WBCHSE Class 11 Physics Notes For Coefficient Of Surface Or Superficial Expansion

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Coefficient Of Surface Or Superficial Expansion

Superficial 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, we get, the coefficient of surface expansion,

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

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

or, \(S_2-S_1=S_1 \beta\left(t_2-t_1\right)\)

or, \(S_2=S_1\left\{1+\beta\left(t_2-t_1\right)\right\}\) ….(2)

If the initial temperature = 0 and the final temperature = t, we may write, St = S0 {1 + βt}……..(3)

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where S0 = surface area at zero temperature.

  1. The 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.
  2. Value of β does not depend on the unit of surface area,
  3. Value of β depends on the unit of temperature.

Unit of β is °C or °F-1. The change in temperature by 1°F = 5/9°C change in temperature.

∴ \(\beta_F=\frac{5}{9} \beta_C \text {, where } \beta_F\) = value of 0 in Fahrenheit scale, and 0C = value of 0 in Celsius scale.

Expansion Of Solid And Liquids – Relation Among The Three Coefficients of Expansion 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? a of brass = 18 x 10-6 °C-1.
Solution:

Increase in surface area = \(S_2-S_1=\beta S_1\left(t_2-t_1\right)\)

Here, \(\beta=2 \alpha=2 \times 18 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1} \text { and } S_1=\pi \times\left(\frac{8}{2}\right)^2 \mathrm{~cm}^2\)

Increase in temperature = t2– t1 = 80-30 = 50 °C

∴ Increase in surface area, \(S_2-S_1\)=\(2 \times 18 \times 10^{-6} \times \pi \times\left(\frac{8}{2}\right)^2 \times 50\)

= \(36 \times 10^{-6} \times 16 \pi \times 50=0.0905 \mathrm{~cm}^2 .\)

Example 2. A rectangular copper block measures 20 cm x 12 cm x 3 cm. What will be the change in volume of the block when it is heated from 0°C to 800°C? The coefficient of linear expansion of copper is 0. 16 x 10-4 °C-1.
Solution:

Initial volume of the block, V0 =20 X 12 X 3 = 720 cm³, increase in temperature = t2 – t1 = 800 – 0 = 800°C.

γ = 3α = 3 x 0.16 x 10-4 °C-1

Cubical expansion, \(V_{800}-V_0=V_0 \times \gamma \times(800-0)\)

= 720 x 3 x 0.16 x 10-4 x 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, the coefficient of volume expansion of lead, \(\gamma=\frac{V_t-V_0}{V_0 t}\)

Given, \(V_t-V_0=0.021 \mathrm{~cm}^3, V_0=2.5 \mathrm{~cm}^3 \text { and } t=98^{\circ} \mathrm{C}\)

∴ \(\gamma=\frac{0.021}{2.5 \times 98}=85.7 \times 10^{-6}{ }^{\circ} \mathrm{C}^{-1}\)

∴ Coefficient of linear expansion of lead \(\alpha=\frac{\gamma}{3}=\frac{8.57 \times 10^{-6}}{3}=2.86 \times 10^{-6 \circ} \mathrm{C}^{-1}\)

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 = 23x 10-6 °C-1.
Solution:

The initial volume of the aluminium sphere,

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

Value of γ for aluminium =3 x α =3x23x 10-6 °C-1.

Hence, increase in volume, \(V_t-V_0=V_0 \gamma t=\frac{4}{3} \pi \times 10^3 \times 3 \times 23 \times 10^{-6} \times 100\)

= 28.9 cm³.

Example 5. A piece of metal weighs 46 g xg in air. When immersed in a liquid of relative density 1.24, kept at 27°C, its weight is 30 g x g. When the temperature of the liquid is raised to 42°C, the metal piece in it weighs 30.5 g x 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 x g;

Thus the volume of the displaced liquid at 27 °C = \(\frac{46-30}{1.24}=\frac{16}{1.24} \mathrm{~cm}^3\) = volume of the metal piece at 27°C(= V1).

Similarly, the volume of the metal piece at 42 °C (= V2)

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

∴ Coefficient of volume expansion of the metal,

⇒ \(\gamma=\frac{V_2-V_1}{V_1\left(t_2-t_1\right)}=\frac{1}{\left(t_2-t_1\right)}\left(\frac{V_2}{V_1}-1\right)\)

= \(\frac{1}{42-27}\left(\frac{15.5}{1.2} \times \frac{1.24}{16}-1\right)=\frac{1}{15}\left(\frac{961}{960}-1\right)\)

= \(\approx 6.94 \times 10^{-5 \circ} \mathrm{C}^{-1}\)

∴ The coefficient of linear expansion of the metal piece \(\alpha=\frac{\gamma}{3}=\frac{6.94 \times 10^{-5}}{3}{ }^{\circ} \mathrm{C}^{-1}=23.15 \times 10^{-6{ }^{\circ}} \mathrm{C}^{-1}\)

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