Ice for Cooling a Drink Calculator
Written by Thierno Sadou Diallo, formula verified per our methodology • Last checked on 10/11/2026
A drink's final temperature after cooling with ice depends both on the heat given up by the drink and the energy absorbed by the melting ice. For 300 g of drink at 20°C with 30 g of ice, the final temperature is about 10.9°C.
Explanation
Cooling a drink with ice cubes involves two distinct physical phenomena, unlike simply mixing two liquids at different temperatures (see our liquid mixing temperature calculator): the ice must first melt entirely, which consumes a fixed amount of energy called the latent heat of fusion (334 joules per gram of ice, regardless of temperature), and then the water from that melting, starting at 0°C, warms up to the final common temperature with the rest of the drink. This energy needed for melting is given up by the drink itself, which cools by that same amount. The full calculation therefore adds up these two energy needs (melting the ice, then warming the melt water) and sets them equal to the energy released by the drink cooling down, to work out the final equilibrium temperature. An important edge case arises when there's too much ice relative to the drink being cooled: the drink then doesn't have enough energy to melt all the ice, and the final equilibrium settles at 0°C with leftover unmelted ice floating in the drink, rather than artificially dropping below that point. This calculator detects this case and shows 0°C instead of a physically impossible result.
Example: 300 g of drink at 20°C, 30 g of ice
Inputs
Drink: 300 g at 20°C. Ice added: 30 g.
Calculation
Available energy = 300 × 4.186 × 20 = 25,116 J. Melting energy needed = 30 × 334 = 10,020 J. Final temperature = (25,116 − 10,020) ÷ (4.186 × 330) ≈ 10.93°C.
Result
This drink reaches a final temperature of about 10.9°C once the ice has fully melted.
Frequently asked questions
Why does ice cool more effectively than cold water of the same mass?
Because ice absorbs extra energy to melt (the latent heat of fusion, 334 J/g) before it even starts warming up, while water already liquid at 0°C only has that second step to go through. For the same mass at 0°C, ice therefore draws much more heat from the drink than liquid water does, which explains why ice cubes cool more effectively than simply adding cold water.
What happens if I add too much ice?
If the amount of ice exceeds what the drink's heat can melt, not all the ice melts: the drink settles at 0°C, with leftover ice cubes floating without melting further (as long as no outside heat is added). This is exactly what happens with a bucket of ice cubes, where a drink stays roughly at a constant temperature for as long as unmelted ice remains.
Does this calculation assume the ice cubes come out of the freezer at 0°C?
Yes, as a simplification: ice cubes straight out of a home freezer are actually colder (often around -18°C), which lets them absorb a bit more energy before even starting to melt, a nuance a full specific heat calculator would capture. This gap generally stays small compared to the latent heat of fusion itself, and the 0°C ice assumption gives a reliable estimate for everyday use.