CBSE Class 12 Physics Thermal Expansion Theory and Examples

CBSE Class 12 Physics Thermal Expansion Theory and Examples. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

 

Class 12 Physics: Thermal Expansion (Theory and Examples)

1. Introduction

Solids are made of atoms and molecules. At a given temperature, the atoms and molecules are placed at some equilibrium distance. When heat is supplied to a solid, the interatomic separation increases, by which there is an expansion of solids. This expansion can be in terms of length, area or volume.

From the above tree, it is clear that the thermal expansion of a solid is classified into three categories:

1. Linear expansion of solids
2. Superficial expansion of solids
3. Cubical expansion of solids

Now we shall discuss these topics one by one.

2. Linear expansion of solid

1. Almost all solids expand on heating. On increasing the temperature of a solid, its length increases. This change in length of a solid on heating is called linear expansion.
2. Coefficient of linear expansion is defined as the fractional increase in length per °C rise in temperature. If \(\ell\) is the length of the rod at \(T\) K and as the temperature is changed to \(T+\Delta T\) its length becomes \(\ell+\Delta\ell\), so the coefficient of linear expansion is given by \[ \alpha=\frac{\Delta\ell/\ell}{\Delta T}=\frac{d\ell}{\ell\cdot dT} \]
3. Unit of \(\alpha\) is per kelvin or per °C. It is positive for metals except carbon. The value of \(\alpha\) is negative for plastic because in plastic when the temperature increases, length decreases.
4. The numerical value of \(\alpha\) is the same in both the units, i.e. in per kelvin or per °C.
5. If \(\alpha\) is the coefficient of linear expansion at \(t_1\)°C, \(\ell_1\) is the length of the rod at \(t_1\)°C and \(\ell_2\) is the length of the rod at \(t_2\)°C, then \[ \ell_2=\ell_1\left[1+\alpha(t_2-t_1)\right] \] \(t\) may be in any unit, °C or K, because in the formula there is a difference of temperature, which remains the same for °C or K.
6. If the length of the rod is \(\ell_0\) at 0°C and \(\ell_t\) at \(t\)°C then \[ \ell_t=\ell_0\left[1+\alpha(t-0)\right]=\ell_0(1+\alpha t) \] where \(\alpha\) is the temperature coefficient of linear expansion at 0°C. Here \(t\) should be in °C only, because the initial temperature is taken as 0°C.

3. Application of linear expansion

3.1 Differential expansion of two solid rods

1. Suppose there are two rods of lengths \(\ell_1\) and \(\ell_2\). The first rod is kept on the other such that the initial separation between the free ends of the rods is \[ S=\ell_2-\ell_1 \] Both the rods are initially at a temperature of \(t_1\)°C.

On heating the entire system, the temperature increases to \(t_2\) such that the length of both the rods increases. So, if the new lengths of the rods are \(\ell_1^{\prime}\) and \(\ell_2^{\prime}\) then \[ \ell_1^{\prime}=\ell_1\left[1+\alpha_1(t_2-t_1)\right],\qquad \ell_2^{\prime}=\ell_2\left[1+\alpha_2(t_2-t_1)\right] \] Here \(\alpha_1\) and \(\alpha_2\) are the coefficients of linear expansion of the first and the second rod.

Now the separation between the free ends of the rods also changes such that \[ S^{\prime}=\ell_2^{\prime}-\ell_1^{\prime}=\ell_2\left[1+\alpha_2(t_2-t_1)\right]-\ell_1\left[1+\alpha_1(t_2-t_1)\right]=(\ell_2-\ell_1)+(\alpha_2\ell_2-\alpha_1\ell_1)(t_2-t_1) \]

If the new separation \(S^{\prime}\) of the composite rod is equal to the original separation, then \(S^{\prime}=S\), so \[ (\alpha_2\ell_2-\alpha_1\ell_1)(t_2-t_1)=0\ \Rightarrow\ \alpha_2\ell_2=\alpha_1\ell_1 \]

