Chilled Drink Calculator

Find out how much ice you need to chill a drink to a target temperature, using real calorimetry — the specific heat of water and the latent heat of fusion of ice.

Quick Facts

Method
Energy balance: heat lost by the drink = heat to melt ice + heat to warm the meltwater
Uses c = 4.186 J/g·°C (water) and L = 334 J/g (ice, at 0 °C). Assumes ice starts at 0 °C and no heat from the surroundings.

Your Results

Calculated
Ice needed
-
Mass of ice at 0 °C that fully melts
Approx. ice by volume
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Loose ice cubes (~0.92 g/mL)
Heat removed
-
Energy pulled from the drink

Ready

Enter your drink volume and temperatures, then calculate.

How much ice does it take to chill a drink?

Cooling a warm drink with ice is a calorimetry problem. Heat flows out of the warm liquid and into the ice, and the ice absorbs that heat in two ways: first by melting (a phase change that soaks up a large amount of energy at a constant 0 °C), and then by warming the resulting meltwater up to the drink's final temperature. This calculator finds the mass of ice needed to bring your drink down to a target temperature, assuming the ice starts at 0 °C and no extra heat leaks in from the glass or the room.

The formula

Setting heat lost by the drink equal to heat gained by the ice gives a clean energy balance:

mdrink · c · (Ti − Tf) = mice · L + mice · c · (Tf − 0)

Solving for the ice mass:

mice = mdrink · c · (Ti − Tf) / (L + c · Tf)

where c = 4.186 J/g·°C is the specific heat of liquid water, L = 334 J/g is the latent heat of fusion of ice, Ti is the drink's starting temperature, and Tf is the target temperature. Volume in millilitres is treated as grams because water is very close to 1 g/mL.

Why the ice does most of the work

The striking part of this problem is how much energy melting absorbs. Turning 1 g of ice into 1 g of 0 °C water takes 334 joules — the same energy would warm that gram of water by about 80 °C. That is why a small amount of ice cools a much larger volume of drink: most of the heat is consumed by the phase change, not by warming the meltwater. For a 355 mL soda at 25 °C chilled to 4 °C, you need only about 89 g of ice — roughly a quarter of the drink's volume, or two to three standard cubes.

Common reference points

  • 355 mL (12 oz can), 25 °C → 4 °C: ≈ 89 g of ice
  • 500 mL bottle, 20 °C → 5 °C: ≈ 88 g of ice
  • 250 mL glass, 22 °C → 4 °C: ≈ 54 g of ice
  • A standard ice cube from a home tray: ≈ 25–30 g, so 89 g is about three cubes
  • Latent heat of fusion of ice: 334 J/g (≈ 80 cal/g) — the energy released when water freezes, absorbed when it melts

Frequently Asked Questions

Does this account for the ice being colder than 0 °C?
No. Freezer ice is often around −18 °C, and warming it from there to 0 °C absorbs a little extra heat (about 2.1 J/g·°C × 18 °C ≈ 38 J/g, roughly 11% more cooling power per gram). This calculator assumes ice at 0 °C, which is a slightly conservative estimate — you may need marginally less freezer ice than the number shown.
Will all the ice melt?
The formula gives the exact mass of ice that fully melts to reach the target temperature with no ice left over. If you add more ice than this, some will remain unmelted and the drink will settle near 0 °C (diluted). If you add less, the drink won't reach the target. For a drink you want to stay cold, adding extra ice is fine — it just holds the temperature near freezing at the cost of dilution.
Why does a bottle at a lower temperature sometimes need almost the same ice?
Because ice needed scales with both the volume and the temperature drop. A 500 mL drink going 20 → 5 °C (15° drop) and a 355 mL drink going 25 → 4 °C (21° drop) both land near 88–89 g: the larger volume of the first offsets its smaller temperature drop. The calculator handles that trade-off exactly.

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