{M}\mathrm{V}=3\\ & \text{ at }{\theta }_2{}^{\circ }\mathrm{C}\mathrm{M}\mathrm{V}=2.2\end{array}$

${\alpha }_{\mathrm{S}}=\text{ linear expansion coefficient of measuring instrument. }$

Case (iii): If both expanded simultaneously MV $=\left\{1+\left({\alpha }_0-{\alpha }_s\right)\left({\theta }_2-{\theta }_1\right)\right.$
(i) If ${\alpha }_0>{\alpha }_{\mathrm{s}}$ , then measured value is more then the actual value at ${\theta }_1{}^{\circ }\mathrm{C}$
(ii) If ${\alpha }_0<{\alpha }_{\mathrm{s}}$ , then measured value is less then the actual value at ${\theta }_1{}^{\circ }\mathrm{C}$

$\text{ at }\begin{array}{ll}& {\theta }_1{}^{\circ }\mathrm{C}\mathrm{M}\mathrm{V}=3.4\\ & {\theta }_2{}^{\circ }\mathrm{C}\mathrm{M}\mathrm{V}=4.1\end{array}$

Measured value $=$ calibrated value $\times \{1+\alpha \mathrm{\Delta }\theta \}$
where $\alpha ={\alpha }_0-{\alpha }_{\text{s }}$
${\alpha }_0=$ coefficient of linear expansion of object material, ${\alpha }_S=$ coefficient of linear expansion of scale material

                                                    $\mathrm{\Delta }\theta =\theta -{\theta }_{\mathrm{C}}$  

$\theta =$ temperature at the time of measurement $\mathrm{}{\theta }_C=$ temperature at the time of calibration.
For scale, true measurement $=$ scale reading $\left[1+\alpha \left(\theta -{\theta }_0\right)\right.$ ]
If $\theta >{\theta }_0$ true measurement $>$ scale reading
$\theta <{\theta }_0$  true measurement < scale reading

When a solid is heated and its area increases, then the thermal expansion is called superficial or areal expansion. Consider a solid plate of area $A_0$ . When it is heated, the change in area of the plate is directly proportional to the original area $A_0$ and the change in temperature $\mathrm{\Delta }T$ .    

$\begin{array}{rr}& \mathrm{d}\mathrm{A}=\beta A_0\mathrm{d}\mathrm{T}\text{ or }\mathrm{}\mathrm{\Delta }A=\beta A_0\mathrm{\Delta }T\\ & \beta =\frac{\mathrm{\Delta }A}{A_0\mathrm{\Delta }T}\text{ Unit of }\beta \overline{\stackrel{¯}{\text{ is }}{}^{\circ }{C}^{-1}}\text{ or }{K}^{-1}.\\ & A=A_0(1+\beta \mathrm{\Delta }T)\end{array}$

where $A$ is area of the plate after heating,

When a solid is heated and its volume increases, then the expansion is called volume expansion or cubical expansion. Let us consider a solid or liquid whose original volume is $V_0$ . When it is heated to a new volume, then the change $\mathrm{\Delta }V$

$\begin{array}{rr}& \mathrm{d}\mathrm{V}=\gamma {\mathrm{V}}_0\mathrm{d}\mathrm{T}\mathrm{}\text{ or }\mathrm{}\mathrm{\Delta }\mathbf{V}=\gamma {\mathrm{V}}_0\mathrm{\Delta }\mathbf{T}\\ & \gamma =\text{ Unit of }\gamma \text{ is }{}^{\circ }{\mathrm{C}}^{-1}\text{ or }{\mathrm{K}}^{-1}.\\ & \mathbf{V}={\mathbf{V}}_0(1+\gamma \mathrm{\Delta }\mathbf{T})\end{array}$

where V is the volume of the body after heating

Relation between $\mathit{\alpha },\mathit{}\mathbf{\beta }$ and $\mathit{\gamma }$

(i)  For isotropic solids: $\alpha :\beta :\gamma =1:2:3$ or $\frac{\alpha }{1}$  = $\frac{\beta }{2}$ = $\frac{\gamma }{3}$

(ii) For non-isotropic solid $\beta ={\alpha }_1+{\alpha }_2$ and $\gamma ={\alpha }_1+{\alpha }_2+{\alpha }_3$ . Here ${\alpha }_1,{\alpha }_2$ and ${\alpha }_3$ are coefficient of linear expansion in $X,Y$ and $Z$ direction.

As we known that mass = volume × density.

Mass of substance does not change with change in temperature so with increase of temperature, volume increases so density decreases and vice-versa.

                  . $d=\frac{d_0}{(1+\gamma \Delta T)}$

For solids values of  are generally small so we can write $d=d_0(1-\gamma \mathrm{\Delta }T)$ (using binomial expansion).
Note: (i) $\gamma$ for liquids are in order of ${10}^{-3}$ .
(ii) Anamolous expansion of water:

For water density increases from $0^{\circ }\mathrm{C}$ to $4^{\circ }\mathrm{C}$ so $\gamma$ is negative and for $4^{\circ }\mathrm{C}$ to higher temperature $\gamma$ is positive. At $4^{\circ }\mathrm{C}$ density is maximum. This anamolous behaviour of water is due to presence of three types of molecules i.e. ${\mathrm{H}}_2\mathrm{O},{\left({\mathrm{H}}_2\mathrm{O}\right)}_2$ and ${\left({\mathrm{H}}_2\mathrm{O}\right)}_3$ having different volume/mass at different temperatures. This anomalous behaviour of water causes ice to form first at the surface of a lake in cold weather. As winter approaches, the water temperature decreases initially at the surface. The water there sinks because of its increase density. Consequently, the surface reaches $0^{\circ }\mathrm{C}$ first and the lake becomes covered with ice. Aquatic life is able to survive the cold winter as the lake bottom remains unfrozen at a temperature of about $4^{\circ }\mathrm{C}$ .

