Fourier's Law of Heat Conduction
Calculate heat transfer through a material using Fourier's law.
Thermal conductivity, temperature gradient, and area.
The Formula
Fourier's law describes how heat flows through a solid material by conduction. Heat always moves from hot regions to cold regions. The rate of flow depends on the material's thermal conductivity, the cross-sectional area, and the temperature difference.
The negative sign indicates that heat flows in the direction of decreasing temperature. In most practical calculations, we work with the magnitude and simply note the direction.
Variables
| Symbol | Meaning | Unit |
|---|---|---|
| Q | Rate of heat transfer | watts (W) |
| k | Thermal conductivity of the material | W/(m·K) |
| A | Cross-sectional area perpendicular to heat flow | m² |
| ΔT | Temperature difference across the material | K or °C |
| Δx | Thickness of the material | m |
Example 1
A glass window (k = 0.8 W/m·K) is 0.005 m thick and 1.5 m² in area. Inside is 22°C, outside is −5°C.
ΔT = 22 − (−5) = 27°C = 27 K
Q = 0.8 × 1.5 × (27 / 0.005)
Q = 0.8 × 1.5 × 5400
Q = 6,480 W — the window loses about 6.5 kW of heat
Example 2
A copper rod (k = 401 W/m·K) has cross-section 0.001 m² and length 0.5 m. One end is at 100°C, the other at 25°C.
ΔT = 100 − 25 = 75 K
Q = 401 × 0.001 × (75 / 0.5)
Q = 401 × 0.001 × 150
Q = 60.15 W — copper conducts heat very efficiently
Common Thermal Conductivities
| Material | k (W/m·K) | Category |
|---|---|---|
| Copper | 401 | Excellent conductor |
| Aluminum | 237 | Good conductor |
| Steel | 50 | Moderate conductor |
| Glass | 0.8 | Poor conductor |
| Wood | 0.15 | Insulator |
| Styrofoam | 0.03 | Excellent insulator |
When to Use It
- Calculating heat loss through walls, windows, and insulation
- Designing heat sinks and thermal management systems
- Choosing insulation materials for buildings
- Engineering heat exchangers and cooling systems
- Estimating energy costs related to thermal losses