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Wize University Physics Textbook (Master) > Heat and Temperature

Thermal Expansion

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Thermal Expansion

For most material, raising their temperature results as an increase in their size.

Linear Expansion

  • The change in the length of a one-dimensional body due to the change in its temperature is called linear thermal expansion and could be obtained as:
ΔL=αL0ΔT\Delta L=\alpha L_0 \Delta T
  • L0L_0 is the initial length of the body
  • ΔT\Delta T could be positive or negative
  • α\alpha is the coefficient of linear expansion which is different for different material
  • The final length is: L=L0(1+αΔT)L=L_0 (1+\alpha\Delta T)
  • The coefficient of linear expansion of some materials is shown in Table 1 below
Materialα[K−1 or (°C)−1]Aluminum2.4×10−5Brass2.0×10−5Copper1.7×10−5Glass0.4−0.9×10−5Steel1.2×10−5\begin{array}{|l|c|}\hline Material & \alpha[K^{-1}\text{ or }(\degree C)^{-1}]\\[7pt]\hline Aluminum & 2.4 \times 10^{-5}\\[7pt]\hline Brass & 2.0 \times 10^{-5}\\[7pt]\hline Copper & 1.7 \times 10^{-5}\\[7pt]\hline Glass & 0.4 - 0.9 \times 10^{-5}\\[7pt]\hline Steel & 1.2 \times 10^{-5}\\[7pt]\hline \end{array}
The Coefficient of Linear Expansion For Different Material
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Area Expansion

  • The change in the temperature of a two-dimensional body leads to area expansion
  • The coefficient of area expansion is approximately equal to 2α2\alpha. (α\alpha for every linear dimension of an object)
  • Example The holes expand as the objects expand
  • Alternatively, one can find the linear expansion of the diameter or radius of a circular surface and find the expanded area using the new diameter or radius
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Example: Thermal Expansion
The diameter of a circular hole in a brass plate is 2.50 cm2.50\ cm at −2°C-2\degree C. What is the area of the hole at 48°C48 \degree C? (αbrass=2.0×10−5 K−1\alpha_{brass}=2.0\times10^{-5}\ K^{-1})

D=D0(1+αbrassΔT)=2.5025 cmD=D_0(1+\alpha_{brass}\Delta T)=2.5025\ cm
A=π(D2)2=π(2.50252)2=4.919 cm2A=\pi\bigg(\dfrac{D}{2}\bigg)^2=\pi\bigg(\dfrac{2.5025}{2}\bigg)^2=4.919\ cm^2
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Volumes Expansion
The increase in volume due to change in temperature of an object is:
ΔV=βV0ΔT\Delta V=\beta V_0 \Delta T
  • β\beta is the coefficient of volume expansion and is approximately equal to 3α3\alpha. (α\alpha for every linear dimension)
  • The coefficient of volume expansion for some material is shown in Table 2 below
Materialβ[K−1 or (°C)−1]Aluminum7.2×10−5Brass6.0×10−5Copper5.1×10−5Glass1.2−2.7×10−5Steel3.6×10−5Ethanol75×10−5Mercury18×10−5\begin{array}{|l|c|}\hline Material & \beta [K^{-1}\text{ or }(\degree C)^{-1}]\\[7pt]\hline Aluminum & 7.2 \times 10^{-5}\\[7pt]\hline Brass & 6.0 \times 10^{-5}\\[7pt]\hline Copper & 5.1 \times 10^{-5}\\[7pt]\hline Glass & 1.2-2.7 \times 10^{-5}\\[7pt]\hline Steel & 3.6 \times 10^{-5}\\[7pt]\hline Ethanol & 75 \times 10^{-5}\\[7pt]\hline Mercury & 18 \times 10^{-5}\\[7pt]\hline \end{array}
The Coefficient of Volume Expansion for Different Materials



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A 250.0 ml250.0\ ml volumetric flask at 20°C20\degree C is completely full of mercury. When heating the flask and mercury to 30°C30\degree C, 0.38 ml0.38\ ml of mercury overflows. If the coefficient of volume expansion of mercury is 18×10−5 K−118 \times 10^{-5}\ K^{-1}, what is the coefficient of volume expansion of the flask?
Hint: Both volumetric flask and mercury expand when temperature rises.