Audited 31 Jul 2026·Last updated 28 Oct 2026·5 citations·Tier 1·0 uses

Heat Transfer Calculator

Calculate steady heat flow in watts through a wall, window or heat-exchanger surface from its U-value, area and temperature difference: Q̇ = U × A × ΔT.

Heat Transfer Calculator

W/(m²·K)
m²
K
Heat-transfer rate
200
The steady rate of heat flow through the surface in watts, equal to U × A × ΔT.
Model scope
Steady UAΔT magnitude with one entered driving difference; it does not derive U, choose arithmetic versus log-mean ΔT, include transients, or resolve radiation and fouling separately.

Background.

Heat leaks from warm to cold through every surface that separates them, and in steady conditions the leak rate depends on just three things. How easily heat passes through the barrier, how much barrier there is, and how big the temperature gap is across it. Engineers fold the first into a single number, the overall heat-transfer coefficient U, and the rate becomes Q̇ = U × A × ΔT.

The same equation sizes very different things. For a building, U is the U-value of a wall, roof or window, and the result is the heat a heating system must replace to hold the room temperature. For a heat exchanger, U describes the metal wall and the fluid films on either side, and the result is how much heat moves from the hot stream to the cold one. In both cases the answer comes out in watts, joules per second, so it converts directly into energy and cost: a steady 200 W running for a day is 4.8 kWh.

The proportionality is the useful part. Halve the U-value with better insulation and the loss halves. Double the area of glazing and the loss through it doubles. A cold night with twice the indoor-outdoor difference doubles the heating load. It also tells you where to look first: because each element contributes its own U × A, the weakest surfaces dominate. A window with several times the U-value of the wall around it can account for a large share of a room's loss while covering a small share of its area, and a few square metres of uninsulated ceiling can outweigh a well-insulated wall.

What the formula does not do is supply U or ΔT for you. U comes from adding up the thermal resistances of each layer and surface film, or from a manufacturer's rating. In a heat exchanger the temperature difference changes along the length of the unit, and the right value to use is the log-mean temperature difference, not the simple average. The equation also assumes conditions have settled; warming up a cold building or a thick wall takes extra energy that a steady rate leaves out.

What is heat transfer calculator?

The overall heat-transfer rate is the power, in watts, flowing through a surface that separates two regions at different temperatures, once conditions are steady. It is the product of the overall heat-transfer coefficient U, which combines conduction through the solid layers with convection (and often radiation) at each surface, the area A through which heat flows, and the temperature difference ΔT that drives the flow.

Two related quantities come up constantly. The product U × A, in watts per kelvin, is the heat-loss coefficient of an element or of a whole building: the power needed for every degree it is held warmer than its surroundings. The reciprocal of U, R = 1/U in m²·K/W, is the thermal resistance of one square metre, the figure insulation is usually sold by. Resistances in layers add, so R is the natural way to build up a wall, and U the natural way to work out a heat flow. For a sense of scale, well-insulated walls and roofs sit well below 1 W/(m²·K), a single pane of glass is close to 6, and liquid-to-liquid heat exchangers reach the hundreds or thousands, because metal walls and moving liquids pass heat far more readily than still air.

How to use this calculator.

  1. Enter the overall heat-transfer coefficient U in W/(m²·K). For building elements this is the U-value from a product data sheet or energy assessment; for a heat exchanger, use the design or measured overall coefficient including any fouling allowance.
  2. Enter the area in square metres through which the heat flows, for example the wall area net of windows, or the tube surface area on the side U is referred to.
  3. Enter the driving temperature difference in kelvin. A difference of 20 °C is the same as 20 K, so you can use the gap between indoor and outdoor temperatures in Celsius directly. For a heat exchanger, enter the log-mean temperature difference.
  4. When sizing a heater rather than estimating a typical day, use the design difference: your indoor set point minus the cold-weather outdoor design temperature that local building guidance publishes for your area, not an average winter temperature.
  5. Read the heat-transfer rate in watts. Multiply by hours of operation and divide by 1,000 for kilowatt-hours.

The formula.

Q̇ = UAΔT

Q̇ = U × A × ΔT. Q̇ is the heat-transfer rate in watts. U, in W/(m²·K), is the reciprocal of the total thermal resistance of one square metre of the barrier: for a flat wall, 1/U = 1/h_in + Σ(L/k) + 1/h_out, where h are the surface coefficients and each layer of thickness L and conductivity k adds L/k. A, in m², is the area heat passes through, and ΔT, in K, is the temperature difference between the two sides. The product U × A, in W/K, is the heat lost per degree of difference.

The equation joins two older laws. Fourier's law of conduction gives the flow through a single solid layer as Q̇ = kAΔT/L, so a bare layer on its own has U = k/L: 0.2 m of insulation with k = 0.04 W/(m·K) gives 0.2 W/(m²·K) before the air films are counted. Newton's law of cooling gives the flow from a surface into the air or fluid beside it as h × A × ΔT. In steady state the same heat passes through every layer in turn, so the temperature drops across the layers add up to the whole ΔT and their resistances add like electrical resistors in series, which is where the 1/U sum comes from.

