Audited 05 Aug 2026·Last updated 28 Oct 2026·3 citations·Tier 2·0 uses

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

L
C
C
W
Idealized cooling time
174.4167
Minutes needed to remove the heat at the constant average rate you entered, assuming that rate holds the whole way down.
Sensible heat removed
20,930kJ

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.

  1. 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.
  2. 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.
  3. 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.
  4. 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.

Q=m*c*deltaT with water density 1 kg/L and c=4.186 kJ/kg K; time=Q/power

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.

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.

water Litres100
average Cooling Power Watts2,000
start Temperature C80
target Temperature C30

Frequently asked questions.

Why does water left on the counter cool more slowly than this predicts near the end?
Passive cooling follows Newton's law of cooling: heat leaves in proportion to the gap between the water and the room. Hot water sheds heat quickly, but as it nears room temperature the gap and the rate both shrink, so the last few degrees take a long time. A single constant rate cannot capture that, so for passive cooling the answer is optimistic.
How can I find the heat-removal rate if I have no chiller rating?
Measure it. Note how far the temperature falls over a timed interval, then compute litres × 4.186 × degrees dropped × 1,000 ÷ seconds. If 100 L falls 2 °C in 10 minutes, the rate is about 1,395 W. Measure near the middle of the range you care about, since passive rates fall as the water cools.
Can I use this to find how long water takes to freeze?
No. Turning water at 0 °C into ice releases about 334 kJ per kilogram, nearly 80 times the heat given up by cooling the same water one degree. That latent heat is not included, so a target below 0 °C produces a time far shorter than reality. Keep both temperatures above freezing.
How much ice would cool the water by the same amount?
Each kilogram of ice at 0 °C absorbs about 334 kJ as it melts, then more as the meltwater warms to the final temperature. For the 100 L example ending at 30 °C, that is 334 + 4.186 × 30 = 459.58 kJ per kilogram, so roughly 45.5 kg of ice, ignoring losses and remembering the meltwater adds to the volume.
Does the pot or tank itself change the answer?
It can. Only the water's heat is counted, but a thick steel pot, glass carboy or masonry tank is hot too and releases its own stored heat as it cools, which lengthens the real time. The container's shape matters through the rate instead: a wide, shallow vessel loses heat faster than a tall narrow one.

How this page was produced

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Quanta Calculator
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

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