Heat Loss Calculator

Calculate a room's heat loss in BTU per hour through walls, ceiling, floor, windows and air leakage from R-values, U-factor and design temperatures to size a heater.

By CalcBuilt Editorial TeamUpdated September 9, 2026Formulas checked against manufacturer specs, see how we calculate.

Quick answer: To estimate heat loss, multiply each surface's area by the indoor-to-outdoor temperature difference and divide by its R-value (or multiply by the U-factor for windows), then add infiltration at 0.018 x room volume x air changes per hour x the temperature difference. A 20 x 15 ft room with 8 ft ceilings, R-13 walls, an R-38 ceiling, an R-19 floor, 40 sq ft of U-0.3 windows and a 60 F design difference loses 5,837 BTU per hour, so it needs a 6,500 to 7,100 BTU heater.

Your measurements

One wall: area is wall length times height, minus each door and window.

Room lengthftRoom widthftCeiling heightftWall R-valueR-13 for a 2 x 4 wall, R-19 to R-21 for 2 x 6, about R-4 uninsulated. Enter 0 if the wall faces heated space.Ceiling R-valueR-38 to R-60 attic; 0 if a heated room is above.Floor R-valueR-19 to R-30 over a crawlspace or garage; 0 for a slab on grade or a heated room below (ignored).Window and door areasq ftTotal glazing plus exterior doors; subtracted from the wall area.Window U-factorFrom the NFRC label: single pane about 1.0, double pane 0.5, low-e double 0.3, triple 0.2.Indoor design temperature (F)Outdoor design temperature (F)The 99% winter design temperature for your area, roughly 40 F on the Gulf Coast, 10 F in the Midwest, -10 F in Minnesota.Air changes per hour0.35 for tight new construction, 0.5 average, 1.0 or more for a drafty older room.

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Results

Total heat loss: 5,837 BTU per hour; At a 60 F indoor-to-outdoor design difference
Walls: 2,400 BTU per hour; 520 sq ft of wall at R-13
Windows and doors: 720 BTU per hour; 40 sq ft at U-0.3
Ceiling and floor: 1,421 BTU per hour; Ceiling 474 at R-38, floor 947 at R-19
Air infiltration: 1,296 BTU per hour; 2,400 cu ft at 0.5 air changes per hour
Heater size: 6,500 to 7,100 BTU per hour; 10 to 20% above the load, rounded up (1.91 to 2.08 kW electric). Planning estimate only: a Manual J calculation by an HVAC contractor governs equipment size

Heat Loss formula

Temperature difference = Indoor design temperature - Outdoor design temperature (F)
Conduction (BTU/h) = Area (sq ft) x Temperature difference / R-value  (windows: Area x U-factor x difference)
Wall area = 2 x (Length + Width) x Height - window and door area
Infiltration (BTU/h) = 0.018 x Volume (cu ft) x air changes per hour x Temperature difference

Worked example

A 12 x 12 ft bedroom with 9 ft ceilings on a slab, R-19 walls, an R-49 attic above, 25 sq ft of U-0.35 windows, kept at 68 F when it is 0 F outside, with 0.35 air changes per hour: the difference is 68 F. Wall area is 2 x (12 + 12) x 9 - 25 = 407 sq ft, so the walls lose 407 x 68 / 19 = 1,457 BTU per hour. The ceiling loses 144 x 68 / 49 = 200. The windows lose 25 x 0.35 x 68 = 595. The slab is ignored.

The room is 144 x 9 = 1,296 cubic feet, so infiltration is 0.018 x 1,296 x 0.35 x 68 = 555 BTU per hour. Total: 1,457 + 200 + 595 + 555 = 2,807 BTU per hour. Add 10 to 20% and round up for a 3,100 to 3,400 BTU heater; a 1,000 W electric baseboard delivers 3,412 BTU per hour and covers it.

