Thermal Expansion

When a body is heated, it expands in terms of length, area & volume and temperature rises. In a solid, molecules can only have thermal agitation (random vibrations). As temperature of a body increases, the vibrations of molecules will become fast and due to this the rate of collision among neighbouring molecules increases, it develops a thermal stress in the body and due to this the intermolecular separation increases which results in thermal expansion of body. At ordinary temperatures, the atoms in a solid oscillate about their equilibrium position with an amplitude of approximately 10^–11 m. The average spacing between the atoms is about 10^–10 m. As the temperature of solid increases, the atoms oscillate with greater amplitudes, as a result the average separation between them increases. Consequently the object expands.

In the similar way the block diagram shown in Fig. explains the way how thermal expansion takes place.

Thermal expansion of a substance can be classified in three broad categories, these are (i) Linear Expansion (ii) Superficial Expansion and (iii) Cubical Expansion or Volume Expansion

(i) Linear Expansion: Consider a rod of length l₁ at a temperature T₁. Let it be a heated to a temperature T₂ and the increased length of the rod be l₂ then

$l_2=l_1\left(1+\alpha \Delta t\right)$

$\alpha$ = Coefficient of linear expansion and $\Delta \text{t}=T_2-T_1$

Illustration 1. The density of substance at 0°C is 10 g cm^-3 and at 100°C, its density is 9.7g cm^-3. The coefficient of linear expansion of substance is

Solution: $d_0=d_t(1+3\alpha t)$ \begin{align}  10=9.7(1+3\alpha \times 100) \\  or\left[ \dfrac{10}{9.7}-1 \right]\dfrac{1}{300}=\alpha \,\,or\,\,\alpha =\dfrac{0.3}{9.7\times 300}=0.0001/{}^\circ C \\ \end{align}

(ii) Superficial Expansion (expansion in surface area): If A₁ is the area of solid at T₁ °C and A₂ is the area at T₂°C . Then

A₂ = A₁(1 + $\beta$$\Delta$t )

$\beta$ Coefficient of superficial (areal) expansion and

$\Delta \text{t}=T_2-T_1$

(iii) Volume expansion (Part A: expansion in solids): If V₁ is the volume of solid at T₁°C and V₁ is the volume at T₂ °C then

$V_2=V_1\left(1+\gamma \Delta t\right)$

$\gamma$ coefficient of cubical (volume) expansion and $\Delta \text{t}=T_2-T_1$

Note: For isotopic solids: Relation between expansion coefficient are

$\beta =2\alpha ,\gamma =3\alpha \left[6\alpha =3,\beta =2\beta \right]$

As temperature increases, density of solid decreases. If d₁ is the density at T₁°C, d₂ is the density at T₂°C then

${\text{d}}_{\text{2}}\text{=}\frac{{\text{d}}_{\text{1}}}{\left(\text{1+}\gamma \Delta \text{t}\right)}\text{ or }{\text{d}}_{\text{2}}\text{=}{\text{d}}_{\text{1}}\left(\text{1-}\gamma \Delta \text{t}\right)\text{where}\Delta \text{t}=T_2-T_1$

(iv) Expansion of gases (Part B: expansion in gases):

Pressure coefficient of a gas is the ratio of increase in pressure for 1°C rise in temperature to the pressure at 0°C, provided the volume of the gas is kept constant.

$\gamma$_p =$\frac{p_t-p_o}{p_{o}t}$

Where $\gamma$_p = pressure coefficient

P_t­ = pressure at t°C

P_o= pressure at 0°C

Volume coefficient of a gas is similarly defined as (pressure being kept constant)

$\gamma$_v = $\frac{V_t-V_o}{V_{o}t}$

Where g_v = volume coefficient

V_t = volume at t°C

V_o= volume at 0°C

Experiments have shown the value of $\gamma$_p (or $\gamma$_v) is the same for all gases and equal to 1/273 per degree celsius, i.e.

