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 S1 and S2 be the surface areas of a solid at temperatures t1 and t2 respectively, where t2 > t1.
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, S2 – S1 = S1β(t2 – t1)
or, S2 = S1{1 + β(t2 – t1)} ……….. (2)
If the initial temperature = 0 and the final temperature = t, we may write, St = S0{1 + βt} …… (3)
where S0 = 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.
∴ βF = \(\frac{5}{9}\)βC where βF = value of β in Fahrenheit scale, and βC = 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 V1 and V2 be the volumes of a solid at temperatures t1 and t2 respectively, where t2 > t1.
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, V2 – V1 = γV1(t2 – t1)
or, V2 = V1{1 + γ(t2 – t1)} …… (2)
When initial temperature = 0 and the final temperature = t, then Vt = V0{ 1 + γt}, where V0 = 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.
∴ γF = \(\frac{5}{9}\)γC
where γF = value of γ in Fahrenheit scale, and γC = 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 l0 at its initial temperature, So = \(l_0^2\)
Let the temperature of the plate be increased by t so that the length of each side changes to lt. If the coefficient of linear expansion of the material of the plate is α then, lt = 4(1 + αt).
The surface area of the plate now becomes,
St = \(l_t^2\) = {l0(1 + αt)}2
= \(l_0^2\)(1 + 2at + α2t2) = S0(1 + 2at) ……… (1)
[neglecting α2t2 as α \(\ll\) 1]
But from the definition of the coefficient of surface expansion β for the material of the plate, ;
St = S0(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 l0, at the initial temperature. So, the volume of the cube at that temperature, V0 = \(l_0^3\)
Let the temperature of the cube be increased by t, so that the length of each side changes to lt. If the coefficient of linear expansion of the material of the cube is α, then,
lt = l0(1 + αt)
The volume of the cube now becomes,
Vt = \(l_t^3\) = \(l_0^3\)(1 + αt)3 = V0(1 + 3αt + 3α2t2 + α3t3)
= V0(1 + 3αt) [neglecting α2t2 and α3t3 as α \(\ll\) 1] …….. (4)
But from the definition of the coefficient of volume expansion γ for the material of the cube,
Vt = V0(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 = l2 ……. (9)
and volume of the cube,
V = l3 …… (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 = S2 – S1 = βS1(t2 – t1)
Here, β = 2α = 2 × 18 × 10-6 × π × \(\left(\frac{8}{2}\right)^2\) × 50
Increase in temperature = t2 – t1 = 80 – 30 = 50°C
∴ Increase in surface area,
S2 – S1 = 2 × 18 × 10-6 × π × \(\left(\frac{8}{2}\right)^2\) cm2
= 36 × 10-6 × 16π × 50 = 0.0905 cm2.
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, V0 = 20 × 12 × 3 = 720 cm3, increase in temperature = t2 – t1 = 800 – 0 = 800°C.
γ = 3α = 3 × 0.16 × 10-4°C-1
Cubical expansion,
V800 – V0 = V0 × γ × (800 – 0)
= 720 × 3 × 0.16 × 10-4 × 800 = 27.65 cm3.
Example 3.
A lead bullet has a volume of 2.5 cm3 at 0°C. Its volume increases by 0.021 cm3 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, Vt = \(\frac{V_t-V_0}{V_0 t}\)
Given, Vt – V0 = 0.021 cm3, V0 = 2.5 cm3 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 cm3
Value of γ for aluminium = 3 × α = 3 × 23 × 10-6°C-1.
Hence, increase in volume,
Vt – V0 = V0γt = \(\frac{4}{3}\)π × 103 × 3 × 23 × 10-6 × 100
= 28.9 cm3.
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 = \(\frac{46-30}{1.24}\) = \(\frac{16}{1.24}\)cm3 = volume of the metal piece at 27°C(=V1).
Similarly. the volume of the metal piece at 42°C (=V2)
= \(\frac{46-30.5}{1.20}\) = \(\frac{15.5}{1.20}\) cm3
∴ 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.