Differential expansion of two solid rods: examples

The length of the steel rod which would have the same difference in length with a copper rod of length 24 cm at all temperatures (\(\alpha_{copper}=18\times10^{-6}\ \text{K}^{-1}\), \(\alpha_{steel}=12\times10^{-6}\ \text{K}^{-1}\)) is

1. 20 cm
2. 18 cm
3. 24 cm
4. 36 cm

Answer: (D)

By linear expansion of solids, \(\Delta\ell=\ell\alpha\Delta T\). So \( \ell_{steel}\alpha_{steel}\Delta T=\ell_{copper}\alpha_{copper}\Delta T \), which gives \[ \ell_{steel}=\frac{\ell_{copper}\alpha_{copper}}{\alpha_{steel}}=\frac{24\times18\times10^{-6}}{12\times10^{-6}}=36\ \text{cm} \]

3.2 Radius of bimetallic strip

If two strips of different metals are welded together to form a bimetallic strip, when heated uniformly it bends in the form of an arc; the metal with the greater coefficient of linear expansion lies on the convex side. The radius of the arc thus formed by the bimetal is \[ R=\frac{d}{(\alpha_2-\alpha_1)(t_2-t_1)}\quad\text{or}\quad R=\frac{d}{(\alpha_2-\alpha_1)\,t} \]

Where \(t\) = temperature difference between the two ends, \(d\) = thickness of each strip, and \(\alpha_1\), \(\alpha_2\) = coefficients of linear expansion.

A metal sheet with a circular hole is heated. The hole

1. Gets larger
2. Gets smaller
3. Remains of the same size
4. Gets deformed

Answer: (A)

When a body is heated, the distance between any two points on it increases. So due to thermal expansion of solids, the hole gets larger.

Two identical rectangular strips, one of copper and the other of steel, are riveted together to form a bimetallic strip (\(\alpha_{copper}>\alpha_{steel}\)). On heating, this strip will

1. Remain straight
2. Bend with copper on convex side
3. Bend with steel on convex side
4. Get twisted

Answer: (B)

When a bimetallic strip of different metals is heated, the strip bends due to unequal linear expansion of the two metals. The strip will bend with the metal of greater \(\alpha\) on the convex side, i.e. the outer side.

3.3 Thermal stress

1. When a rod is heated or cooled, it expands or contracts. This is termed free expansion of the rod.

Actually no strain is developed, because on increasing the temperature the length of the rod increases, so at \(t_2\)°C, \(\ell_2\) becomes the natural length of the rod.

1. Now if the ends of the rod are rigidly fixed so as to prevent it from expansion or contraction, then stress is produced in the rod. By virtue of this thermal stress the rod exerts a large force on the supports.

The first figure indicates the rod kept at a temperature \(t_1\)°C having length \(\ell_1\). The rod is between two rigid supports. If the supports were not there then on increasing the temperature the free expansion occurs, by which the length becomes \(\ell_2\) at the higher temperature \(t_2\)°C. The third figure indicates that on increasing the temperature to \(t_2\)°C, the length remains the same but a compressive strain is induced in the rod.

If \(\ell_1\) is the length of the rod at \(t_1\)°C and on increasing the temperature to \(t_2\)°C its length becomes \(\ell_2\), then \[ \ell_2=\ell_1\left[1+\alpha(t_2-t_1)\right]=\ell_1+\ell_1\alpha(t_2-t_1)\ \Rightarrow\ \frac{\ell_2-\ell_1}{\ell_1}=\alpha(t_2-t_1) \] The above relationship indicates the thermal strain developed in the rod, i.e. \[ \text{Thermal strain}=\frac{\ell_2-\ell_1}{\ell_1}=\alpha(t_2-t_1) \]

In elasticity, Young's modulus = stress/strain, so \[ \text{Stress}=Y\alpha(t_2-t_1) \] As force = stress × area, \[ \text{Force}=YA\alpha(t_2-t_1) \]

1. If the rod is in its natural length at \(t_1\)°C while at \(t_2\)°C it is in a compressed state, then strain = change in length / original length.