Initially container was full. When temperature change by $\mathrm{\Delta }T$ ,

Volume of liquid ${\mathrm{V}}_{\mathrm{L}}={\mathrm{V}}_0(1+\gamma \mathrm{L}\mathrm{\Delta }\mathrm{T})$

Volume of container ${\mathrm{V}}_{\mathrm{c}}={\mathrm{V}}_0\left(1+{\gamma }_c\mathrm{\Delta }\mathrm{T}\right)$

So overflow volume of liquid relative to container $\mathrm{\Delta }\mathrm{V}={\mathrm{V}}_{\mathrm{L}}-{\mathrm{V}}_{\mathrm{c}}$

$\mathrm{\Delta }\mathbf{V}={\mathbf{V}}_0\left({\gamma }_L-{\gamma }_C\right)\mathrm{\Delta }T$

So, coefficient of apparent expansion of liquid w.r.t. container

In case of expansion of liquid + container system:

if $\gamma \mathrm{L}>\gamma \overline{\stackrel{¯}{c}}⟶$ level of liquid rise

if ${\gamma }_{\mathrm{L}}<\gamma \overline{\stackrel{¯}{c}}⟶$ level of liquid fall

Increase in height of liquid level in tube when bulb was initially not completely filled

$\begin{array}{rr}& h=\frac{\text{ v}\text{olume}\text{of liquid }}{\text{ area of tube }}=\frac{V_0\left(1+{\gamma }_L\mathrm{\Delta }T\right)}{A_0\left(1+2{\alpha }_S\mathrm{\Delta }T\right)}=h_0\left\{1+\left(\gamma L-2{\alpha }_S\right)\mathrm{\Delta }T\right\}\\ & h=h_0\left\{1+\left({\gamma }_L-2{\alpha }_S\right)\mathrm{\Delta }T\right\}\end{array}$

where $h_0=$ original height of liquid in container
$\alpha s=$ linear coefficient of expansion of container.

If body is submerged completely inside the liquid

For solid, Buoyancy force

$\begin{array}{rr}& {\mathrm{F}}_{\mathrm{B}}={\mathrm{V}}_0\mathrm{d}\mathrm{g}\mathrm{g}\\ & {\mathrm{V}}_0=\mathrm{V}\text{ olume of the solid inside liquid, }\\ & {\mathrm{d}}_{\mathrm{L}}=\text{ density of liquid }\end{array}$

Volume of body after increase its temperature $\mathrm{V}={\mathrm{V}}_0[1+\gamma s\mathrm{\Delta }\theta ]$ ,

Density of body after increase its temperature $d^{\text{'}}L=\frac{d_L}{\left[1+{\gamma }_L\mathrm{\Delta }\theta \right]}$ .

Buoyancy force of body after increase its temperature, $F^{\text{'}}=V{d}^{\text{'}}Lg,\frac{F_B^{\text{'}}}{F_B}=\frac{\left[1+{\gamma }_S\mathrm{\Delta }\theta \right]}{\left[1+{\gamma }_L\mathrm{\Delta }\theta \right]}$ ,

if ${\gamma }_S<{\gamma }_{\bar{L}}$ then ${\mathrm{F}}^{\text{'}}<{\mathrm{F}}_{\mathrm{B}}$

(Buoyant force decreases) or

apparent weight of body in liquid gets increased $\left[W-F^{\text{'}}>W-F_B\right]$

It two strip of different metals are welded together to form a bimetallic strip, when heated uniformly it bends in form of an arc, the metal with greater coefficient of linear expansion lies on convex side. The radius of arc thus formed by bimetal is :

A bimetallic strip, consisting of a strip of brass and a strip of steel welded together, at temperature T₀ in figure (a) and figure (b). The strip bends as shown at temperatures above the reference temperature. Below the reference temperature the strip bends the other way. Many thermostats operate on this principle, making and breaking an electrical circuit as the temperature rises and falls.

(a) A small gap is left between two iron rails of the railway.

(b) Iron rings are slipped on the wooden wheels by heating the iron rings

(c) Stopper of a glass bottle jammed in its neck can be taken out by heating the neck.

(d) The pendulum of a clock is made of invar [an alloy of zinc and copper].

The following elements have an impact on thermal expansion:

• Strong atomic bonds prevent separation caused by heat. Consequently, there is little thermal expansion.
• Atomic vibrations vary in direction in crystals. Thermal expansion is hence directionally dependent.
• When a softer material is filled with rigid fibers, its growth is inhibited. 
• Expansion depends on the temperature range. Therefore, thermal expansion rates can change at different temperatures.
•  Liquids and gases expand more than solids because their forces are weaker. Accordingly, thermal expansion is much greater in fluids.
•  Fixed parts cannot lengthen freely. Consequently, thermal expansion causes internal stress instead of visible growth.
•  Manufacturing stresses change atomic spacing. As a result, thermal expansion deviates from simple predictions.
• Materials that absorb moisture or chemicals also change size. In turn, thermal expansion appears inaccurate.