Estimate a single heating event for a fixed volume of liquid water. This is useful for an immersion heater or a tank reheating comparison when no water is being drawn. It does not model a continuously refilling tank, heat-pump performance, or steady-state heat loss over long periods.
How to use
- Enter the water volume and its initial and desired temperatures.
- Enter actual heating input power and an estimated fraction of power reaching the water.
- Review useful heat, purchased electricity, time and the cost at your tariff.
How the calculation works
t = mcΔT ÷ (Pη), using consistent energy and power units. c = 4.186 kJ/kg·°C; η is efficiency as a fraction. Purchased electricity = useful heat ÷ η.
Example
Heating 50 liters from 15°C to 60°C requires about 2.616 kWh of useful heat. A 3 kW resistance heater at an assumed 90% efficiency takes about 58 minutes and uses 2.907 kWh.
Factors affecting the result
- Incoming cold water and simultaneous hot-water use extend tank recovery.
- Tank metal, pipework and changing heat losses are combined only approximately in the efficiency input.
- Heat pumps require a COP-based model; their delivered heat may exceed electrical input.
Frequently asked questions
Can this size a tankless heater?
No. A tankless unit is sized for a continuous flow and temperature rise. This calculator heats a fixed batch of water.
Is this suitable for heating ice or making steam?
No. Melting and vaporization require latent heat, which is not included. Use it only for warming liquid water within its liquid range.
Sources & methodology
Calculation basis: Physics-based calculation with estimated efficiency. Methodology
- Specific Heat Capacity and Water — U.S. Geological Survey
Last updated: September 15, 2026