$\gamma$_p = $\gamma$_v = $\frac{1}{273}$ per°C or $\frac{1}{273}$ per K,

Where K stands for absolute of kelvin temperature.

Illustration 2. A grid iron pendulum consists of 5 iron rods and 4 brass rods. What will be the length of each brass rod if the length of each iron rod is 1 m ? ($\alpha$_Fe= 12 x10^-6/°C, $\alpha$_brass = 20 x 10^-6/°C)

Solution: There are 5 iron rods and 4 brass rods in the pendulum. The number of iron in one half of the pendulum including the central rod is 3. The number of brass rods is 2. Since the pendulum is compensated,

\begin{align}  \therefore \Delta {{l}_{brass}}=\Delta {{l}_{iron}} \\  or2{{l}_{2}}{{\alpha }_{2}}t=3{{l}_{1}}{{\alpha }_{1}}t\,\,\,or\,\,{{l}_{2}}=\dfrac{3{{l}_{1}}{{\alpha }_{1}}}{2{{\alpha }_{2}}} \\ =\dfrac{3\times 1\times 12\times {{10}^{-6}}}{2\times 20\times{{10}^{-6}}}m=0.9\,m \\ \end{align}

ERROR in measurament DUE TO THERMAL EXPANSION

HEATING A METALLIC SCALE

A metallic scale (linear) expands in length when heated. As a result all the markings are displaced from their usual (correct) positions.

A reading of l unit on a heated scale is equivalent to an actual length of l x$\left(1+\alpha \Delta t\right)$.., where$\alpha$ is coefficient of linear expansion of material of scale, and $\Delta$t is the temperature of the heated scale.

$\Rightarrow$If the reading is x, actual length = x$\left(1+\alpha \Delta t\right)$

actual length = reading $\left(1+\alpha \Delta t\right)$

DIFFERENCE OF LENGTHS OF TWO RODS

Consider two rods 1 and 2 of lengths l₁ and l₂. Let they be heated through a temperature $\Delta t$. If l'₁ and l'₂ are their expanded lengths, then:

l'₂ = l₂ $\left(1+{\alpha }_2\Delta t\right)$ where ${\alpha }_2$ is coefficient of linear expansion of rod 2

l'₁ = l₁$\left(1+{\alpha }_1\Delta t\right)$ where ${\alpha }_1$ is coefficient of linear expansion of rod 1

Since the difference of length of two rods is constant

\begin{align}  \Rightarrow \,\,\,l{{'}_{2}}-l{{'}_{1}}_{{}}={{l}_{2}}-{{l}_{1}} \\  \Rightarrow \,\,\,\,{{l}_{1}}{{\alpha }_{1}}={{l}_{2}}{{\alpha }_{2}} \\ \end{align}

Illustration 3. Two rods of length l₁ and l₂ are made of materials whose coefficients of linear expansion are ${\alpha }_1$ and ${\alpha }_2$ respectively. If the difference between the two lengths is independent of temperature, then relation between length and linear expansion coefficients.

Solution : $\Delta l_1=l_1{\alpha }_1\Delta t$

\begin{align}  \Delta {{l}_{2}}={{l}_{2}}{{\alpha }_{2}}\Delta t \\  \,\,\,\,\,\,\,\Delta {{l}_{1}}-\Delta {{l}_{2}}=0 \\  \therefore {{l}_{1}}{{\alpha }_{1}}\Delta t={{l}_{2}}{{\alpha }_{2}}\Delta t\,\,or\,\,\dfrac{{{l}_{1}}}{{{l}_{2}}}=\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}} \\ \end{align}

TIME PERIOD OF PENDULUM

Time period (T) of a sample pendulum of length l is given by

$T=2\pi \sqrt{\frac{I}{g}}$

If there is a rise in temperature by $\Delta$t, length of the pendulum increases and hence the time period increase. As a result the clock slows down.

