ClarityHow to use this tool
- Enter thermal conductivity, area, temperature difference, thickness.
- Select Calculate to view the result.
- Check the method and assumptions below before using the result.
Input reference and sample values
Use the units and choices shown for each field. Sample values are provided for practice, not as recommendations.
| Input | Format and units | Sample |
|---|---|---|
| Thermal conductivity | Numeric value. Unit: W/(m·K). Must be greater than zero. | 0.04 |
| Area | Numeric value. Unit: m². Must be greater than zero. | 0.5 |
| Temperature difference | Numeric value. Unit: K. | 20 |
| Thickness | Numeric value. Unit: m. Must be greater than zero. | 0.1 |
Result for these sample inputs
4 W
Select Try the example in the form to load these values and calculate them again. This replaces the current form inputs. Compare the result with the method and worked explanation below.
The method, explained
Heat-transfer rate = conductivity × area × temperature difference/thickness.
Using thermal conductivity = 0.04 W/(m·K), area = 0.5 m², temperature difference = 20 K, thickness = 0.1 m, the result is 4 W. Change these example inputs to match your task; use the method above to check each step.
What to keep in mind
One-dimensional steady state; ignores contact resistance and thermal bridges. An idealized educational model using the stated SI units. Real systems may need additional forces, losses or experimental measurements. Not a safety or equipment specification.
Methodology maintained by ClarityKit. How these tools are built and checked.