How it works
A gallon of water weighs about 8.34 lb, and one BTU raises one pound of water one degree Fahrenheit. Multiply the gallons, 8.34, and the temperature rise for the heat the water needs. One kilowatt-hour is 3,412.14 BTU.
Divide that heat by the rate the heater actually delivers (input rating times efficiency) to get the time. The estimate ignores heat lost through the tank and pipes while heating, so real times run somewhat longer.
Worked example
40 gallons heated from 50°F to 120°F with a 4,500 W element:
- Heat = 40 gal × 8.34 lb/gal × (120 − 50) °F = 23,352 BTU
- In kWh = 23,352 ÷ 3,412.14 = 6.84 kWh
- Time = 23,352 ÷ (15,354 BTU/hr × 98%) = 1.55 hours
| Input | Value |
|---|---|
| Water to heat | 40 gal |
| Starting temperature | 50 °F |
| Target temperature | 120 °F |
| Heater input rating | 15354 BTU/hr |
| Efficiency | 98 % |
| Result | Value |
|---|---|
| Heating time | 1.55 hours |
| Heat needed in the water | 23,352 BTU |
| Same energy in kWh | 6.84 kWh |
| Energy the heater must consume | 23,829 BTU |
Assumptions and limits
- Plain water with a constant specific heat of 1 BTU/lb·°F; it varies slightly with temperature.
- Heat loss from the tank and pipes is ignored, so real heating takes a bit longer.
- The heater delivers its rated input continuously. Thermostat cycling, element wear, and supply voltage affect real output.
Common questions
How many BTU does it take to heat a gallon of water?
8.34 BTU for each degree Fahrenheit. Raising a gallon by 70°F takes about 584 BTU.
How many kWh is that?
Divide the BTU by 3,412.14. Heating 40 gallons by 70°F takes about 6.8 kWh.
How do I find the element wattage in BTU/hr?
Multiply watts by 3.412. A 4,500 W element is about 15,354 BTU/hr.
Why does my heater take longer?
Tank and pipe losses, lower supply voltage, and worn elements slow it down.
Sources
- Specific heat of water: 1 BTU/lb·°F; 8.34 lb per US gallon.
- 1 kWh = 3,412.14 BTU (NIST Handbook 44 Appendix C).
Updated 2026-09-30