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Heat Loss Calculator

Estimate how fast your home loses heat on the coldest days, through walls, windows, ceiling, floor and air leaks, and the furnace size that covers it.

Level: Grades 6–7

  • Common Core standard 6.EE.A.2.cEvaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations).
Printable worksheet

Heat Loss Calculator

Inputs

A cold but typical winter low for your area, as used in heating design. Not the record low.

Windows are rated by U-factor (lower is better). Single glazing is about 1.0; modern double-pane low-e about 0.3.

Enter 0 for a slab or a floor over a heated basement.

Floor area × ceiling height, for example 1,500 ft² × 8 ft = 12,000 ft³.

About 0.35 for a tight newer home, 0.5–1 for average, 1–2 for a leaky older house.

The furnace's AFUE rating (yearly efficiency). Use 100% for electric resistance heat.

Result

Design heat loss

17,627 BTU/h

At 10 °F outside the house loses about 17,627 BTU/h, so the heating system needs an input of about 18,555 BTU/h.

Heating input needed
18,555 BTU/h
Heat loss in kilowatts
5.17 kW
Through walls
5,538 BTU/h
Through windows and doors
3,240 BTU/h
Through the ceiling
2,368 BTU/h
Through the floor
0 BTU/h
Air leakage
6,480 BTU/h
Where the heat goes
  • Walls5,538.4631.4%
  • Windows and doors3,24018.4%
  • Ceiling2,368.4213.4%
  • Air leakage6,48036.8%

See the step-by-step solution ↓

Step-by-step solution

  1. Find the temperature gap (ΔT): indoor minus outdoor. A bigger gap pushes heat out faster

    70−10=60 °F

  2. Walls: area times the gap, divided by the R-value (more insulation, less loss)

    1,200×6013=5,538.5 BTU/h

  3. Windows and doors: the U-factor times the area times the gap

    0.3×180×60=3,240 BTU/h

  4. Ceiling: area times the gap, divided by the R-value

    1,500×6038=2,368.4 BTU/h

  5. Air leaks: 0.018 times the volume, the air changes per hour and the gap (0.018 BTU heats 1 ft³ of air by 1 °F)

    0.018×12,000×0.5×60=6,480 BTU/h

  6. Add up every part to get the total heat loss

    5,538.5+3,240+2,368.4+6,480=17,626.9 BTU/h

  7. Divide by the 95% efficiency to get the input rating you need, since output is input times efficiency

    17,626.90.95=18,554.6 BTU/h

Test yourself

20 quick questions built from this calculator's formula: work out results, predict what happens when an input changes, and work backward from an answer. Every answer comes with the working.

What this calculator does

Find out how much heat your home loses on the coldest days, and how big a heater you need to keep up. On those days, a heating system has one job: replace the heat as fast as the house loses it. That rate is the design heat loss, measured in BTU per hour (BTU/h). Heat soaks out through every surface between heated and unheated space (conduction). It also escapes as warm air leaks out through gaps and cold outside air takes its place.

The calculator works it out surface by surface. R-value is a rating of how well something resists heat; higher is better. Each wall, ceiling and floor loses its area times the temperature difference, divided by its R-value. Windows and doors are rated by U-factor instead, how easily heat passes through. They lose U times area times the difference. Leaking air loses 0.018 BTU for every cubic foot per degree.

The total is the heat your system must put out. Dividing by its efficiency gives the input rating to look for on the label.

Where it's used

Why this math matters outside the classroom:

  • Furnace replacement. A homeowner finds a 17,600 BTU/h loss and sees that a 95% furnace needs about 18,600 BTU/h of input, far below the old unit's rating.
  • Window upgrades. Swapping 180 ft² of single glazing at U-1.0 for U-0.30 windows cuts the window loss at ΔT 60 °F from 10,800 to 3,240 BTU/h.
  • Air sealing. Energy auditors show that 12,000 ft³ at 0.5 air changes per hour and ΔT 60 °F loses 6,480 BTU/h through leaks alone.
  • Science class. Physics students evaluate Q = A × ΔT ÷ R for a 1,200 ft² wall at R-13 and get about 5,538 BTU/h.
  • Heat pump sizing. Converting a 17,627 BTU/h design load at 3,412.14 BTU/h per kW gives about 5.17 kW of heating capacity needed.

Grade level and standards: Grades 6–7; Common Core State Standards for Mathematics 6.EE.A.2.c. © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.

How to use it

  1. Enter the indoor temperature you want and the outdoor design temperature for your area.
  2. Enter the wall area (minus windows and doors) and its R-value. Then enter the total window and door area with an average U-factor, and the ceiling area and R-value.
  3. If a floor sits over a crawlspace or garage, enter its area and R-value. Otherwise leave it at 0.
  4. Enter the heated volume, an air-change rate for leaks and the heating system's efficiency. Then read the heat loss and the input needed.