If l₀ be the length of the pendulum and corresponding time period be T₀ then

$T_0=2\pi \sqrt{\frac{l_0}{g}}$

If the pendulum be heated by $\Delta$t (rise in temperature), the new time period T₁ is:

$T_1=2\pi \sqrt{\frac{l_1}{g}}$

$\Rightarrow \frac{T_1}{T_0}=\sqrt{\frac{l_1}{l_0}}=\sqrt{\frac{l_0\left(1+\alpha \Delta t\right)}{l_n}}$

$\Rightarrow$ $\frac{T_1}{T_0}=\sqrt{1+\alpha \Delta t}=1+\frac{1}{2}\alpha \Delta t$ as a is very small

$\Rightarrow$ $\frac{T_1-T_0}{T_0}=\frac{1}{2}\alpha \Delta t\Rightarrow \frac{\Delta T}{T_0}=\frac{1}{2}\alpha \Delta t$

The above relation gives the time lost per second by the pendulum clock. If $\Rightarrow$t is the fall in temperature, same equation will give the time gained by the clock per second as its oscillation will become faster due to reduction in its length.

Illustration 4: A clock with a metallic pendulum is 5 second fast each day at a temperature of $15^0$ and 10 seconds slow each day at a temperature of $30^0$. Find coefficient of linear expansion for the metal.

Solution: Time lost or gained per second by a pendulum clock is given by

$\delta t=\frac{1}{2}\alpha \Delta T$ ( Here $\Delta T$ is difference of temperature)

Here temperature is higher then graduation temperature thus clock will loose time and if it is lower then graduation temperature will gain time.

$\delta t=\frac{1}{2}\alpha \Delta T\times 86400[as1day=86400\sec .]$

If graduation temperature of clock is T₀ then we have.

At 15°C, clock is gaining time, thus

5 = $\frac{1}{5}\alpha \left(T_0-15\right)\times 86400$ ……(1)

At 30°C clock is loosing time, thus

10 = $\frac{1}{5}\alpha \left(30-T_0\right)\times 86400$ ……(2)

Dividing equation (2) by (1), we get

2(T₀ – 15) = (30 – T₀)

or T₀ = 20°C

Thus from equation (1)

\begin{align}  \text{5 = }\dfrac{1}{2}\times \alpha \times \left[ 20-15 \right]\times 86400 \\  \alpha \,=\,2.31\times {{10}^{-5}}^{\circ }C \\ \end{align}

STRESS IN OBJECTS DUE TO THERMAL EXPANSION

If cross–sectional area of wire is A and F be the tension developed in wire due to stretching. Thus the stress developed in the wire due to this tension F is given as

Strees = $\frac{Force}{Area}$= F / A …..(1)

Strain produced in wire due to its elastic properties is

Strees = $\frac{\Delta L}{L}=\alpha \Delta T$ …..(2)

If young's modulus of the material of wire is Y, we have

$Y=\frac{strees}{strain}$

or $Y=\frac{F/A}{\alpha \Delta T}$

or $F=YA\alpha \Delta T$ …..(3)

Equation - (3) gives the expression for tension in the wire due to decrease in its temperature by $\Delta$T. This result gives the tension in wire if initially wire is just taught between clamps. If it already has some tension in it then this expression will give the increment in tension in the wire.

Illustration 5. A uniform metal rod of 2 m m² cross-section is heated from 0°C to 20°C. The coefficient of liner expansion of the rod is 12 x 10^-6 per °C, Y=10¹¹ N/m². The energy stored per unit volume of the rod is

solution: Energy per unit volume

\begin{align}  =\dfrac{1}{2}\times stress\times strain=\dfrac{1}{2}(Y\alpha t)(\alpha t)=\dfrac{1}{2}Y{{\alpha }^{2}}{{t}^{2}} \\  =\dfrac{{{10}^{11}}\times 144\times {{10}^{-12}}\times 400}{2}J\,\,{{m}^{-3}} \\  =288\times 10\,J\,\,{{m}^{-3}}=2880\,J\,\,{{m}^{-3}} \\ \end{align}