๐ง Ice to Cool a Drink Calculator
An energy balance using water's specific heat (4.186 J/(gยทยฐC)) and ice's latent heat of fusion (334 J/g).
Enter a drink's mass, current temperature, and target temperature to calculate how much ice is needed, based on an energy balance: the heat the drink loses equals the heat absorbed by the ice melting plus the heat absorbed warming the melted water.
How to use
- Enter the drink's mass (g; roughly 1 mL โ 1 g).
- Enter the drink's current temperature (ยฐC).
- Enter the target temperature you want to cool it to (ยฐC).
- The amount of ice needed is calculated automatically.
How the calculation works
Calculate how much ice you need to cool a drink to a target temperature, using a heat balance. Heat the drink loses = drink weight ร 4.186 ร (current temperature โ target) Heat 1 g of ice absorbs = 334 (to melt) + 4.186 ร target temperature (to warm the meltwater) Ice needed (g) = heat the drink loses รท heat per gram of ice 4.186 is the specific heat of water and 334 the heat ice absorbs as it melts (the latent heat of fusion). The ice is taken as 0 ยฐC and the drink as having the properties of water. The target must be at least 0 ยฐC and below the current temperature.
Worked example
300 g of drink, 25 ยฐC โ 5 ยฐC: about 71 g of ice 500 g of drink, 30 ยฐC โ 4 ยฐC: about 155 g of ice The heat absorbed by melting ice (334 J/g) is as much as it takes to warm water by 80 ยฐC, which is why a little ice cools a lot.
Things to be aware of
- Ice straight from the freezer is colder than 0 ยฐC, so slightly less is needed than calculated.
- Heat also comes in from the glass and the air, so in practice you may need a little more.
- Melted ice dilutes the drink. To avoid that, use chilled glasses or ice packs as well.
FAQ
What's the formula?
The heat lost by the drink (mass ร specific heat of water ร temperature drop) is set equal to the heat absorbed by the ice melting (mass ร latent heat of fusion) plus the heat absorbed warming the melted water to the target temperature. This uses the well-known physical constants: specific heat of water = 4.186 J/(gยทยฐC), latent heat of fusion of ice = 334 J/g.
Does this account for the ice's starting temperature?
This tool assumes the ice starts at 0ยฐC (its melting point). Ice straight from a freezer is colder than that, so the actual amount needed may be slightly less.
Can the target temperature be below 0ยฐC?
No. The formula assumes the ice has fully melted (result at or above 0ยฐC), so enter a target temperature at or above 0ยฐC and below the current temperature.