Free Heat Transfer Calculator
Choose conduction through a flat layer or the energy needed to heat or cool a mass. Enter your values to see the result, unit conversions and worked steps.
Enter your values or load an example. Calculations stay in this browser.
Choose the heat-transfer calculation you need
Use conduction when you know the two surface temperatures of a solid layer. Use specific heat when you know a material’s mass and its change in temperature. The first gives a rate; the second gives an amount of energy. Neither mode is a complete building heat-loss model.
Conduction: P = kAΔT / L
Enter conductivity k in W/(m·K), area A, thickness L and the temperature difference across the layer. The result P is thermal power in watts. Duration lets you convert that steady rate into transferred energy using Q = Pt.
Example: a layer with k = 0.04 W/(m·K), area 10 m², thickness 100 mm and temperature difference 20 K conducts 80 W. Over one hour that is 288,000 J, or 288 kJ. Doubling the thickness halves the rate under the same conditions.
Heating or cooling: Q = mcΔT
Enter mass m, specific heat capacity c and final temperature minus initial temperature. For 1 kg with c = 4,184 J/(kg·K) warmed by 20 K, Q is 83,680 J, or 83.68 kJ. Cooling by the same amount gives −83.68 kJ.
This model describes a temperature change without melting, boiling or another phase change. During a phase change, latent heat must be included separately. Heat lost to the surroundings and energy absorbed by a container are also outside this simple calculation.
Units, signs and assumptions to check
- Watts measure energy per second; joules and kilojoules measure energy. Do not label an energy result in watts.
- A temperature difference of 1 °C equals 1 K. Multiply a Fahrenheit difference by 5/9; do not subtract 32.
- For conduction, ΔT is side 1 minus side 2. A negative result means heat flows toward side 1.
- Use surface temperatures across the layer, not surrounding air temperatures unless the surface temperature is known to match the air.
- Conductivity and specific heat depend on the material and temperature. Use a value suited to your problem; the example values are illustrative.
When this model needs more detail
Composite walls need the resistance of every layer. Curved pipes need a cylindrical model. Transient heating, radiation and moving fluids need additional equations. Start with the mechanism and assumptions before interpreting a number as a prediction.
References
Related free tools
Frequently Asked Questions
Does this calculate conduction or specific heat?
Both. Choose conduction for P = kAΔT/L, or heating and cooling for Q = mcΔT. The input fields and worked steps change with the selected mode.
What is the difference between heat and heat-transfer rate?
Heat transferred is energy in joules. Heat-transfer rate is power in watts, where one watt is one joule per second. For a constant rate, multiply watts by seconds to get joules.
Can the heat-transfer result be negative?
Yes. In conduction, a negative result means flow from side 2 toward side 1. In the specific-heat mode, it means energy leaves the material as it cools.
Can I use Fahrenheit temperatures?
Enter the difference between the two temperatures and select °F. The calculator multiplies that difference by 5/9 to obtain kelvin. Do not enter an absolute temperature in the difference field.
Can this calculate melting or boiling energy?
No. The specific-heat calculation covers temperature changes within a phase. Melting or boiling requires latent heat, calculated separately from the heating or cooling segments.
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