Heating and melting: the fusion crust
During entry the surface of a meteoroid is heated to melting in seconds. Specific heat capacity and specific latent heat explain why the crust is so thin and the interior stays cold.
- Use ΔE = mcΔθ and E = mL in a two-stage heating calculation.
- Explain the formation of a fusion crust using the particle model.
- Explain why most of a meteoroid's mass is lost by ablation.
Links to: GCSE physics, energy and the particle model: specific heat capacity, specific latent heat and changes of state; maths: substitution and standard form.
Two stages of heating
To melt a solid, energy is needed first to raise its temperature to the melting point, and then to change its state without changing its temperature.
The first stage uses ΔE = mcΔθ, where c is the specific heat capacity, the energy needed to raise 1 kg by 1 °C. For rock, c is about 800 J/kg °C. The second uses E = mL, where L is the specific latent heat of fusion, the energy needed to melt 1 kg with no temperature change. For silicate rock it is roughly 4 × 10⁵ J/kg. During melting, the energy breaks bonds between particles instead of making them vibrate faster.
Plenty of energy, little time
A meteoroid has far more kinetic energy per kilogram than is needed to melt it. A rock moving at 15 km/s carries about 1 × 10⁸ J of kinetic energy per kilogram, roughly 70 times the energy needed to heat and melt it. That is why most of the original mass is lost: it is melted and stripped away in ablation.
The heating only lasts a few seconds, and rock conducts heat slowly, so a thin layer melts and is blown away again and again. When the meteoroid slows and the heating stops, the last molten film freezes into a fusion crust, typically under 1 mm thick.
Melting the surface
Estimate the energy needed to heat a 0.050 kg surface layer of rock from 20 °C to its melting point of about 1,500 °C and then melt it. Use c = 800 J/kg °C and L = 4.0 × 10⁵ J/kg.
- Temperature change: Δθ = 1,500 − 20 = 1,480 °C.
- Heating: ΔE = mcΔθ = 0.050 × 800 × 1,480 = 59,200 J.
- Melting: E = mL = 0.050 × 4.0 × 10⁵ = 20,000 J.
- Total ≈ 59,200 + 20,000 = 79,200 J ≈ 7.9 × 10⁴ J.
Answer: About 8 × 10⁴ J, roughly 1.6 × 10⁶ J for every kilogram. A kilogram moving at 15 km/s carries about 70 times that much kinetic energy.
Look at photographs of fusion crust on the glossary and famous meteorite pages. Describe its colour, texture and any flow lines. Then draw particle diagrams showing the outer layer of a meteoroid as a solid, while melting, and as the frozen crust, with an energy label for each change.
Quiz
5 questions. Choose an answer to see whether you are right and why. Finish the quiz to complete the lesson.
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Question 1Which equation gives the energy to melt a solid at its melting point?
Latent heat, E = mL, is the energy for a change of state with no change in temperature.
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Question 2What happens to the energy supplied while a solid is melting?
During a change of state, energy changes the arrangement of particles rather than their average kinetic energy.
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Question 3How much energy raises 2.0 kg of rock (c = 800 J/kg °C) by 10 °C?
ΔE = mcΔθ = 2.0 × 800 × 10 = 16,000 J.
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Question 4Why is a fusion crust so thin?
Only a thin surface layer is ever molten at once, and the last film to freeze becomes the crust.
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Question 5Why is most of a meteoroid's mass lost during entry?
Each kilogram carries around 70 times the energy needed to melt it, so ablation removes most of the mass.