Quick reference

Heat loss per 100 sq ft of surface at a 60 F temperature difference
Surface	R-value	U-factor	BTU per hour per 100 sq ft
Single-pane window	R-1	1.00	6,000
Double-pane window	R-2	0.50	3,000
Low-e double-pane window	R-3.3	0.30	1,800
Uninsulated 2 x 4 wall	R-4	0.25	1,500
R-13 wall	R-13	0.077	462
R-19 wall or floor	R-19	0.053	316
R-30 attic	R-30	0.033	200
R-38 attic	R-38	0.026	158
R-49 attic	R-49	0.020	122

How to calculate heat loss

Heat leaves a room two ways: conduction through every surface that touches the cold, and air that leaks in and has to be warmed. For conduction, take each surface's area, multiply by the difference between the indoor and outdoor design temperatures, and divide by the R-value. Windows and doors are rated by U-factor instead, so multiply area by U by the temperature difference. Only count surfaces that face outdoors, an unheated attic, garage or crawlspace; a wall shared with a heated room loses nothing, so enter 0.

For infiltration, multiply the room volume by the air changes per hour to get cubic feet of cold air per hour, then by 0.018 and the temperature difference. Air leakage is often a third of the total in an older room, which is why sealing gaps around windows, outlets and the rim joist pays off before adding insulation.

Sizing the heater

The result is the heat a room loses on a design cold day, so a heater that delivers that many BTU per hour holds the set temperature. Round up 10 to 20 percent for recovery after a setback and for windier days than the design assumes, but no more; an oversized furnace or heat pump short-cycles. Electric heat converts at 3,412 BTU per hour per kilowatt, so a 5,837 BTU load is about 1.7 kW, or a 2,000 W baseboard.

This is a single-room, steady-state estimate. It ignores solar gain, internal gains from people and appliances, thermal bridging through studs, duct losses and the way rooms share heat, all of which a full ACCA Manual J calculation accounts for. Use it to compare insulation upgrades and to size a space heater or a single mini split head; let an HVAC contractor's Manual J and local code set the size of a whole-house system.

Key facts

One kilowatt equals 3,412 BTU per hour, so a 1,500 W space heater delivers about 5,100 BTU per hour.
U-factor is the inverse of R-value: an R-13 wall has a U of 0.077, while a single-pane window at U-1.0 loses 13 times as much heat per square foot.
The infiltration constant 0.018 is the density of air (0.075 lb per cubic foot) times its specific heat (0.24 BTU per lb per degree F).
Energy Star Version 7 (2023) requires a window U-factor of about 0.22 in the northern zone, down from 0.27 under the previous version; older double-pane windows are about 0.50 and single-pane about 1.0.
ASHRAE's 99% heating design temperature is the outdoor temperature exceeded 99% of the hours in a year, which is how Manual J picks the design day.

Frequently asked questions

How many BTU do I need to heat a 300 sq ft room?

With R-13 walls, an R-38 ceiling, an R-19 floor, 40 sq ft of low-e windows and a 60 F design difference, about 5,837 BTU per hour, so a 6,500 to 7,100 BTU heater. In a poorly insulated room with single-pane windows the loss can be two to three times higher.

What is the formula for heat loss through a wall?

Q = Area x temperature difference / R-value. A 520 sq ft R-13 wall at a 60 F difference loses 520 x 60 / 13 = 2,400 BTU per hour. Upgrading to R-21 drops that to about 1,486.

How do I convert heat loss to kilowatts?

Divide BTU per hour by 3,412. A 5,837 BTU per hour loss is 1.71 kW, so a 2 kW electric heater covers it with a small margin.

What outdoor temperature should I use?

The 99% winter design temperature for your city, published by ASHRAE and in Manual J: roughly 35 to 40 F on the Gulf Coast, 15 to 20 F in the mid-Atlantic, 0 to 10 F in the Midwest and -10 to -20 F in the northern plains. Do not use the record low.

Sources and references

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