What the inputs mean

Outdoor design temperature
Heating design uses a cold temperature that your area stays above for all but a few hours a year. It's not the record low. Local building departments and HVAC references publish it by city.
Wall R-value
Use the whole-wall R-value if you know it. The insulation's rating alone overstates it, because the studs (wall framing) let more heat through. A wall with R-13 batts is often closer to R-11 overall.
Window and door U-factor
U-factor is the flip side of R-value: 1 divided by R. It's printed on window labels. An average across windows and doors is fine for an estimate.
Floor area over unheated space
Enter 0 for a slab or a floor over a heated basement.
Heated volume
Floor area × ceiling height, for example 1,500 ft² × 8 ft = 12,000 ft³.
Air changes per hour (leakage)
Air changes per hour (ACH) is how many times the house's air is swapped each hour. A blower-door test (a fan that measures leaks) gives the best figure. Without one, use the typical ranges in the field hint.
Heating system efficiency
The furnace's AFUE rating (yearly efficiency). Use 100% for electric resistance heat.

Formula and method

ΔT = indoor temperature − outdoor design temperature
Q surface = area × ΔT ÷ R   (or U × area × ΔT for windows and doors)
Q air = 0.018 × volume × ACH × ΔT   (= 1.08 × CFM × ΔT)
Q total = Q walls + Q windows + Q ceiling + Q floor + Q air
heating input = Q total ÷ efficiency;  kW = BTU/h ÷ 3,412.14

In the formulas, ΔT ("delta T") is the temperature difference between inside and outside. Q is the heat lost, in BTU per hour. CFM is airflow in cubic feet per minute.

The heat soaking through a layer grows with its area and with the temperature difference across it. It shrinks as the layer's R-value goes up. In US units, R is in ft²·°F·h/BTU, so area × ΔT ÷ R comes out in BTU per hour.

Air carries about 0.018 BTU per cubic foot per degree Fahrenheit. That comes from air's weight, near 0.075 lb/ft³, times the heat it takes to warm it, near 0.24 BTU/lb·°F. Multiply by the house volume and by how many times its air is replaced each hour to get the leakage loss.

Worked example

A 1,500 ft² single-story house is kept at 70 °F with a design low of 10 °F. It has 1,200 ft² of R-13 wall, 180 ft² of windows and doors at U-0.30, and 1,500 ft² of R-38 ceiling. Its 12,000 ft³ of air changes 0.5 times per hour, and its furnace is 95% efficient.

  1. ΔT = 70 − 10 = 60 °F.
  2. Walls: 1,200 × 60 ÷ 13 = 5,538.5 BTU/h. Windows and doors: 0.30 × 180 × 60 = 3,240 BTU/h.
  3. Ceiling: 1,500 × 60 ÷ 38 = 2,368.4 BTU/h.
  4. Air leakage: 0.018 × 12,000 × 0.5 × 60 = 6,480 BTU/h.
  5. Total: 17,626.9 BTU/h, or 5.17 kW. Input: 17,626.9 ÷ 0.95 = 18,554.6 BTU/h.

Result: The house loses about 17,627 BTU/h on a design day, so the furnace needs an input of roughly 18,555 BTU/h.

Understanding the result

The breakdown shows where upgrades pay off. In the example, air leakage is the largest single part. So sealing air leaks would cut more heat loss than adding ceiling insulation.

Furnaces come in fixed sizes. The usual practice is to choose the smallest one whose output covers the design heat loss. Don't add a large safety margin, because an oversized furnace switches on and off more often.

Assumptions and limitations

  • Conditions hold steady. The outside temperature stays at the design value, and sunshine, people and appliances add no heat.
  • Each surface has one R-value or U-factor. The unheated attic, crawlspace or garage is treated as being at the outdoor temperature.
  • Basement walls and slabs lose heat to the ground, which this simple method doesn't model.
  • Heat lost from ducts in unheated spaces, and ventilation equipment, aren't included.
  • This is an estimate for planning and comparing upgrades. It's not a formal load calculation for sizing equipment. A licensed professional does that, using a recognized method.

Frequently asked questions

What is the difference between R-value and U-factor?

Each is 1 divided by the other: U = 1 ÷ R. R-value measures how well heat is blocked, so higher is better. U-factor measures how easily heat passes, so lower is better. Insulation is sold by R-value, and windows by U-factor.

Why does air leakage matter so much?

Every cubic foot of warm air that escapes is replaced by cold air that must be heated. Swapping half the house's air each hour can match the loss through all the walls. That's why sealing air leaks is often the cheapest improvement.

Is the heat loss the furnace size?

The heat loss is the output needed. Furnaces are labeled by input, and their output is the input times the efficiency. So divide the heat loss by the efficiency to compare with the input rating.

How do I get BTU/h into kilowatts for a heat pump?

Divide by 3,412.14. The example's 17,627 BTU/h is about 5.17 kW. A heat pump's output also drops as it gets colder. So check its capacity at your design temperature.

How this calculator is checked

Method: Steady-state conduction Q = A·ΔT/R (U·A·ΔT for fenestration) plus sensible infiltration load 1.08 × CFM × ΔT, the standard US HVAC air factor. The calculation is covered by automated tests with known reference values and edge cases. Last updated 2026-10-18. Found a problem? Tell us.

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