Thermal Conductivity Converter - Convert W/m·K, BTU/hr·ft·°F & More Units
Result:
1 W/m·K = 0.57778921 BTU/hr·ft·°F
How Thermal Conductivity Conversion Works
Input Value
Enter thermal conductivity value
Select Units
Choose from and to units
Convert
Apply conversion formula
Thermal Conductivity Formulas
Basic Heat Conduction (Fourier's Law)
q = Heat flux (W)
k = Thermal conductivity (W/m·K)
A = Cross-sectional area (m²)
dT/dx = Temperature gradient (K/m)
Thermal Resistance
R = Thermal resistance (K/W)
L = Thickness (m)
k = Thermal conductivity (W/m·K)
A = Area (m²)
Unit Conversion
k₁ = Original value
k₂ = Converted value
CF = Conversion factor
Example: 1 W/m·K = 0.578 BTU/hr·ft·°F
Heat Transfer Rate
Q = Heat transfer rate (W)
ΔT = Temperature difference (K)
L = Material thickness (m)
Used for steady-state conduction
Thermal Conductivity Conversion Table
| W/m·K | BTU/hr·ft·°F | cal/s·cm·°C | kcal/hr·m·°C | BTU·in/hr·ft²·°F | W/cm·K |
|---|---|---|---|---|---|
| 0.1 | 0.0578 | 0.000239 | 0.086 | 0.69 | 0.001 |
| 0.5 | 0.2889 | 0.001195 | 0.430 | 3.47 | 0.005 |
| 1 | 0.5778 | 0.002390 | 0.860 | 6.93 | 0.010 |
| 2 | 1.1556 | 0.004780 | 1.720 | 13.87 | 0.020 |
| 5 | 2.8889 | 0.011950 | 4.299 | 34.67 | 0.050 |
| 10 | 5.7779 | 0.023901 | 8.598 | 69.33 | 0.100 |
| 20 | 11.5558 | 0.047801 | 17.197 | 138.67 | 0.200 |
| 50 | 28.8895 | 0.119503 | 42.992 | 346.67 | 0.500 |
| 100 | 57.7789 | 0.239006 | 85.985 | 693.35 | 1.000 |
| 150 | 86.6684 | 0.358509 | 128.977 | 1040.02 | 1.500 |
| 200 | 115.5578 | 0.478011 | 171.969 | 1386.69 | 2.000 |
| 300 | 173.3368 | 0.717017 | 257.954 | 2080.04 | 3.000 |
| 400 | 231.1157 | 0.956023 | 343.938 | 2773.39 | 4.000 |
| 500 | 288.8946 | 1.195029 | 429.923 | 3466.73 | 5.000 |
| 1000 | 577.7892 | 2.390057 | 859.845 | 6933.47 | 10.000 |
Thermal Conductivity Units Progression Chart
0.1 W/m·K
1 W/m·K
10 W/m·K
50 W/m·K
100 W/m·K
400 W/m·K
Practice Problems
Problem 1:
Convert 25 W/m·K to BTU/hr·ft·°F
Solution: 25 ÷ 1.730735 = 14.44 BTU/hr·ft·°F
Problem 2:
Convert 0.5 cal/s·cm·°C to W/m·K
Solution: 0.5 × 418.4 = 209.2 W/m·K
Problem 3:
Convert 100 kcal/hr·m·°C to W/m·K
Solution: 100 × 1.163 = 116.3 W/m·K
Problem 4:
Convert 50 W/m·K to W/cm·K
Solution: 50 ÷ 100 = 0.5 W/cm·K
Problem 5:
Convert 10 BTU/hr·ft·°F to cal/s·cm·°C
Solution: (10 × 1.730735) ÷ 418.4 = 0.0414 cal/s·cm·°C
Daily Uses of Thermal Conductivity
Building insulation materials selection for energy efficiency
Cookware design determines heat distribution and cooking performance
Electronic device heat sinks prevent component overheating
Automotive radiator materials optimize engine cooling systems
Thermal interface materials improve heat transfer in computers