# Thermal expansion

By Levi Clancy for Student Reader on *updated *

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## Linear expansion

Δℓ = α ℓ_{c} ΔT

ℓ_{h} = ℓ_{c} ( 1 + α ΔT )

When considering only length, then this equation is used to measure thermal expansion. Note that linear expansion has no meaning for liquids or gases since they do not have fixed shapes.

## Volume expansion

ΔV = β V_{c} ΔT

When considering volume, then this equation is used to measure thermal expansion.

ΔV ≈ 3 α V_{c} ΔT

However, β ≈ α in many cases. Thus, when there is not tremendous thermal expansion, that is, if α · ΔT << 1, then the equation above is used.

Linear Expansion Coefficient (α) | Each solid has its own linear expansion coefficient α that represents how much its length will increase or decrease with temperature changes in °C or °K. | |
---|---|---|

Volume Expansion Coefficient (β) | Every material has its own volume expansion coefficient β that represents the change in volume with respect to temperature changes in °C or °K. | |

Length (ℓ) | ℓ_{c} = Length of material when coolerℓ _{h} = Length of material when warmerΔℓ = Change in length due to temperature change | |

Volume (V) | V_{c} = Volume of material when coolerV _{h} = Final volume after warmerΔV = Change in volume due to temperature change | |

Temperature (T) | °C or °K | T_{c} = Cooler temperature of material in °C or °KT _{h} = Warmer temperatureΔT = Temperature change |

Thermal stress | N m^{-2} | Stress = F / A = α E ΔT Δℓ = ( F · ℓ E = Young's modulus (a constant) for the object undergoing ΔT If the stress meets or exceeds the absolute strength, tension limit or shear limit of a material, then it will likely fracture under that level of stress. |

###### There are some special cases with volume expansion due to temperature.

For instance, consider a ring that is heated. Will the hole get smaller or larger as temperature increases? An initial guess may be that the material expands to fill the hole, making the hole smaller. This is incorrect. In fact, the entire ring gets larger, including the hole. Thus, many textbooks include problems whereby a ring must be heated a certain ΔT so it can fit onto a rod.

###### Most materials expand and contract linearly with respect to changes in temperature. However, water behaves anomalously below 4°C.

If water is being cooled, then between 4°C and 0°C it actually expands; if it is being heated, then between 0°C and 4°C it actually decreases in volume. This is due to the fact that exceptionally, solid water (ice) has a greater volume than liquid water.

## Practice problems

This is an example of linear thermal expansion.

However, the temperature at 15.00 cm is not immediately attainable.

You have all the variables except for α -- so first find α.

Δℓ = α ℓ_{c} ΔT

ℓ_{c} = 10.00 cm = .1000 m

ℓ_{h} = 20.00 cm = .2000 m

Δℓ = .2000 m - .1000 m = .1000 m

T_{c} = 10.00°C

T_{h} = 110.0°C

ΔT = 100.0°C

.1000 = α · .1000 · 100.00

1 = α · 100.00

.0100 = α

Now use the linear thermal expansion equation again.

Δℓ = α ℓ_{c} ΔT

ℓ_{c} = 10.00 cm = .1000 m

ℓ_{h} = 15.00 cm = .1500 m

Δℓ = .2000 m - .1000 m = .0500 m

T_{c} = 10.00°C

T_{h} = ?

ΔT = T_{h} - 10.00°C

.0500 = .0100 · .100 · ΔT

5 = .100 · ΔT

50 = ΔT

50 = T_{h} - 10.00

40 = T_{h} → The temperature when the column is 15.00 cm is 40°C.

When dealing with rings, or rivets, or other circular items, consider the diameters as ℓ values.

Δℓ = α ℓ_{c} ΔT

ℓ_{c} = 1.666 cm = 1.666 · 10^{-2} m

ℓ_{h} = 1.667 cm = 1.667 · 10^{-2} m

Δℓ = 1.667 · 10^{-2} m - 1.666 · 10^{-2} m = .001000 · 10^{-2} m = 1.000 · 10^{-5} m

T_{c} = ?

T_{h} = 25.0°C

ΔT = 24.0°C - T_{c}α = 12 · 10^{-6} (according to Giancoli, p 460)

1.000 · 10^{-5} = 12 · 10^{-6} · 1.666 · 10^{-2} · ΔT

.05000 · 10^{3} = ΔT

50.00°C = ΔT

50.00°C = 24.0°C - T_{c}T_{c} = -26.0°C → The rivet must be cooled to -26.0°C in order to fit the hole.

Note that if you are unclear whether ΔT equals T_{h} - T_{c} or T_{c} - T_{h} then simply take a conceptual approach.

You know the difference between T_{h} and T_{c} is 50.0°C, so T_{c} must be 74.0°C or -26.0°C.

But the answer cannot be 74.0°C -- that is *hotter* than 24.0°C and the rivet must be *cooler* in order to fit.

So the answer must be -26.0°C.

^{-4}(via Giancoli, p 460) and at 60°C has a density of 0.9880393 g mL

^{-1}(via Fred Senese). What is the coefficient of volume expansion of the container?

First find how much the wine expands.

ΔV = β V_{c} ΔT

V_{c} = 750 mL = .750 L

V_{h} = ?

ΔV = ?

ΔT = 60.0°C - 20.0°C = 40.0°C

β = 2.10 · 10^{-4}

ΔV = 2.10 · 10^{-4} · .750 · 40.0

ΔV = 6.3 · 10^{-3} L

The wine expands by 6.3 · 10^{-3} L when heated.

However, only .25 grams of wine actually overflow.

Volume = Mass ⁄ Density = 5.04 g / ( .9880393 g mL^{-1} ) = 5.1 mL = 5.1 · 10^{-3} L

The wine expanded by 6.3 · 10^{-3} L, but only 5.1 · 10^{-3} overflowed.

Thus, the carafe must have expanded by 6.3 · 10^{-3} - 5.1 · 10^{-3} = 1.2 · 10^{-3} L

Now we may find the volume expansion coefficient (β) for the carafe:

ΔV = β V_{c} ΔT

V_{c} = .750 L

ΔV = 1.200 · 10^{-3}β = ?

ΔT = 40.0°C

1.200 · 10^{-3} = β · .750 · 40

β = 40 · 10^{-6} → The carafe appears to be made of gold.