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Thermodynamics - Thermal Expansion

Grade 11ICSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Thermal expansion is the increase in the dimensions (length, area, or volume) of a body due to an increase in its temperature, caused by the increased amplitude of atomic vibrations.

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Coefficient of Linear Expansion (α\alpha): Defined as the fractional change in length per unit change in temperature: α=ΔLL0ΔT\alpha = \frac{\Delta L}{L_0 \Delta T}. Its unit is K−1K^{-1} or ∘C−1^{\circ}C^{-1}.

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Coefficient of Superficial Expansion (β\beta): Defined as the fractional change in surface area per unit change in temperature: β=ΔAA0ΔT\beta = \frac{\Delta A}{A_0 \Delta T}.

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Coefficient of Volume Expansion (γ\gamma): Defined as the fractional change in volume per unit change in temperature: γ=ΔVV0ΔT\gamma = \frac{\Delta V}{V_0 \Delta T}.

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For isotropic solids, the relationship between the coefficients is given by α:β:γ=1:2:3\alpha : \beta : \gamma = 1 : 2 : 3. Specifically, β=2α\beta = 2\alpha and γ=3α\gamma = 3\alpha.

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Variation of density with temperature: As temperature increases, volume increases, causing density (ρ\rho) to decrease. The relation is ρt=ρ01+γΔT\rho_t = \frac{\rho_0}{1 + \gamma \Delta T}.

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Anomalous expansion of water: Water contracts when heated from 0∘C0^{\circ}C to 4∘C4^{\circ}C and expands when heated above 4∘C4^{\circ}C. Water has its maximum density at 4∘C4^{\circ}C (1000 kg/m31000 \, kg/m^3).

📐Formulae

ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T

Lt=L0(1+αΔT)L_t = L_0 (1 + \alpha \Delta T)

At=A0(1+βΔT)A_t = A_0 (1 + \beta \Delta T)

Vt=V0(1+γΔT)V_t = V_0 (1 + \gamma \Delta T)

α=β2=γ3\alpha = \frac{\beta}{2} = \frac{\gamma}{3}

ρt≈ρ0(1−γΔT)\rho_t \approx \rho_0 (1 - \gamma \Delta T)

💡Examples

Problem 1:

A steel railway track is 20 m20 \, m long at 20∘C20^{\circ}C. If the temperature rises to 50∘C50^{\circ}C, find the increase in its length. (Given αsteel=1.2×10−5 ∘C−1\alpha_{steel} = 1.2 \times 10^{-5} \, ^{\circ}C^{-1})

Solution:

Given: L0=20 mL_0 = 20 \, m, T1=20∘CT_1 = 20^{\circ}C, T2=50∘CT_2 = 50^{\circ}C, α=1.2×10−5 ∘C−1\alpha = 1.2 \times 10^{-5} \, ^{\circ}C^{-1}.
Change in temperature ΔT=50−20=30∘C\Delta T = 50 - 20 = 30^{\circ}C.
Increase in length ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T.
ΔL=20×1.2×10−5×30=0.0072 m=7.2 mm\Delta L = 20 \times 1.2 \times 10^{-5} \times 30 = 0.0072 \, m = 7.2 \, mm.

Explanation:

The change in length is directly proportional to the original length and the rise in temperature, governed by the coefficient of linear expansion.

Problem 2:

Find the change in volume of an aluminum sphere of radius 10 cm10 \, cm when it is heated from 0∘C0^{\circ}C to 100∘C100^{\circ}C. (Given αAl=2.3×10−5 ∘C−1\alpha_{Al} = 2.3 \times 10^{-5} \, ^{\circ}C^{-1})

Solution:

Initial volume V0=43πr3=43×π×(10 cm)3≈4188.79 cm3V_0 = \frac{4}{3} \pi r^3 = \frac{4}{3} \times \pi \times (10 \, cm)^3 \approx 4188.79 \, cm^3.
γ=3α=3×2.3×10−5=6.9×10−5 ∘C−1\gamma = 3\alpha = 3 \times 2.3 \times 10^{-5} = 6.9 \times 10^{-5} \, ^{\circ}C^{-1}.
ΔT=100−0=100∘C\Delta T = 100 - 0 = 100^{\circ}C.
ΔV=V0γΔT=4188.79×6.9×10−5×100≈28.90 cm3\Delta V = V_0 \gamma \Delta T = 4188.79 \times 6.9 \times 10^{-5} \times 100 \approx 28.90 \, cm^3.

Explanation:

Since the sphere is a 3D object, we use the coefficient of volume expansion γ\gamma, which is thrice the coefficient of linear expansion α\alpha.

Thermal Expansion - Revision Notes & Key Formulas | ICSE Class 11 Physics