Materials Science · Heat transport, expansion and thermal response
Thermal diffusivity and response times
Useful background: Energy, stability and environmental context; Measuring and checking material claims. Each worked case explains the quantities it uses.
Back to the chapterAfter reading, you can
- Estimate a conduction timescale from diffusivity and length.
A temperature disturbance takes time to spread
Thermal diffusivity α = k/(ρc) combines conductivity k with volumetric heat capacity ρc, where ρ is mass density. Its SI unit is m²/s. In simple conduction problems, a characteristic diffusion time scales as L²/α. The numerical prefactor depends on geometry, boundaries and the chosen response criterion.
Size has a squared effect
Doubling a conduction distance makes the characteristic time roughly four times larger when the same model applies. Surface heat transfer, contact resistance and internal heat generation can impose other limits. A scaling estimate is not an exact complete heating schedule.
Worked example
Work through the case
L = 0.01 m and α = 10⁻⁵ m²/s. Estimate L²/α.
- Square length: 10⁻⁴ m².
- Divide by 10⁻⁵ m²/s.
- The units reduce to seconds.
10 s as a characteristic conduction scale, not an exact equilibration time.
Try the reasoning
How does this scale change if L triples?
Show the answer and reasoning
It increases ninefold.
Time scales with L².
Can convection change the overall heating time?
Show the answer and reasoning
Yes.
The boundary can limit heat delivery.
What to carry forward
- Diffusivity controls spreading relative to storage.
- State boundaries when converting a scale into a prediction.
References
Teaching resources and research supporting this lesson’s explanations and examples.