Playground
Interactive heat conduction through a bar: adjust conductivity, temperature difference, and length to see heat flow rate with animated particles.
Variables
| Symbol | Name | SI | Dimension | Range |
|---|---|---|---|---|
| Heat flow rateoutput Rate of heat energy transfer through the material | W | M·L²·T⁻³ | 0 – 10000 | |
| Thermal conductivity Material's ability to conduct heat | W/(m·K) | M·L·T⁻³·Θ⁻¹ | 0.01 – 500 | |
| Cross-sectional area Area perpendicular to heat flow direction | m² | L² | 0.0001 – 0.01 | |
| Temperature difference Temperature difference across the conductor (T_H - T_C) | K | Θ | 1 – 500 | |
| Length Length of the conducting path | m | L | 0.01 – 5 |
Deep dive
Derivation
Fourier postulated that heat flux is proportional to the negative temperature gradient: q = −k(dT/dx). For a slab of area A, thickness L, with faces at T_H and T_C: Q̇ = kA(T_H − T_C)/L. This follows from integrating q over area A with a linear temperature profile in steady state.
Experimental verification
Originally verified by Fourier using heated metal bars with thermometers along their length. Modern methods use laser-flash analysis and guarded hot-plate apparatus (ASTM C177) to measure k to ±1% accuracy for materials from aerogel (k ≈ 0.015) to diamond (k ≈ 2200).
Common misconceptions
- The negative sign does not mean heat flows backward — it ensures Q̇ is positive when heat flows from hot to cold (down the gradient)
- Thermal conductivity k is not constant — it varies with temperature, especially in metals and semiconductors
- Fourier's law applies to conduction only, not convection or radiation — real heat transfer often involves all three simultaneously
Real-world applications
- Building insulation design (R-value = L/kA)
- CPU heat sink engineering
- Cooking — why copper pans heat more evenly than steel
- Spacecraft thermal protection systems
Worked examples
Copper rod heat flow
Given:
- k:
- 385
- A:
- 0.001
- T_H:
- 373
- T_C:
- 293
- L:
- 0.5
Find: Q_dot
Solution
Q̇ = kA(T_H − T_C)/L = 385 × 0.001 × 80 / 0.5 = 61.6 W
Styrofoam wall insulation
Given:
- k:
- 0.033
- A:
- 10
- T_H:
- 293
- T_C:
- 263
- L:
- 0.05
Find: Q_dot
Solution
Q̇ = kAΔT/L = 0.033 × 10 × 30 / 0.05 = 198 W
Scenarios
What if…
- scenario:
- What if you replace copper (k = 385) with wood (k = 0.15)?
- answer:
- Heat flow drops by a factor of ~2567. Wood conducts 0.15/385 ≈ 0.04% as well as copper — this is why wooden spoon handles stay cool while copper ones burn.
- scenario:
- What if you double the rod length?
- answer:
- Heat flow halves. Q̇ ∝ 1/L — longer conduction paths reduce heat transfer. This is why thick walls insulate better than thin ones.
- scenario:
- What if ΔT = 0 but heat is being generated inside?
- answer:
- Fourier's law gives zero flux at that point, but the heat equation (∂T/∂t = α∇²T + q̇/ρc) handles internal generation. Steady state requires the generated heat to flow out through boundaries.
Limiting cases
- condition:
- k → 0 (perfect insulator)
- result:
- Q_dot → 0
- explanation:
- No conduction through a perfect insulator.
- condition:
- Delta_T → 0
- result:
- Q_dot → 0
- explanation:
- No temperature gradient means no heat flow.
- condition:
- L → 0
- result:
- Q_dot → ∞
- explanation:
- Infinitely thin conductor implies infinite flux (unphysical; contact resistance intervenes).
Context
Joseph Fourier · 1822
Published in 'Théorie analytique de la chaleur.' Fourier invented Fourier series specifically to solve the heat equation — a mathematical revolution born from a physics problem.
Hook
Why can you touch a wooden spoon in a hot pot but not a metal one?
A copper rod (k = 385 W/m·K) is 0.5 m long with cross-section 0.001 m². One end is at 100°C, the other at 20°C. Find the heat flow rate.
Dimensions: [Q̇] = [k][A][ΔT]/[L] → M·L²·T⁻³ = (M·L·T⁻³·Θ⁻¹)(L²)(Θ)/(L) = M·L²·T⁻³ ✓
Validity: Valid for steady-state conduction in isotropic, homogeneous solids. Breaks down for radiative or convective heat transfer, and at nanoscale where ballistic phonon transport dominates.