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03 // Bidirectional STEM Calculators // PHYSICSQ = (k × A × ΔT) / L

Fourier's Law Conductive Heat Transfer Calculator

Fourier's Law of Heat Conduction states that steady-state conductive heat transfer rate Q through a planar barrier is proportional to thermal conductivity k, cross-sectional area A, and temperature difference ΔT, and inversely proportional to thickness L.

Q = (k × A × ΔT) / L
Pick which value you want to calculate (or type into any field):Solving for: Q — Heat Transfer Rate
Calculated

Steady-state thermal power conducted in Watts

0.11,000,000.00

Material conductivity in W/(m·K)

0.01500
unit

Heat transfer surface area in m²

0.0110,000.00

Hot-to-cold surface temperature delta in K or °C

0.51,000.00
unit

Conduction path length / wall thickness in meters

0.00110
Calculated Result · Heat Transfer Rate (Q)
320W
Converted units:

Step-by-Step Calculation

01Canonical Law: Q = (k × A × ΔT) / L
02Given Inputs (Base Units): k (W/m·K) = 0.04, A = 40 m², ΔT (K) = 20, L = 0.1 m
03Solve for Q (Heat Transfer Rate): Q = 320 W

Fourier's Law of Heat Conduction states that steady-state conductive heat transfer rate Q through a planar barrier is proportional to thermal conductivity k, cross-sectional area A, and temperature difference ΔT, and inversely proportional to thickness L.

Real-World Reference Values

Mineral Wool / Fiberglass Battk = 0.035 – 0.042 W/(m·K)

High-R building envelope insulation

Structural Concrete / Glassk = 1.4 / 0.96 W/(m·K)

Moderate thermal bridging materials

Pure Copper Heatsinkk = 398 W/(m·K)

High-flux electronics thermal spreader

20-Row Comparison Table

Heat Transfer Rate (Q) · Thermal Conductivity (k (W/m·K))

Fourier's Law Conductive Heat Transfer Calculator — Thermal Conductivity → Heat Transfer Rate
#Thermal Conductivity (k (W/m·K))Heat Transfer Rate (W)Δ %
010.04 Current320 W0.00%
023.6299 29,039.42 W+8974.82%
037.2499 57,998.84 W+18024.64%
0410.8698 86,958.26 W+27074.46%
0514.4897 115,917.68 W+36124.28%
0618.1096 144,877.11 W+45174.10%
0721.7296 173,836.53 W+54223.91%
0825.3495 202,795.95 W+63273.73%
0928.9694 231,755.37 W+72323.55%
1032.5893 260,714.79 W+81373.37%
1136.2093 289,674.21 W+90423.19%
1239.8292 318,633.63 W+99473.01%
1343.4491 347,593.05 W+108522.83%
1447.0691 376,552.47 W+117572.65%
1550.689 405,511.89 W+126622.47%
1654.3089 434,471.32 W+135672.29%
1757.9288 463,430.74 W+144722.11%
1861.5488 492,390.16 W+153771.92%
1965.1687 521,349.58 W+162821.74%
2068.7886 550,309.00 W+171871.56%
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Frequently Asked Questions

How is Thermal R-Value related to k and L?+

Metric R-value (m²·K/W) equals thickness divided by conductivity: R_val = L / k. Thus Q = A × ΔT / R_val.

How do I find the insulation thickness L needed to cap heat loss at 200 W?+

Click SOLVE FOR on L (Material Thickness) and set Q = 200 W.

Can I use this for CPU heatsink thermal pads?+

Yes. Switch Area A to cm² or mm² and Thickness L to mm, then enter the TIM pad conductivity (e.g., k = 6 to 12 W/m·K).