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Energy - Thermal Expansion

Grade 9IB

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

🔑Concepts

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Thermal expansion is the increase in the size (length, area, or volume) of a substance when its temperature increases. This occurs because particles move faster and push further apart as they gain kinetic energy.

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Linear Expansion: In solids, expansion primarily occurs along the length. The change in length ΔL\Delta L depends on the original length L0L_0, the change in temperature ΔT\Delta T, and the material's specific coefficient of linear expansion α\alpha.

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Volume Expansion: For liquids and gases (and 3D solids), we consider the change in volume ΔV\Delta V. Gases expand much more than liquids or solids for the same increase in temperature because their intermolecular forces are negligible.

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Bimetallic Strip: Made of two different metals (e.g., brass and iron) bonded together. Since they have different coefficients of expansion, the strip bends when heated, which is used in thermostats and circuit breakers.

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Anomalous Expansion of Water: Water behaves differently between 0∘C0^{\circ}C and 4∘C4^{\circ}C. When water is heated from 0∘C0^{\circ}C to 4∘C4^{\circ}C, it actually contracts and its density increases, reaching maximum density at 4∘C4^{\circ}C.

📐Formulae

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

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

ΔV=γV0ΔT\Delta V = \gamma V_0 \Delta T

γ≈3α\gamma \approx 3\alpha

ΔT=Tfinal−Tinitial\Delta T = T_{final} - T_{initial}

💡Examples

Problem 1:

A steel railway track has a length of 20.0 m20.0\text{ m} at a temperature of 15∘C15^{\circ}C. On a hot summer day, the temperature rises to 45∘C45^{\circ}C. Calculate the increase in the length of the track. (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:

ΔT=45∘C−15∘C=30∘C\Delta T = 45^{\circ}C - 15^{\circ}C = 30^{\circ}C ΔL=αL0ΔT\Delta L = \alpha L_0 \Delta T ΔL=(1.2×10−5)×20.0×30\Delta L = (1.2 \times 10^{-5}) \times 20.0 \times 30 ΔL=0.000012×600\Delta L = 0.000012 \times 600 ΔL=0.0072 m\Delta L = 0.0072\text{ m}

Explanation:

First, calculate the temperature difference ΔT\Delta T. Then, substitute the values into the linear expansion formula. The track expands by 0.0072 m0.0072\text{ m} (or 7.2 mm7.2\text{ mm}).

Problem 2:

Calculate the final volume of 500 cm3500\text{ cm}^3 of ethanol when it is heated from 10∘C10^{\circ}C to 60∘C60^{\circ}C. Given the coefficient of volume expansion γ=7.5×10−4 ∘C−1\gamma = 7.5 \times 10^{-4}\text{ }^{\circ}C^{-1}.

Solution:

ΔT=60∘C−10∘C=50∘C\Delta T = 60^{\circ}C - 10^{\circ}C = 50^{\circ}C ΔV=γV0ΔT\Delta V = \gamma V_0 \Delta T ΔV=(7.5×10−4)×500×50\Delta V = (7.5 \times 10^{-4}) \times 500 \times 50 ΔV=0.00075×25000\Delta V = 0.00075 \times 25000 ΔV=18.75 cm3\Delta V = 18.75\text{ cm}^3 Vfinal=V0+ΔV=500+18.75=518.75 cm3V_{final} = V_0 + \Delta V = 500 + 18.75 = 518.75\text{ cm}^3

Explanation:

To find the final volume, we first calculate the change in volume ΔV\Delta V using the volumetric expansion formula and then add it to the initial volume V0V_0.