Water heating time calculator
How long it takes to heat water at a given power.
Fill in the fields and the result will appear here automatically.
Relate liquid-water volume, temperature rise and heater power to heating time. For 100 L from 10 to 60 °C, useful heat is 5.814 kWh; at 95% efficiency the source supplies 6.120 kWh, so 2 kW takes 3.060 hours. The two energy quantities are shown separately. The retained model takes 1 L≈1 kg and constant heat capacity 4186 J/(kg·°C); ice, steam, tank heating and changing losses are excluded.
How it works
Formula and logic
For liquid water assume 1 L≈1 kg and constant c≈4186 J/(kg·K). Useful heat Q=V×c×(T_end−T_start), η=efficiency/100 and useful power=P×1000×η watts. Time=Q/(P×1000×η) seconds. Energy reports Q/3600000 kWh; Source energy reports Q/(η×3600000), the value for costing. Domain 0≤T_start<T_end≤100°C at approximately ordinary pressure, without melting, vaporisation or vessel heating.
Example
A hundred litres from 10 to 60 degrees at 2 kW and 95 % takes 3.06 hours. Useful heat is 5.813889 kWh; at 95% the source supplies 6.119883 kWh. At 100% they coincide.
Fields and units
- Water volume, L — L
- Start temperature, °C — °C
- Target temperature, °C — °C
- Heater power, kW — kW
- Efficiency, % — unitless
How to use
- — Enter litres of liquid water; the model approximates 1 L as 1 kg.
- — Set temperatures from 0 to 100 °C, final above initial; freezing and steam are excluded.
- — Enter source power and a consistent fixed efficiency, not a percentage chosen solely by fuel type.
- — Use source energy for cost, not useful heat; temperature maintenance is excluded.
Method and limitations
- Calculation method
- Formula and logic
- Data or methodology source
- OpenStax College Physics 2e: Q=mcΔT without phase change; 4186 J/(kg·K) as an approximate water value NIST: water freezing near 0 °C and boiling near 100 °C; K and °C temperature differences match
- Limitation
- Domain 0≤T_start<T_end≤100°C at approximately ordinary pressure, without melting, vaporisation or vessel heating.
FAQ
Why is heating water so energy-hungry?
The model uses 4186 J per kg per degree. Thus 100 L and 50 °C require 20.93 MJ of useful heat, or 5.814 kWh. This is a chosen constant for liquid water, not its exact heat capacity at every temperature.
What does twice the power buy?
With the same volume, temperature rise and efficiency, model time halves exactly. Useful heat and source energy stay unchanged. Actual losses can depend on duration; the tool does not assess connection requirements.
Should cooling be accounted for?
Losses over time are not calculated separately. Efficiency applies one fixed fraction of power throughout heating. Do not assume it is sufficient for an open vessel; tank heat, cooling and temperature maintenance may need a separate model.
Why is a gas heater less efficient?
Some energy may leave with exhaust or through the housing, but the fraction depends on the device and operating conditions. No universal fuel-based efficiency is assigned; condensation and fuel heating value are not modelled. Enter consistent data for the actual heating run.