How much heat does your house lose?

Common questions

How do you calculate heat loss in a house?

Heat loss through each part of the building envelope is Q = U × A × ΔT, where U is the assembly's thermal conductance (1 ÷ R-value), A is the area and ΔT is the indoor–outdoor temperature difference. You calculate that for walls, ceiling, windows and floor separately, then add infiltration heat loss (0.33 × ACH × volume × ΔT using the ASHRAE sensible infiltration formula), and sum all four paths. The calculator above does all of that automatically once you enter the floor area, insulation level and temperature difference.

What is the formula for heat loss through a wall?

Q (watts) = U × A × ΔT, where U = 1 ÷ R (the R-value in SI is RSI; in imperial R the U-value is 5.678 ÷ R), A is the net wall area in m², and ΔT is the temperature difference in kelvin (equivalent to °C). In imperial: Q (BTU/hr) = (1 ÷ R) × A (ft²) × ΔT (°F). A 2×4 stud wall insulated to about R-13 and 20 m² of net wall area with a 20°C temperature difference loses roughly 2×4 stud wall: U = 5.678 ÷ 13 ≈ 0.437 W/m²·K, so Q ≈ 0.437 × 20 × 20 ≈ 175 W.

What is a typical heat loss for a house?

A typical 150 m² (1,600 sq ft) two-storey house with average insulation and a 20–30°C (35–55°F) indoor–outdoor temperature difference loses roughly 4–9 kW (14,000–30,000 BTU/hr). Older or poorly insulated houses can reach 10–15 kW; well-insulated newer homes may need only 3–5 kW. The dominant variables are floor area, climate (temperature difference), and how well the envelope is insulated against conduction and air infiltration.

How does R-value affect heat loss?

Heat loss through a building assembly is directly proportional to its U-value, which is 1 ÷ R-value. Doubling the R-value halves the U-value and halves the conductive heat loss through that assembly. So going from R-13 to R-26 walls cuts wall conduction losses in half — but because infiltration, windows and ceiling all have separate heat-loss paths, the saving on the whole-house total is smaller than halving one component alone. Use the calculator above to see the breakdown and how each change affects the total.

What size furnace or heat pump do I need?

A rough starting point is your total calculated heat loss multiplied by a safety factor of about 1.2, rounded up to the next available equipment size. So a house with 10 kW of calculated heat loss points to a 12–12.5 kW unit. For an accurate sizing, a licensed HVAC contractor performs a Manual J load calculation, which accounts for local climate data, window orientation, shading, thermal mass and duct losses — the calculator here gives a planning estimate, not a design figure.