Contents

Physics Topics can help us understand the behavior of the natural world around us.

## Define the Coefficient of Surface Expansion

Definition: The increase in surface area for a unit rise in temperature for a unit surface area of a solid is called the coefficient of surface expansion of the material of that solid.

Let S_{1} and S_{2} be the surface areas of a solid at temperatures t_{1} and t_{2} respectively, where t_{2} > t_{1}.

Proceeding in a way similar to that in section 5.2, we get, the coefficient of surface expansion,

β = \(\frac{S_2-S_1}{S_1\left(t_2-t_1\right)}\)

\(=\frac{\text { increase in area }}{\text { initial area } \times \text { rise in temperature }}\) ……… (1)

or, S_{2} – S_{1} = S_{1}β(t_{2} – t_{1})

or, S_{2} = S_{1}{1 + β(t_{2} – t_{1})} ……….. (2)

If the initial temperature = 0 and the final temperature = t, we may write, S_{t} = S_{0}{1 + βt} …… (3)

where S_{0} = surface area at zero temperature.

i) Coefficient of surface expansion β is not a constant.

For precise measurements of β, the surface area at 0°C is to be taken as the initial surface area.

ii) Value of β does not depend on the unit of surface area.

iii) Value of β depends on the unit of temperature. Unit of β is °C^{-1} or °F^{-1}. The change in temperature by 1°F = \(\frac{5}{9}\)°C change in temperature.

∴ β_{F} = \(\frac{5}{9}\)β_{C} where β_{F} = value of β in Fahrenheit scale, and β_{C} = value of β in Celsius scale.

### Coefficient of Volume Expansion

Definition: The increase in volume for unit rise in temperature for a unit volume of a solid is called the coefficient of volume expansion of the material of that solid.

The coefficient of volume expansion is denoted by γ.

Let V_{1} and V_{2} be the volumes of a solid at temperatures t_{1} and t_{2} respectively, where t_{2} > t_{1}.

Proceeding in a way similar to that in Section 5.2, we get, the coefficient of volume expansion,

γ = \(\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}\)

\(=\frac{\text { increase in volume }}{\text { initial volume } \times \text { rise in temperature }}\) ……… (1)

Hence, V_{2} – V_{1} = γV_{1}(t_{2} – t_{1})

or, V_{2} = V_{1}{1 + γ(t_{2} – t_{1})} …… (2)

When initial temperature = 0 and the final temperature = t, then V_{t} = V_{0}{ 1 + γt}, where V_{0} = volume at zero temperature.

i) Coefficient of volume expansion γ is not a constant.

For precise measurements, the volume at 0°C is to be taken as the initial volume.

ii) Value of γ does not depend on the unit of volume.

iii) Value of γ depends on the unit of temperature. Unit of γ is °C^{-1} or °F^{-1}. The change in temperature by 1°F = \(\frac{5}{9}\)°C change in temperature.

∴ γ_{F} = \(\frac{5}{9}\)γ_{C}

where γ_{F} = value of γ in Fahrenheit scale, and γ_{C} = value of γ in Celsius scale.

### Relation Among The Three Coefficients of Expansion

Relation between α and β: Consider a square metal plate with each side of length l_{0} at its initial temperature, S_{o} = \(l_0^2\)

Let the temperature of the plate be increased by t so that the length of each side changes to l_{t}. If the coefficient of linear expansion of the material of the plate is α then, l_{t }= 4(1 + αt).

The surface area of the plate now becomes,

S_{t} = \(l_t^2\) = {l_{0}(1 + αt)}^{2}

= \(l_0^2\)(1 + 2at + α^{2}t^{2}) = S_{0}(1 + 2at) ……… (1)

[neglecting α^{2}t^{2} as α \(\ll\) 1]

But from the definition of the coefficient of surface expansion β for the material of the plate, ;

S_{t} = S_{0}(1 + βt) ……. (2)

∴ Comparing equation (1) and (2), we get,

β = 2α ….. (3)

### Relation between α and γ :

Let us consider, a metal cube with each side of length l_{0}, at the initial temperature. So, the volume of the cube at that temperature, V_{0} = \(l_0^3\)

Let the temperature of the cube be increased by t, so that the length of each side changes to l_{t}. If the coefficient of linear expansion of the material of the cube is α, then,

l_{t} = l_{0}(1 + αt)

The volume of the cube now becomes,

V_{t} = \(l_t^3\) = \(l_0^3\)(1 + αt)^{3} = V_{0}(1 + 3αt + 3α^{2}t^{2} + α^{3}t^{3})

= V_{0}(1 + 3αt) [neglecting α^{2}t^{2} and α^{3}t^{3} as α \(\ll\) 1] …….. (4)

But from the definition of the coefficient of volume expansion γ for the material of the cube,

V_{t} = V_{0}(1 + γt) ….. (5)

∴ From equations (4) and (5),

γ = 3α ……. (6)

So, for the same material,

α = \(\frac{\beta}{2}\) = \(\frac{\gamma}{3}\) …… (7)

This is the relation among α, β and γ. The relation can also be expressed as

α : β : γ = 1 : 2 : 3

To find the relation among α, β and γ using calculus: Suppose the length of a solid changes from l to (l + dl) when it is heated from a temperature θ to a temperature (θ + dθ). From the definition of coefficient of linear expansion,

