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Thermal Physics - Thermal expansion of solids, liquids and gases

Grade 11IGCSEPhysics

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

🔑Concepts

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Thermal expansion occurs when the internal energy of a substance increases, causing its particles to move faster and take up more space. In solids, particles vibrate with larger amplitudes around fixed positions; in liquids and gases, particles move further apart.

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The relative degree of expansion varies by state: Gases>Liquids>Solids\text{Gases} > \text{Liquids} > \text{Solids} for the same increase in temperature.

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Different materials expand at different rates. This is the principle behind the bimetallic strip, where two metals with different coefficients of expansion are bonded together, causing the strip to bend when heated.

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Liquids expand significantly more than solids, which is the operational principle of liquid-in-glass thermometers (e.g., mercury or alcohol thermometers).

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Gases expand the most because intermolecular forces are negligible, allowing for a much larger increase in volume for a given ΔT\Delta T.

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Precautions in engineering: Bridges are built with expansion joints (rollers or gaps) and railway tracks have small gaps to prevent buckling due to thermal expansion in summer.

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The temperature change ΔT\Delta T can be expressed in either degrees Celsius (∘C^{\circ}C) or Kelvin (KK), as the magnitude of a degree is the same in both scales: ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}.

📐Formulae

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

ΔV=V0βΔT\Delta V = V_0 \beta \Delta T

Lfinal=L0(1+αΔT)L_{final} = L_0 (1 + \alpha \Delta T)

Vfinal=V0(1+βΔT)V_{final} = V_0 (1 + \beta \Delta T)

💡Examples

Problem 1:

A steel railway track is 20.0 m20.0\text{ m} long at a temperature of 15∘C15^{\circ}C. If the temperature rises to 45∘C45^{\circ}C, calculate the increase in its length. The coefficient of linear expansion for steel is α=1.2×10−5 ∘C−1\alpha = 1.2 \times 10^{-5}\text{ }^{\circ}C^{-1}.

Solution:

Using the formula ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T:

  1. Identify values: L0=20.0 mL_0 = 20.0\text{ m}, α=1.2×10−5 ∘C−1\alpha = 1.2 \times 10^{-5}\text{ }^{\circ}C^{-1}, ΔT=45−15=30∘C\Delta T = 45 - 15 = 30^{\circ}C.
  2. Calculate: ΔL=(1.2×10−5)×20.0×30\Delta L = (1.2 \times 10^{-5}) \times 20.0 \times 30.
  3. ΔL=0.0072 m\Delta L = 0.0072\text{ m} or 7.2 mm7.2\text{ mm}.

Explanation:

The change in length is directly proportional to the original length and the change in temperature. Although 7.2 mm7.2\text{ mm} seems small, the cumulative expansion across many tracks can cause significant stress or buckling if gaps are not provided.

Problem 2:

Explain why a glass jar with a tight metal lid can often be opened by running hot water over the lid.

Solution:

Metal has a higher coefficient of thermal expansion than glass. When hot water is applied, the metal lid expands more than the glass rim of the jar (αmetal>αglass \alpha_{\text{metal}} > \alpha_{\text{glass}}).

Explanation:

The differential expansion increases the diameter of the lid more than the diameter of the glass opening, thereby loosening the seal and making it easier to unscrew.

Thermal expansion of solids, liquids and gases Revision - Grade 11 Physics IGCSE