Electric Space Heating Calculator
Enter room size, insulation and design temperatures to size an electric heater and its branch circuit.
Step-by-step
- Enter valid values to begin.
The insulation coefficients are typical planning values for a whole room (conduction plus infiltration), not measured U-values; a proper load calculation sums every surface and air change. The area rule ignores ceiling height and climate. The heater is assumed to be 100% efficient, resistive and a continuous load. Breaker and conductor results are typical copper 75 °C figures from the 15/20/30/40/55 A column; verify against NEC 310.16, termination ratings, ambient and local code before installing.
Electric Space Heating: Sizing the Heater and the Circuit
Core Engineering Principles
A room loses heat by conduction through walls, roof, floor and glass, and by infiltration of cold air through gaps. Both scale with the indoor-outdoor temperature difference, so the heater must replace ΔT times a loss coefficient. We fold the envelope into one number, watts per cubic metre per kelvin: 1.0 for poor, 0.7 average, 0.45 good, 0.30 excellent. Those are typical planning values, not measured U-values. A real calculation sums U × area for every surface plus air changes, and big windows or a leaky floor push a room above our table.
Resistance heat is simple: every watt drawn becomes a watt of heat, so kW in equals kW out, and BTU/hr is 3.412 times the watts. What varies is delivery. Baseboards are limited by watt density and clearances, and thermostats must be rated for line voltage and current. For the circuit the rule is blunt: heating is a continuous load, so the branch circuit carries 125% of the heater current. A heat pump moves two to four times more heat per kWh, so resistance heat suits small, occasional or backup duty.
I = P / Vsupply (single-phase) • Icircuit = 1.25 × I
Area rule: P ≈ 100 W/m² × floor area (about 10 W/ft²)
NEC & Standard References
NEC Article 424 covers fixed electric space heating. 424.3(B) requires branch-circuit conductors and overcurrent devices at not less than 125% of the heater load, and 424.22 addresses overcurrent protection of the equipment. NEC 210.19 and 210.20 set conductor and overcurrent sizing for continuous loads. Conductors come from NEC 310.16; the copper 75 °C picks here (14/12/10/8/6 AWG at 15/20/30/40/55 A) are typical and need checking against terminations. IEC 60335-2-30 covers household room heaters and ASHRAE the load-calculation methods. Check the adopted NEC edition.1. Volume = 5 × 4 × 2.6 = 52 m³; ΔT = 21 − (−5) = 26 K.
2. P = 52 × 26 × 0.7 = 946.4 W.
3. With margin: 946.4 × 1.10 = 1041 W, which is 1.04 kW or 3552 BTU/hr.
4. I = 1041 / 230 = 4.53 A.
5. Circuit at 125% = 4.53 × 1.25 = 5.66 A, so the next standard breaker is 15 A and 14 AWG copper is the typical minimum.
6. Area rule check: 20 m² × 100 W/m² = 2000 W, nearly double.
- Do not skip the 125%. A heater runs for hours, and a breaker loaded to 100% runs hot and trips on cold mornings.
- Check thermostat ratings. A low-voltage or 120 V thermostat on a 240 V heater fails at the contacts.
- Keep clearances. Curtains and furniture too close to a heater are a common ignition source.
- Design for the cold snap. Undersized heaters lag at the worst moment, and cold-load pickup after an outage is large.
- Consider a heat pump if the heater runs most of the winter.