α \(=\frac{\text { increase in length }}{\text { initial length } \times \text { rise in temperature }}\) = \(\frac{1}{l} \cdot \frac{d l}{d \theta}\)

Similarly, if initial surface area and initial volume are S and V respectively,

coefficient of surface expansion, β = \(\frac{1}{S} \cdot \frac{d S}{d \theta}\)

and coefficient of volume expansion, γ = \(\frac{1}{V} \cdot \frac{d V}{d \theta}\)

Let us consider a cube of length l at temperature θ. Therefore, area of each surface of the cube,

S = l^{2} ……. (9)

and volume of the cube,

V = l^{3} …… (10)

Differentiating equation (9) with respect to θ, we get

Hence, from equation (11) and (12)

α = \(\frac{\beta}{2}\) = \(\frac{\gamma}{3}\) …… (13)

### Numerical Examples

**Example 1.**

At 30°C the diameter of a brass disc is 8 cm. What will be the Increase in surface area if it is heated to 80°C? α of brass = 18 × 10^{-6}°C^{-1}.

**Solution:**

Increase in surface area = S_{2} – S_{1} = βS_{1}(t_{2} – t_{1})

Here, β = 2α = 2 × 18 × 10^{-6} × π × \(\left(\frac{8}{2}\right)^2\) × 50

Increase in temperature = t_{2} – t_{1} = 80 – 30 = 50°C

∴ Increase in surface area,

S_{2} – S_{1} = 2 × 18 × 10^{-6} × π × \(\left(\frac{8}{2}\right)^2\) cm^{2}

= 36 × 10^{-6} × 16π × 50 = 0.0905 cm^{2}.

**Example 2.**

A rectangular copper block measures 20cm × 12cm × 3cm. What will be the change in volume of the block when it Is heated from 0°C to 800°C? Coefficient of linear expansion of copper is 0.16 × 10^{-4}°C^{-1}.

**Solution:**

Initial volume of the block, V_{0} = 20 × 12 × 3 = 720 cm^{3}, increase in temperature = t_{2} – t_{1} = 800 – 0 = 800°C.

γ = 3α = 3 × 0.16 × 10^{-4°}C^{-1}

Cubical expansion,

V_{800} – V_{0} = V_{0} × γ × (800 – 0)

= 720 × 3 × 0.16 × 10^{-4} × 800 = 27.65 cm^{3}.

**Example 3.**

A lead bullet has a volume of 2.5 cm^{3} at 0°C. Its volume increases by 0.021 cm^{3} when heated to 98°C. Find the coefficient of linear expansion of lead.

**Solution:**

By definition, coefficient of volume expansion of lead, γ = \(\frac{V_t-V_0}{V_0 t}\)

Given, V_{t} = \(\frac{V_t-V_0}{V_0 t}\)

Given, V_{t} – V_{0} = 0.021 cm^{3}, V_{0} = 2.5 cm^{3} and t = 98°C

∴ γ = \(\frac{0.021}{2.5 \times 98}\) = 85.7 × 10^{-6}°C^{-1}

∴ Coefficient of linear expansion of lead,

α = \(\frac{\gamma}{3}\) = \(\frac{8.57 \times 10^{-6}}{3}\) = 2.86 × 10^{-6}°C^{-1}

**Example 4.**

An aluminium sphere of diameter 20 cm is heated from 0°C to 100°C. What Will be its change in volume? Coefficient of linear expansion of aluminium = 23 × 10^{-6}°C^{-1}.

**Solution:**

Initial volume of the aluminium sphere,

= \(\frac{4}{3} \pi\left(\frac{20}{2}\right)^3\) = \(\frac{4}{3}\)π(10)^{3} cm^{3}

Value of γ for aluminium = 3 × α = 3 × 23 × 10^{-6}°C^{-1}.

Hence, increase in volume,

V_{t} – V_{0} = V_{0}γt = \(\frac{4}{3}\)π × 10^{3} × 3 × 23 × 10^{-6} × 100

= 28.9 cm^{3}.

**Example 5.**

A piece of metal weighs 46g × g in air. When immersed in a liquid of relative density 1.24, kept at 27°C, its weight is 30 g × g. When the temperature of the liquid is raised to 42°C, the metal piece in it weighs 30.5 g × g. At 42°C, the relative density of the liquid is 1.20. Find the coefficient of linear expansion of the metal.

**Solution:**

The apparent loss in weight of the metal at 27°C = weight of an equal volume of the liquid = (46 – 30) g × g;

Thus the volume of the displaced liquid at 27°C = \(\frac{46-30}{1.24}\) = \(\frac{16}{1.24}\)cm^{3} = volume of the metal piece at 27°C(=V_{1}).

Similarly. the volume of the metal piece at 42°C (=V_{2})

= \(\frac{46-30.5}{1.20}\) = \(\frac{15.5}{1.20}\) cm^{3}

∴ Coefficient of volume expansion of the metal,

∴ The coefficient of linear expansion of the metal piece,

α = \(\frac{\gamma}{3}\) = \(\frac{6.94 \times 10^{-5}}{3}\)°C^{-1} = 23.15 × 10^{-6}°C^{-1}.