Two arrangements need care. Where heat has parallel routes, such as timber studs through insulation or a frame around a pane, the U × A values of the routes add instead of their resistances, so a small area of poor material carries a disproportionate share. Around a pipe or tube the area grows with radius, so U has to be quoted against a stated area, inside or outside; U × A comes out the same either way.

All three inputs enter linearly, so a 10% change in any one changes the result by 10%. In practice U and ΔT carry the most uncertainty, and U is not perfectly constant: surface coefficients in still air rise somewhat as the temperature difference grows. The equation assumes steady state and a single representative ΔT, entered as a positive difference; the result is the size of the flow from the warmer side to the cooler one.

A worked example.

Example

Take a surface with an overall coefficient U of 1 W/(m²·K), an area of 10 m² and a 20 K difference between the two sides. First U × A = 1 × 10 = 10 W/K: every degree of difference drives 10 watts through the surface. With a 20 K difference, the rate is 10 × 20 = 200 W. Held steady for 24 hours, that is 200 × 24 = 4,800 Wh, or 4.8 kWh of heat to be replaced. A real winter does not hold a steady 20 K, which is what heating degree-days account for: each day adds the number of degrees by which its mean outdoor temperature fell below a chosen base. In a climate with 2,000 degree-days a year, the same 10 W/K surface loses 10 × 2,000 × 24 = 480,000 Wh, or 480 kWh, over the season, before any free heat from sunshine, people or appliances is credited.

transfer Area M210
temperature Difference K20
overall Coefficient Wm2 K1

Frequently asked questions.

Can I enter the temperature difference in degrees Celsius?
Yes. A kelvin and a degree Celsius are the same size, so a difference of 20 °C is exactly 20 K. Only absolute temperatures differ by 273.15 between the scales. Subtract the cold-side temperature from the warm-side one in Celsius and enter the result.
How do I work out U for a wall made of several layers?
Add the thermal resistances and take the reciprocal. Each layer contributes its thickness divided by its conductivity, and each surface adds a film resistance. With 0.2 m of insulation at 0.04 W/(m·K), plus surface coefficients of 10 and 25 W/(m²·K), the total is 0.1 + 5 + 0.04 = 5.14 m²·K/W, so U is about 0.195 W/(m²·K).
Why should a heat exchanger use the log-mean temperature difference?
Because the gap between the hot and cold streams shrinks or grows along the exchanger, and heat flow follows that changing gap. The log-mean, (ΔT₁ − ΔT₂) ÷ ln(ΔT₁/ΔT₂), is the exact average for this. With end differences of 40 K and 10 K it is about 21.64 K, whereas the simple average of 25 K would overstate the duty.
Does the result include radiation?
Only to the extent that U does. Building U-values normally use surface resistances that already combine convection and radiation, so the result covers both. For hot surfaces where radiation dominates, such as a furnace wall or a pipe at several hundred degrees, radiant exchange depends on the difference of the fourth powers of the absolute temperatures, and a single fixed U becomes a poor approximation.
Why does my heating system use more energy than this when it first switches on?
The formula gives the steady rate once temperatures have settled. Bringing a cold room, its furniture and its walls up to temperature takes extra energy stored in their thermal mass, and that warm-up period is not included. It describes the holding load, not the start-up load.
How does fouling change the heat-transfer rate?
Scale, rust or biological film on heat-exchanger surfaces adds an extra thermal resistance in series, which lowers U and therefore the heat transferred for the same temperature difference. Designers include fouling resistances when choosing U, so enter a U that already accounts for them if you want the performance of a unit in service.
How much heat-exchanger area do I need for a given duty?
Rearrange the equation: A = Q̇ ÷ (U × ΔT). For a 50 kW duty with a U of 500 W/(m²·K) and end temperature differences of 40 K and 10 K, the log-mean difference is about 21.64 K and the area comes out at about 4.62 m². The simple average of 25 K would have suggested 4 m², an exchanger too small for the job. If the U you used is a clean-surface value, the unit will also fall short once fouling builds up.
How do I work out the heat loss of a whole room?
Apply the equation to each element separately and add up the U × A products. Say 40 m² of wall at 0.3 W/(m²·K), 6 m² of window at 1.4 and 20 m² of roof at 0.2: that is 12 + 8.4 + 4 = 24.4 W/K, or 488 W at a 20 K difference. The windows are under a tenth of the area but over a third of the loss. Air leaking in and out carries heat as well, and that ventilation loss has to be added separately.
My insulation is rated R-13. How do I turn that into a U-value?
R-13 is a US rating in ft²·°F·h/Btu. Divide by 5.678 to reach SI units: 13 ÷ 5.678 = 2.289 m²·K/W, so the insulation layer alone has U ≈ 0.437 W/(m²·K); framing, sheathing and surface films change the figure for the whole wall. US window labels give a U-factor in Btu/(h·ft²·°F) instead, which you multiply by 5.678, so 0.30 becomes about 1.70 W/(m²·K).
Why does a single pane of glass lose so much heat?
Because the glass itself is barely a barrier; the thin layers of air clinging to each face do the insulating. With the surface resistances building standards use for walls and windows, 0.13 m²·K/W inside and 0.04 outside, and 4 mm of glass at about 1.0 W/(m·K), the total is 0.174 m²·K/W and U ≈ 5.75 W/(m²·K). The glass supplies about 2% of that resistance.

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Q̇ = UAΔT
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