In this case the first figure represents the rod of length \(\ell\) at \(t_1\)°C; when the temperature is lowered to \(t_2\)°C the length of the rod remains the same, but a tensile strain is developed in the rod. \[ \text{Strain}=\frac{(\ell+\ell\alpha\Delta t)-\ell}{\ell+\ell\alpha\Delta t}=\frac{\ell\alpha\Delta t}{\ell+\ell\alpha\Delta t}=\frac{\ell\,\Delta t}{\ell+\ell\,\Delta t}\simeq\alpha\,\Delta t \]

1. When the temperature of the rod is increased, a compressive stress is developed, while on decreasing the temperature of the rod a tensile stress is developed.

Thermal stress: examples

A steel rod of length 1 m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed?

Answer: Zero.

In this case the rod rests on a horizontal base, so it is free to expand on heating. Hence no strain is developed in the rod, i.e. strain = 0.

A steel rod of length 50 cm has a cross-sectional area of 0.4 cm². What force would be required to stretch this rod by the same amount as the expansion produced by heating it through 10°C? (\(\alpha=10^{-5}\ \text{K}^{-1}\) and \(Y=2\times10^{11}\ \text{N/m}^2\))

Answer: 800 N

\[ \text{Force}=YA\alpha\Delta\theta=2\times10^{11}\times0.4\times10^{-4}\times10^{-5}\times10=0.8\times10^3=800\ \text{N} \]

3.4 Effect of temperature on a pendulum clock

A pendulum clock consists of a metal rod or wire with the bob at one end. Let \(\ell_1\) be the length of the simple pendulum at \(\theta_1\)°C; then the time period \(T_1\) is given by \[ T_1=2\pi\sqrt{\frac{\ell_1}{g}}\qquad\ldots(1) \] Now when the temperature increases to \(\theta_2\)°C the effective length becomes \(\ell_2\) so that \[ T_2=2\pi\sqrt{\frac{\ell_2}{g}}\qquad\ldots(2) \] Dividing equation (2) by (1), \[ \frac{T_2}{T_1}=\sqrt{\frac{\ell_2}{\ell_1}} \] but \( \ell_2=\ell_1[1+\alpha(\theta_2-\theta_1)] \), so \[ \frac{T_2}{T_1}=\sqrt{\frac{\ell_1[1+\alpha(\theta_2-\theta_1)]}{\ell_1}}=\left[1+\alpha(\theta_2-\theta_1)\right]^{1/2}\approx1+\tfrac12\alpha(\theta_2-\theta_1) \] \[ \frac{T_2-T_1}{T_1}=\tfrac12\alpha(\theta_2-\theta_1) \] Change in time period \( \Delta T=T_2-T_1=\tfrac12\alpha\,\Delta\theta\,T_1 \).

The above expression represents the time lost per oscillation. Thus a pendulum clock loses time in summer and gains time in winter.

Note: if a pendulum clock is giving correct time when the time period is \(T\), then if \(T\) increases, the clock becomes slow; if \(T\) decreases, the clock becomes fast.

Pendulum clock: example

A pendulum clock with a pendulum made of invar (\(\alpha=0.7\times10^{-6}\ ^\circ\text{C}^{-1}\)) has a period of 0.5 s and is accurate at 25°C. If the clock is used in a country where the temperature averages 35°C, what correction is necessary at the end of a month (30 days) to the time given by the clock?

Answer: About 9.1 s (the clock becomes slow by this much).

In time interval \(t\), the clock will become slow by \[ \Delta t=\tfrac12\alpha\,t\,\Delta\theta=\tfrac12\times7\times10^{-7}\times30\times86400\times(35-25)=9.1\ \text{s} \]

4. Superficial expansion of solid

1. On increasing the temperature of a solid, its area increases. This change in area is referred to as superficial expansion of solids.
2. If \(A_0\) is the area of the solid at 0°C, on heating to \(t\)°C the area becomes \(A_t\) so that \( A_t=A_0[1+\beta t] \), where \(\beta\) is the coefficient of superficial expansion at 0°C and \(t\) should be in °C only.
3. If the area of the solid at temperature \(t_1\)°C is \(A_1\) and on heating the area becomes \(A_2\) at \(t_2\)°C, then \( A_2=A_1[1+\beta(t_2-t_1)] \), where \(\beta\) is the coefficient of superficial expansion at \(t_1\)°C.
4. Coefficient of superficial expansion is defined as the fractional increase in area per °C rise in temperature: \[ \beta=\frac{\Delta A/A}{\Delta T}=\frac{dA}{A\cdot dT} \]
5. Unit of \(\beta\) is per °C or per kelvin.

