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Heat Transfer Calculator

Tested

Conductive heat flow through a flat wall.

Tested, not yet published. The formula for this calculator is implemented and passes its automated checks. This page stays non-indexable until it is reviewed and published.

W/(m·K)
m²
K
m

Results

Enter your values and press Calculate. Results appear here.

  • • Result

Quick answer

Conductive heat flow is Q̇ = k A ΔT ÷ d. An insulation layer with k = 0.04 W/m·K, 20 m² area, 22 K temperature difference and 0.1 m thickness loses 176 W, or 600.5 BTU/h.

Enter conductivity, area, temperature difference and thickness. The calculator returns the heat flow in watts and BTU/h. Use this heat transfer calculator to check your own figures.

Related tools: Specific Heat Calculator: Specific heat capacity or heat required from Q = m c ΔT. Heat Pump COP Calculator: Coefficient of performance from heat output and power input, or the Carnot limit. Thermodynamic Processes Calculator: Work, heat and internal energy for an ideal gas process.

How to use the Heat Transfer Calculator

  1. Enter thermal conductivity (W/(m·K)) (above 0).
  2. Enter area (m²) (above 0).
  3. Enter temperature difference (K).
  4. Enter thickness (m) (above 0).
  5. Press Calculate to see the result, and use Reset to start again.
Inputs of the Heat Transfer Calculator
InputWhat to enterAllowed values
Thermal conductivityA number, in W/(m·K)> 0
AreaA number, in m²> 0
Temperature differenceA number, in K—
ThicknessA number, in m> 0

Formula and method

Formula

Q̇ = k A ΔT ÷ d; 1 W = 3.412141633 BTU/h

IEEE-754 double precision; shown to the decimals indicated.

One-dimensional steady conduction.

Worked examples

These examples are verified: each one is an automated test case that the live calculator must reproduce exactly.

Example 1

Inputs: Thermal conductivity = 0.04; Area = 20; Temperature difference = 22; Thickness = 0.1.

Results of example 1
ResultValue
Heat flow176 W
In BTU/h600.5369

Example 2

Inputs: Thermal conductivity = 205; Area = 0.01; Temperature difference = 50; Thickness = 0.005.

Results of example 2
ResultValue
Heat flow20,500 W
In BTU/h69,948.9035

Example 3

Inputs: Thermal conductivity = 0.8; Area = 1.5; Temperature difference = 15; Thickness = 0.2.

Results of example 3
ResultValue
Heat flow90 W
In BTU/h307.0927

The first line is the headline result (heat flow). Below it the heat Transfer Calculator lists in BTU/h, which describe the same case from other angles. IEEE-754 double precision; shown to the decimals indicated.

Worked example

0.04 × 20 × 22 ÷ 0.1 = 176 W. Doubling the thickness halves the loss to 88 W. 176 W × 3.412 = 600.5 BTU/h.

Conduction only

Convection and radiation are not included, and the real loss through a wall also depends on surface films and thermal bridges. Typical k values: mineral wool 0.04, brick 0.7, concrete 1.4, glass 1.0 and copper 400.

How the result changes with thermal conductivity

Holding every other input at the first worked example, this table changes only thermal conductivity. It shows how sensitive the answer is to that one value; each row was produced by running this calculator.

How the result changes with thermal conductivity
Thermal conductivityHeat flowIn BTU/h
0.0288 W300.2685
0.04176 W600.5369
0.08352 W1,201.0739
0.16704 W2,402.1477

What happens with invalid input

These messages are produced by the calculator itself, so you know what to fix.

  • Thermal conductivity: entering "0" shows "Thermal conductivity must be greater than 0." beside that field and no result.
  • Thickness: entering "0" shows "Thickness must be greater than 0." beside that field and no result.

Choosing between the Heat Transfer Calculator and similar calculators

Each row shows what a related calculator on this site asks for and what it returns, so you can pick the one that matches your question.

Choosing between the Heat Transfer Calculator and similar calculators
CalculatorYou enterIt returns
[Specific Heat Calculator](/science/specific-heat-calculator/)solve for, mass and temperature changeSpecific heat
[Heat Pump COP Calculator](/science/heat-pump-cop-calculator/)useCOP
[Thermodynamic Processes Calculator](/science/thermodynamic-processes/)process, gas and amount of gasWork done by the gas
[EER Calculator](/science/energy-efficiency-ratio-calculator/)cooling capacity and power inputEER
[Thermal Expansion Calculator](/science/thermal-expansion-calculator/)coefficient of linear expansion, original length and temperature changeChange in length

Background: heat and temperature

The basic heat relation is Q = m × c × ΔT: heat in joules equals mass times specific heat capacity times temperature change. A positive Q means heat absorbed and a negative Q heat released, and a change in °C is numerically the same as a change in K.

It covers a change in temperature only. During melting or boiling the temperature stays constant while heat still flows (latent heat), which needs a separate term.

Assumptions and limitations

  • Convection and radiation are not included

If a required value is empty, not a number, or outside its allowed range, the form marks that field with a message and shows no result instead of a misleading number. The automated checks deliberately try invalid entries for thermal conductivity and thickness to confirm this.

Frequently asked questions

What is thermal conductivity?

How well a material conducts heat.

How do I cut heat loss?

Add insulation or reduce area.

What is the unit?

Thermal conductivity k is in W/(m·K); the example's 0.04 W/(m·K) is typical of mineral wool.

What does the Heat Transfer Calculator calculate?

Conductive heat flow through a flat wall. It returns Heat flow and In BTU/h.

What formula does the Heat Transfer Calculator use?

Q̇ = k A ΔT ÷ d; 1 W = 3.412141633 BTU/h. One-dimensional steady conduction.

What do I need to enter in the heat Transfer Calculator?

You enter Thermal conductivity (W/(m·K)), Area (m²), Temperature difference (K) and Thickness (m).

How is the result rounded or limited?

IEEE-754 double precision; shown to the decimals indicated. Convection and radiation are not included.

What is an example of a heat Transfer result?

With thermal conductivity 0.04, area 20, temperature difference 22 and thickness 0.1, the heat flow is 176 W.

Sources

  • OpenStax, University Physics Volume 1 (Rice University): mechanics

Content last reviewed 2026-11-04. Formulas and worked examples are checked by automated tests against independently computed values.