ThermodynamicsTopic #18 of 35

Thermal Expansion

Linear and volumetric expansion of solids, liquids, and gases.

Overview

Most materials expand when heated and contract when cooled. This occurs because increased temperature causes atoms to vibrate with greater amplitude, increasing the average distance between them.

Linear Expansion

For solids, the change in length is proportional to original length and temperature change:

ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T L=L0(1+αΔT)L = L_0(1 + \alpha \Delta T)

Where:

  • ΔL\Delta L = change in length
  • L0L_0 = original length
  • α\alpha = coefficient of linear expansion (1/K or 1/°C)
  • ΔT\Delta T = temperature change

Coefficients of Linear Expansion

Materialα\alpha (×10⁻⁶ /°C)
Aluminum24
Brass19
Copper17
Glass (ordinary)9
Glass (Pyrex)3.2
Iron/Steel12
Lead29
Concrete12
Invar0.9

Area Expansion

For a flat surface:

ΔA=2αA0ΔT=γA0ΔT\Delta A = 2\alpha A_0 \Delta T = \gamma A_0 \Delta T A=A0(1+2αΔT)A = A_0(1 + 2\alpha \Delta T)

Where γ=2α\gamma = 2\alpha is the area expansion coefficient

Volume Expansion

For solids:

ΔV=3αV0ΔT=βV0ΔT\Delta V = 3\alpha V_0 \Delta T = \beta V_0 \Delta T V=V0(1+βΔT)V = V_0(1 + \beta \Delta T)

Where β=3α\beta = 3\alpha is the volume expansion coefficient

For liquids (β\beta is given directly):

ΔV=βV0ΔT\Delta V = \beta V_0 \Delta T

Volume Expansion Coefficients (Liquids)

Liquidβ\beta (×10⁻⁴ /°C)
Alcohol (ethyl)11
Gasoline9.5
Mercury1.8
Water (20°C)2.1

Anomalous Expansion of Water

Water has unusual behavior:

  • Contracts when heated from 0°C to 4°C
  • Maximum density at 4°C
  • Expands when heated above 4°C
  • Ice is less dense than water (floats)

This is crucial for aquatic life in cold climates!

Thermal Stress

When expansion is constrained, thermal stress develops:

Stress=σ=YαΔT\text{Stress} = \sigma = Y\alpha\Delta T Force=F=YAαΔT\text{Force} = F = YA\alpha\Delta T

Where YY = Young's modulus

Bimetallic Strip

Two metals with different α\alpha bonded together:

  • Bends when temperature changes
  • Used in thermostats

The radius of curvature:

Rt(α2α1)ΔTR \approx \frac{t}{(\alpha_2 - \alpha_1)\Delta T}

Where tt = thickness of the strip

Examples

Example 1: Railroad Track

A 10 m steel rail installed at 20°C. Find expansion at 40°C.

ΔL=αL0ΔT=(12×106)(10)(20)=2.4×103 m=2.4 mm\Delta L = \alpha L_0 \Delta T = (12 \times 10^{-6})(10)(20) = 2.4 \times 10^{-3} \text{ m} = 2.4 \text{ mm}

Example 2: Ring and Rod

An aluminum ring has inner diameter 5.000 cm at 20°C. A brass rod has diameter 5.010 cm at 20°C. At what temperature will the ring fit the rod?

Need: ΔLAl=ΔLbrass+0.010\Delta L_{Al} = \Delta L_{\text{brass}} + 0.010 cm

αAl×5.000×ΔT=αbrass×5.010×ΔT+0.010\alpha_{Al} \times 5.000 \times \Delta T = \alpha_{\text{brass}} \times 5.010 \times \Delta T + 0.010 (2419×1.002)×106×ΔT=0.0105(24 - 19 \times 1.002) \times 10^{-6} \times \Delta T = \frac{0.010}{5} ΔT=0.0024.96×106403°C\Delta T = \frac{0.002}{4.96 \times 10^{-6}} \approx 403°\text{C} T=20+403=423°CT = 20 + 403 = 423°\text{C}

Example 3: Area Expansion

A copper sheet 2 m × 3 m is heated from 20°C to 120°C.

ΔA=2αA0ΔT=2×17×106×6×100=0.0204 m2=204 cm2\Delta A = 2\alpha A_0 \Delta T = 2 \times 17 \times 10^{-6} \times 6 \times 100 = 0.0204 \text{ m}^2 = 204 \text{ cm}^2

Example 4: Volume of Mercury

A mercury thermometer contains 0.5 cm³ at 20°C. Find volume at 100°C.

ΔV=βV0ΔT=1.8×104×0.5×80=0.0072 cm3\Delta V = \beta V_0 \Delta T = 1.8 \times 10^{-4} \times 0.5 \times 80 = 0.0072 \text{ cm}^3 V=0.5072 cm3V = 0.5072 \text{ cm}^3

Example 5: Thermal Stress

A steel rod fixed between rigid walls at 20°C. Find stress at 60°C. (Y=200Y = 200 GPa)

σ=YαΔT=200×109×12×106×40\sigma = Y\alpha\Delta T = 200 \times 10^9 \times 12 \times 10^{-6} \times 40 σ=96×106 Pa=96 MPa\sigma = 96 \times 10^6 \text{ Pa} = 96 \text{ MPa}

Example 6: Apparent vs Real Expansion

Gasoline in an aluminum tank. If tank expands, how much does gasoline level rise?

Apparent expansion coefficient:

βapparent=βliquidβcontainer\beta_{\text{apparent}} = \beta_{\text{liquid}} - \beta_{\text{container}} βapparent=9.5×1043×24×106=9.5×1040.72×104\beta_{\text{apparent}} = 9.5 \times 10^{-4} - 3 \times 24 \times 10^{-6} = 9.5 \times 10^{-4} - 0.72 \times 10^{-4} βapparent=8.78×104 /°C\beta_{\text{apparent}} = 8.78 \times 10^{-4} \text{ /°C}

Applications

  • Expansion joints in bridges and buildings
  • Gaps in railroad tracks
  • Thermostat switches (bimetallic strips)
  • Mercury/alcohol thermometers
  • Shrink fitting of metal parts