Thermal Conductivity Calculator
Calculate thermal conductivity k from measured heat flow using k = Q·d/(A·ΔT·t).
k (W/m·K): Air=0.026, Glass wool=0.04, Brick=0.8, Glass=1.0, Steel=50, Copper=401, Diamond=2000
What Is Thermal Conductivity Calculator?
The formula Calculate thermal conductivity k from measured heat flow using k = Q·d/(A·ΔT·t). is fundamental to this area of physics.
It follows from first principles and has wide application in science and engineering.
Pay careful attention to units — results are only valid when SI units are used consistently.
See the worked examples for typical values and real-world applications of this formula.
Formula Reference Table
| Solve For | Formula | Notes |
|---|---|---|
| Thermal conductivity | k = Q·d/(A·ΔT·t) | W/(m·K) |
| Heat flow rate | Q/t = k·A·ΔT/d | W (Fourier's Law) |
| Temperature diff | ΔT = Q·d/(k·A·t) | K |
| Thermal resistance | R_th = d/(k·A) | K/W |
| R-value | R'' = d/k | m²·K/W (SI) |
| Thermal diffusivity | α = k/(ρ·Cₚ) | m²/s |
3 Worked Examples
Q=1000 J in t=3600 s, d=50 mm, A=2 m², ΔT=20 K.
- k = Q×d/(A×ΔT×t) = 1000×0.05/(2×20×3600) = 50/144,000 = 3.47×10⁻⁴ W/m·K
- Very low — aerogel-like material!
k=0.8, A=30 m², ΔT=20 K, d=0.22 m. Heat flow rate?
- Q/t = k×A×ΔT/d = 0.8×30×20/0.22 = 2,182 W
- Over 24 hr: Q = 2182×86400 = 1.88×10⁸ J = 52.5 kWh
Q/t=1000 W, k=401, A=0.001 m², d=0.5 m.
- ΔT = (Q/t)×d/(k×A) = 1000×0.5/(401×0.001) = 500/0.401 = 1.247 K
- Copper conducts extremely well — tiny ΔT for large heat flow
Real-World Applications
Common Mistakes to Avoid
SI units.
Correct variant.
Ideal conditions.
Check direction.
Verify.
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Interpretation: This formula tracks heat, temperature, work, entropy or transport in a thermodynamic system. Assumption: Use absolute temperature where required and consistent energy units. Constant properties, equilibrium, ideal gases or negligible losses may be assumed.