Water Cooling Calculator
Work out the heat a volume of water must lose to drop from one temperature to another, and how many minutes that takes at a steady cooling rate in watts.
Water Cooling Calculator
Background.
Hot water stores a surprising amount of heat. Every litre gives up about 4.186 kilojoules for each degree Celsius it falls, so a 100-litre tank cooling from 80 °C to 30 °C has to shed 20,930 kJ, close to 5.8 kilowatt-hours. That is the same energy a 2 kW kettle draws in almost three hours of continuous running.
How long the cool-down takes depends on one thing the water cannot tell you: how fast heat is leaving it. Here you supply that as an average net rate in watts, meaning joules per second actually pulled out. "Net" matters. A chiller rated at 3 kW that is fighting 500 W of heat from a circulating pump and a warm room is only removing 2.5 kW from the water.
The time is then simply energy divided by rate, which assumes the rate stays constant. That is a fair description of a chiller or heat exchanger working at steady capacity. It is a poor description of a pot left to cool on its own, where heat loss shrinks as the water approaches room temperature. With a constant rate, the drop from 80 °C to 55 °C takes exactly as long as the drop from 55 °C to 30 °C; on a kitchen counter in a 25 °C room, the second half would drag on far longer.
A few other things sit outside the model. Steam leaving an open surface carries heat, but the water lost to it is not tracked. The tank or pot holds heat of its own, which also has to come out. Nothing here handles freezing. And water is counted at 1 kg per litre, which overstates the mass slightly for hot water: at 80 °C a litre weighs about 0.97 kg.
What is water cooling calculator?
The quantity is sensible heat: the energy water gives up while its temperature falls and it stays liquid, as opposed to latent heat absorbed or released when it freezes or boils. It equals mass times specific heat times the temperature drop. Dividing that energy by a cooling rate gives an idealized time, idealized because the rate is held fixed for the whole run instead of tapering as the water nears its surroundings.
How to use this calculator.
- Enter the volume of water in litres. Use the amount actually in the vessel, not its nominal capacity; one US gallon is 3.785 litres.
- Enter the starting and target temperatures in degrees Celsius. The target has to be lower than the start, and both should stay above 0 °C, because nothing here accounts for the extra heat released when water freezes.
- Enter the average net heat-removal rate in watts. A chiller rated in BTU per hour converts at 0.293 W per BTU/h, and one ton of refrigeration is about 3,517 W. Subtract any steady heat you know is flowing back in, such as pump heat or a heater left on.
- Read the time in minutes and the heat removed in kilojoules. Divide the kilojoules by 3,600 for kilowatt-hours. If the water is cooling passively toward room temperature, treat the time as a best case.
The formula.
The heat removed is Q = V × ρ × c × (T_start − T_target), where V is the volume in litres, ρ is the density taken as 1 kg per litre, and c = 4.186 kJ per kilogram per kelvin is the specific heat of liquid water. A one-degree Celsius change is the same size as a one-kelvin change, so the temperature difference can be read in either. The true specific heat of liquid water moves by less than 1% between freezing and boiling, so the fixed 4.186 costs very little accuracy; the fixed density matters more for hot water. The time is t = Q × 1,000 ÷ P ÷ 60: multiplying by 1,000 turns kilojoules into joules, dividing by the rate P in watts (joules per second) gives seconds, and dividing by 60 gives minutes. Every input acts proportionally. Twice the water or twice the temperature drop doubles the time; twice the cooling power halves it.
A worked example.
Take 100 litres of water at 80 °C that need to come down to 30 °C, with heat being removed at an average of 2,000 W. At 1 kg per litre the water has a mass of 100 kg, and the temperature has to fall by 80 − 30 = 50 °C. The heat to remove is 100 × 4.186 × 50 = 20,930 kJ, or 20,930,000 joules. At 2,000 joules per second that takes 20,930,000 ÷ 2,000 = 10,465 seconds, and 10,465 ÷ 60 = 174.42 minutes, just under 2 hours 54 minutes. Halve the cooling power to 1,000 W and the same job takes 348.83 minutes. Because the rate is held constant, the first 25 degrees (80 °C to 55 °C) account for exactly half the time, 87.21 minutes; a tank left to cool on its own would cover that first stretch much faster than the second.
Frequently asked questions.
Why does water left on the counter cool more slowly than this predicts near the end?
How can I find the heat-removal rate if I have no chiller rating?
Can I use this to find how long water takes to freeze?
How much ice would cool the water by the same amount?
Does the pot or tank itself change the answer?
References& sources.
- [1]OpenStax, Rice University. University Physics, 2016. Volume 1 mechanics and Volume 2 thermodynamics. Retrieved 2026-08-06.
- [2]National Institute of Standards and Technology. NIST SI Guide, 2008 edition. Appendix B unit relationships. Retrieved 2026-08-06.
- [3]U.S. Department of Energy. Energy Saver current 2026. Household energy guidance. Retrieved 2026-08-06.
How this page was produced
- Published by
- Quanta Calculator
- Overseen by
- Daniel, Quanta Calculator Researcher
- Primary sources
- 3 cited below
- Method
- Q=m*c*deltaT with water density 1 kg/L and c=4.186 kJ/kg K; time=Q/power
- Published
- Last verified
Built with AI assistance and verified by automated tests against the cited sources — every worked example on this page is computed by the same code that runs the calculator. How we build and check calculators.
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