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Thermal Properties of Matter - Thermal Expansion

Grade 11CBSEPhysics

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

🔑Concepts

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Thermal expansion is the increase in the dimensions of a body (length, area, or volume) due to an increase in its temperature. This occurs because the average distance between atoms increases with higher kinetic energy.

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Linear Expansion: The increase in length of a solid. The coefficient of linear expansion is denoted by α\alpha and is defined as the fractional change in length per degree change in temperature.

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Area (Superficial) Expansion: The increase in surface area of a solid. The coefficient of area expansion is denoted by β\beta. For isotropic solids, β=2α\beta = 2\alpha.

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Volume Expansion: The increase in volume of a solid, liquid, or gas. The coefficient of volume expansion is denoted by γ\gamma. For isotropic solids, γ=3α\gamma = 3\alpha.

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The relationship between the three coefficients of expansion for an isotropic solid is α=β2=γ3\alpha = \frac{\beta}{2} = \frac{\gamma}{3}, or α:β:γ=1:2:3\alpha : \beta : \gamma = 1 : 2 : 3.

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Anomalous Expansion of Water: Water exhibits a unique property where it contracts on heating between 0∘C0^{\circ}C and 4∘C4^{\circ}C. Its density is maximum at 4∘C4^{\circ}C (1000 kg/m31000 \text{ kg/m}^3).

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Thermal Stress: If a rod is fixed between two rigid supports and its temperature is changed, the supports prevent expansion/contraction, creating internal 'Thermal Stress'. The stress is given by YαΔTY \alpha \Delta T, where YY is Young's Modulus.

📐Formulae

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T

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

ΔA=βA0ΔT\Delta A = \beta A_0 \Delta T

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

ΔV=γV0ΔT\Delta V = \gamma V_0 \Delta T

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

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

Thermal Stress=FA=YαΔT\text{Thermal Stress} = \frac{F}{A} = Y \alpha \Delta T

Thermal Force=F=YAαΔT\text{Thermal Force} = F = Y A \alpha \Delta T

💡Examples

Problem 1:

A steel railway track has a length of 30 m30 \text{ m} at 20∘C20^{\circ}C. What will be its increase in length on a hot summer day when the temperature rises to 45∘C45^{\circ}C? (Given α\alpha for steel is 1.1×10−5 ∘C−11.1 \times 10^{-5} \text{ }^{\circ}C^{-1})

Solution:

Given: L0=30 mL_0 = 30 \text{ m}, T1=20∘CT_1 = 20^{\circ}C, T2=45∘CT_2 = 45^{\circ}C, α=1.1×10−5 ∘C−1\alpha = 1.1 \times 10^{-5} \text{ }^{\circ}C^{-1}. Change in temperature ΔT=45∘C−20∘C=25∘C\Delta T = 45^{\circ}C - 20^{\circ}C = 25^{\circ}C. Using the formula ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T: ΔL=(1.1×10−5)×30×25\Delta L = (1.1 \times 10^{-5}) \times 30 \times 25 ΔL=1.1×10−5×750\Delta L = 1.1 \times 10^{-5} \times 750 ΔL=8.25×10−3 m=8.25 mm\Delta L = 8.25 \times 10^{-3} \text{ m} = 8.25 \text{ mm}.

Explanation:

The change in length is calculated by multiplying the original length, the coefficient of linear expansion, and the temperature difference.

Problem 2:

A brass rod of length 50 cm50 \text{ cm} and diameter 3.0 mm3.0 \text{ mm} is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250∘C250^{\circ}C, if the original lengths are at 40∘C40^{\circ}C? (Take αbrass=2.0×10−5 ∘C−1\alpha_{brass} = 2.0 \times 10^{-5} \text{ }^{\circ}C^{-1} and αsteel=1.2×10−5 ∘C−1\alpha_{steel} = 1.2 \times 10^{-5} \text{ }^{\circ}C^{-1})

Solution:

Change in temperature ΔT=250∘C−40∘C=210∘C\Delta T = 250^{\circ}C - 40^{\circ}C = 210^{\circ}C. For Brass: ΔLB=αBLBΔT=(2.0×10−5)×50×210=0.21 cm\Delta L_B = \alpha_B L_B \Delta T = (2.0 \times 10^{-5}) \times 50 \times 210 = 0.21 \text{ cm}. For Steel: ΔLS=αSLSΔT=(1.2×10−5)×50×210=0.126 cm\Delta L_S = \alpha_S L_S \Delta T = (1.2 \times 10^{-5}) \times 50 \times 210 = 0.126 \text{ cm}. Total change in length ΔL=ΔLB+ΔLS=0.21+0.126=0.336 cm\Delta L = \Delta L_B + \Delta L_S = 0.21 + 0.126 = 0.336 \text{ cm}.

Explanation:

Since the rods are joined end-to-end, the total expansion is the sum of the individual expansions of the brass and steel rods.

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