5. Volume expansion of solid

1. On increasing the temperature of a rod, its volume changes.
2. If \(V_0\) is the volume of the solid at 0°C and on increasing the temperature the volume becomes \(V_t\), then \( V_t=V_0[1+\gamma t] \), where \(\gamma\) is the coefficient of volume expansion at 0°C. Here also \(t\) should be in °C only.
3. If \(V_1\) is the volume of the solid at \(t_1\)°C and on increasing the temperature to \(t_2\)°C the volume becomes \(V_2\), then \( V_2=V_1[1+\gamma(t_2-t_1)] \), where \(\gamma\) is the coefficient of volume expansion at \(t_1\)°C.
4. Coefficient of volume expansion is defined as the fractional increase in volume per °C rise in temperature: \[ \gamma=\frac{\Delta V/V}{\Delta T}=\frac{dV}{V\cdot dT} \]
5. Unit of \(\gamma\) is per °C or per kelvin.

6. Relation between the coefficients of linear expansion (\(\alpha\)), superficial expansion (\(\beta\)) and cubical expansion (\(\gamma\))

\[ \beta=2\alpha,\qquad \gamma=3\alpha,\qquad \beta=\tfrac23\gamma \]

Superficial and volume expansion of solid: examples

The coefficient of linear expansion of a crystal in one direction is \(\alpha_1\) and that in every direction perpendicular to it is \(\alpha_2\). The coefficient of cubical expansion is

1. \(\alpha_1+\alpha_2\)
2. \(2\alpha_1+\alpha_2\)
3. \(\alpha_1+2\alpha_2\)
4. None of above

Answer: (C)

\( V=V_0(1+\gamma T) \), i.e. \( L^3=L_0(1+\alpha_1T)\,L_0^2(1+\alpha_2T)^2 \), or \( V=V_0(1+\alpha_1T)(1+\alpha_2T)^2 \). Hence \[ 1+\gamma T=(1+\alpha_1T)(1+\alpha_2T)^2\approx(1+\alpha_1T)(1+2\alpha_2T)=1+(\alpha_1+2\alpha_2)T \] Hence \( \gamma=\alpha_1+2\alpha_2 \).

Consider the following statements: (1) The coefficient of linear expansion has dimension K\(^{-1}\). (2) The coefficient of volume expansion has dimension K\(^{-1}\).

1. Both 1 and 2 are correct
2. 1 is correct but 2 is wrong
3. 2 is correct but 1 is wrong
4. 1 and 2 are both wrong

Answer: (A)

Coefficient of linear expansion is given by \( \alpha=\dfrac{d\ell}{\ell\,dt} \) and coefficient of volume expansion is given by \( \gamma=\dfrac{dV}{V\,dt} \). From the above formulae it is clear that both have units of per kelvin.

7. Change in density of solid with temperature

Suppose \(m\) is the mass of a solid which at a given temperature occupies a volume \(V\), so that the density at 0°C is \( d_0=\dfrac mV \). Now if the temperature is increased by \(t\)°C, the mass will remain unchanged but due to thermal expansion the volume increases so that \( V^{\prime}=V(1+\gamma t) \). Now the density is \[ d_t=\frac m{V^{\prime}}=\frac{m}{V(1+\gamma t)}=\frac{d_0}{1+\gamma t} \] Here \(\gamma\) is the coefficient of cubical expansion at 0°C.

 

 

 

 

 

is –
(A) 20 cm (B) 18 cm (C) 24 cm (D) 36 cm