Liquids and solids usually expand (i.e. increase their physical dimensions) when heated.

As temperature increases, atomic vibration amplitude also increases. This tends to cause an increase in bond lengths and therefore leads to thermal expansion.

**Anomalies:**

**Water**

– Water contracts with increasing temperatures between 0°C and 4°C, before expanding at higher temperatures.

**Reason:**

– Hydrogen-bonding leads to an increase in the H_{2}O – H_{2}O distance between 4°C. Ice has an open hydrogen-bonded structure that is less dense than water. That is why it floats on water.

**Calcium Carbonate**

CaCO_{3} expands along one direction and contracts along another with increasing temperature.

**Polyethylene single crystals**

Polyethylene single crystals expand along one direction and contracts along the chain axis direction with increasing temperature.

### Linear Expansion Coefficient (Derivation)

If the temperature change ($\delta T$) is sufficiently small, the fractional change in length ($\frac{\delta L}{L}$) is proportional to the temperature change through the** linear thermal expansion coefficient ($\alpha$)**.

$$\begin{aligned} \frac{\delta L}{L} &\propto \delta T \\ \frac{dL}{L} &= \alpha \, dT \end{aligned}$$

**To integrate, you have to assume α is constant.** (In real life, α is not constant!!) For a finite temperature change, we integrate to obtain:

$$\int\limits_{L_{1}}^{L_{2}} \frac{1}{L} \, dL = \int\limits_{T_{1}}^{T_{2}} \alpha \, dT$$

Typically, the change in L is tiny because α is small, and the practical temperature change cannot be large anyway.

Doing some calculus and approximation magic:

$$\begin{aligned} \int\limits_{L_{1}}^{L_{2}} \frac{1}{L} \, dL &= \text{ln} \, \frac{L_{2}}{L_{1}} \\ &= \text{ln} \, \frac{L_{1} + \Delta L}{L_{1}} \\ &= \text{ln} \, \left( 1 + \frac{\Delta L}{L_{1}} \right) \\ &\approx \frac{\Delta L}{L_{1}} \end{aligned}$$

,where the final approximation is valid for small $\frac{\Delta L}{L_{1}} << 1$.

Therefore, we can write:

$$\Delta L = \int\limits_{T_{1}}^{T_{2}} \alpha L_{1} \, dT$$If α is temperature independent over the temperature range of interest, we can further simplify to: (NOTE: This means this equation is only valid IF α is CONSTANT)

$$\Delta L = \alpha L \, \Delta T$$

**The equation basically means: the change in length is given by the product of linear expansion coefficient with length and the temperature change.**

**Average linear expansion coefficients of some solids near room temperature**

Material | $\alpha$ $\left(10^{-6} \, K^{-1} \right)$ |
---|---|

Quartz (fused) | 0.4 |

Glass (Pyrex) | 3.2 |

Steel | 11 |

Concrete | 12 |

Copper | 17 |

Brass | 19 |

Aluminum | 24 |

The linear expansion coefficients of most materials are in the ten ppm (parts per million) range per degree K change in temperature.

### Isotropic Thermal Expansion

Consider a “doughnut” shaped disk with inner radius a and outer radius b, where $b > a$. If the thermal expansion is the same in all direction (isotropic), the dimensions of the body will change proportionally in all directions and therefore the body will keep the same shape:

What this means?

– All holes and cavities will expand proportionally.

Note: Both the outer and inner parameter expands by the same proportion, and the central hole also becomes bigger!

### Volume Expansion Coefficient

Consider a solid having the dimensions x, y and z at temperature T. Its volume at T is $V = x . y . z$.

At temperature $T + \delta T$, its volume will be expanded to $V + \delta V$ where

$$\begin{aligned}\delta V &= (x + \delta x) (y + \delta y) (z + \delta z) \, – xyz \\ &= (x + \alpha x \, \delta T) (y + \alpha y \, \delta T) (z + \alpha z \, \delta T) \, – xyz \\ &= xyz (1 + \alpha \, \delta T)^{3}\, – xyz \\ &= V [ 1 + 3( \alpha \, \delta T) + 3 (\alpha \, \delta T)^{2} + (\alpha \, \delta T)^{3} ] \, – V \\ &= V [ 3(\alpha \, \delta T) + 3(\alpha \, \delta T)^{2} + (\alpha \, \delta T)^{3} ] \end{aligned}$$

$$\frac{\delta V}{V} = 3(\alpha \, \delta T) + 3(\alpha \, \delta T)^{2} + (\alpha \, \delta T)^{3}$$

For $\alpha \delta T << 1$, the square and cubic terms of ($\alpha \, \delta T$) can be neglected.(Too small to make a difference) And so, we have

$$\frac{\delta V}{V} \approx 3 \alpha \, \delta T$$

The above equation relates linear and volume expansion coefficients.

We can define a new term called volume expansion coefficient, $\beta$. So, we rewrite, $\beta = 3 \alpha$ to give:

$$\frac{\delta V}{V} \approx \beta \, \delta T$$

For a finite temperature change and constant $\beta$, we obtain:

$$\Delta V = \beta V \, \Delta T$$

**Average volume expansion coefficients of some liquids near room temperature:**

Material | $\beta$ $\left( 10^{-4} \, K^{-1} \right)$ |
---|---|

Alcohol | 1.12 |

Benzene | 1.24 |

Acetone | 1.5 |

Mercury | 1.82 |

Water (20$^{\circ}$C) | 2 |

Turpentine | 9.0 |

Liquids have roughly 10-40 times larger volume expansion coefficients than solids. That is why liquid-in-glass thermometers can work.

How to measure thermal expansion coefficient

Because thermal expansion used for heating but it was a little heat of sun

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