Water heating energy and time

Estimate the energy and time to heat a measured water volume, including an explicit heater efficiency. Enter litres, initial and target temperatures, efficiency and input power.

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How it works

Formula

$Heat delivered to water = \frac{Water volume \times 4.186 \times (Target water temperature - Starting water temperature)}{3,600}$
$Input energy required = \frac{Heat delivered to water}{\frac{Heating efficiency}{100}}$
$Estimated heating time = \frac{Input energy required}{Input heater power} \times 60$

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Sources

OpenStax — temperature change and heat capacitySources reviewed: 2026-10-30

Assumptions and limitations

Liquid water is approximated as 1 kg/L with heat capacity 4.186 kJ/(kg·°C). Constant input power and efficiency; no boiling, evaporation, heat-pump COP or vessel heat capacity. Not a thermostat setting recommendation.

How to use this tool

Set up the inputs

Enter litres, initial and target temperatures, efficiency and input power. The target must not be lower than the initial temperature.

Worked examples

Estimate heating time after efficiency losses

50 L × 4.186 × (55 − 15) / 3,600 = 2.325556 kWh; / 0.90 = 2.583951 kWh; / 2 kW × 60 ≈ 77.5185 min.

Frequently asked questions

How should I read the result from Water heating energy and time?

Useful heat goes into the water. Input energy also includes modeled losses; dividing by input power gives the estimated heating time.

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The calculation or transformation runs in your browser. The tool does not send its input values or generated output to an application server. Loading a page still makes infrastructure requests; optional analytics and Google consent messages have separate controls.

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Consult the linked sources for definitions and reference data. Contact Support if you have a question or notice a problem.

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Water heating energy and time

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