Latent Heat: Ice in a Drink

Set the drink, set the ice, and see where the two end up. The heat the drink gives up is split three ways: warming the ice, melting it, warming the meltwater. Watch which part does the work.

Worked example — Latent Heat: Ice in a Drink

Final temperature (°C)	4.5
Total solid ice (g)	0.0
Ice that lands on 0 °C (g)	67.6
Ice that freezes the drink (g)	3,331
Heat spent on melting	85.6%
Same job in 0 °C water (g)	1,045

These are the values the calculator shows when the page opens. Change any input above to recompute them.

Loading interactive simulation...

Between 68 g and 3.3 kg of ice, the answer is the same number 🖖

Drag the ice slider on the opening glass. The temperature falls, reaches 0.0 °C at 67.6 g, and then stays there. At 100 g, at 400 g, at 900 g the card still reads 0.0 °C. Every gram beyond the first 67.6 spends the drink's remaining heat on melting, and melting happens at one fixed temperature, so nothing is left over to move a thermometer with. The plateau closes at 3,331 g, where 300 g of water no longer carries enough warmth to melt anything and some of it freezes instead. Across a factor of 49 in the ice you add, the tool prints the same number. All that changes is how much ice is still floating when you look.

A gram of ice is worth twenty grams of ice-cold water, and the twenty collapses as the drink warms 🖖

Fifty grams of freezer ice takes the opening glass to 4.5 °C. Reaching 4.5 °C with water that is already at 0 °C takes 1,045 g, which the tool prints beside it: twenty to one, and more than three times the mass of the drink itself. The budget bar says why. Melting draws 334 J from every gram of ice, while a gram of 0 °C water can only take 4.186 J for each degree it warms, and here it warms by four and a half. Now push the drink to 80 °C without touching the ice. The equivalent-water figure drops to 129 g, barely twice the ice, because chilled water now has eighty degrees of warming to sell instead of four. Ice wins by its largest margin in exactly the drink you most want to keep cold.

Eighteen degrees of freezer cold buys 1.3 degrees of drink 🖖

Ice at −18 °C feels far colder than ice at 0 °C, and in the glass it is worth almost nothing extra. Switch between the first two presets: 4.5 °C against 5.7 °C. Ice is a poor heat store, 2.09 J per gram per degree against liquid water's 4.186, so hauling 50 g of it from −18 °C up to its melting point absorbs 1,881 J, next to the 16,700 J the melting then absorbs. The first segment of the budget bar is that tenth. The exchange rate is m_ice·c_ice / (m_total·c_water) = 0.071 °C of drink per degree of freezer, and it is a straight line all the way down: at −30 °C, the coldest the slider goes, the glass reaches 3.6 °C instead of 4.5 °C.

Problems solved in full

Freezer ice for a 250 g mug of tea at 80 °C 7 steps
A 250 g mug of tea at 80 °C, and you want to drink it now, at 55 °C. How much freezer ice does that take, and how much does the mug itself change the answer?
1. Start with the tea, which needs nothing about ice. It has to shed a fixed quantity of heat: its mass, times the specific heat capacity of water, times the drop you are asking for.
  
  $Q = m c Δ T = 250 \times 4.186 \times (80 - 55) = 26, 163 J$
2. Now price one gram of ice. It does three jobs on the way to becoming tea at 55 °C: it warms from −18 °C up to its melting point, it melts, and then the meltwater warms all the way to 55 °C. The third job is the one that gets forgotten, and at this temperature it is the second largest of the three.
  
  $2.09 \times 18 + 334 + 4.186 \times 55 = 601.85 J g^{- 1}$
3. Divide the one by the other. That works in a single step only because the final temperature was fixed at 55 °C before we began. Had the question asked for the temperature instead, the meltwater's share would depend on the answer, and the algebra would have to be solved rather than evaluated.
  
  $m_{ice} = \frac{26 , 163}{601.85} = 43.5 g$
4. Check it. Set the tool to 250 g at 80 °C with 43.5 g of ice at −18 °C, and the final-temperature card reads 55.0 °C.
  
  $T_{f} = 55.0^{\circ} C$
5. Compare the lazy alternative. Water already at 0 °C does only the third job, 230.23 J per gram, so it takes 113.6 g of it. Same temperature in the mug, 70 g more water in the tea.
  
  $\frac{26 , 163}{4.186 \times 55} = \frac{26 , 163}{230.23} = 113.6 g$
6. Here is what the page in front of you does not model: it gives the container no heat capacity at all. A 350 g stoneware mug sitting at 80 °C has to come down to 55 °C as well, and ceramic holds about 0.84 J per gram per degree.
  
  $350 \times 0.84 \times 25 = 7, 350 J$
7. So the real mug needs 28% more ice than the tea alone asked for. The stoneware is 294 J per kelvin against the tea's 1,047, which is 22% of the thermal mass on the table and absent from every number above this line.
  
  $m_{ice} = \frac{26 , 163 + 7 , 350}{601.85} = 55.7 g \frac{294}{294 + 1 , 047} = 22%$
Answer

43.5 g of ice in a container that stores no heat, and about 56 g in a real mug. The tool prints the first and cannot print the second. This is why a calorimetry experiment begins by measuring the calorimeter: the mug's 294 J per kelvin is 70 g of water in disguise, and that quantity has a name, the water equivalent of the apparatus. Anyone who has tipped 43 g of ice into strong tea and found it still too hot to drink has measured it the hard way.
A cool box with 900 g of freezer ice, priced in hours instead of degrees 7 steps
Press A cup packed in ice: 900 g of ice at −25 °C around 120 g of water at 4 °C. The panel says −2.3 °C, an ice mass of 1,020 g and a latent share of 0.0%. Work out how much heat the box can absorb before any of that is liquid again, then decide whether the extra freezer cold was worth having.
1. The ice card reads more than you put in, so start there. For the mixture to settle at 0 °C the ice would have to be brought up from −25, which takes 47,025 J. The drink can supply 2,009 J by cooling to 0 and another 40,080 J by freezing solid — and even both together fall short. So it freezes, all 120 g of it, and the ice mass on the card is everything in the box.
  
  $120 \times 4.186 \times 4 + 120 \times 334 = 42, 089 J < 900 \times 2.09 \times 25 = 47, 025 J$
2. Everything is now solid, and the shortfall of 4,936 J has to be paid by the whole 1,020 g cooling below zero. Ice stores 2.09 J per gram per degree, so divide.
  
  $T_{f} = \frac{42 , 089 - 47 , 025}{1 , 020 \times 2.09} = \frac{- 4 , 936}{2 , 131.8} = - 2.3^{\circ} C$
3. That is where the temperature story ends and the useful one begins. Nothing melted, so the latent-share card reads 0.0% and the budget bar has a single segment; and no mass of chilled water could have got a drink below freezing, which is why the equivalent-water card reads ∞. Three cards saying the same thing: this is not a drink, it is a store.
  
  $Q = 1, 020 \times 2.09 \times 2.315 + 1, 020 \times 334 = 4, 936 + 340, 680 = 345, 616 J$
4. So ask the question a cool box is actually for. How much heat can come in before the contents are liquid water at 0 °C and the cooling is over? Warm the solid back to zero, then melt all 1,020 g of it. 345.6 kilojoules.
  
  $900 \times 334 = 300, 600 (87.0%); 900 \times 2.09 \times 25 = 47, 025 (13.6%); - 120 \times 4.186 \times 4 = - 2, 009 (- 0.6%)$
5. Split it up and the ranking is the point. The latent heat of the ice you packed is 87% of the store. The 25 degrees of freezer cold is 13.6%. The drink itself is negative — it arrived warm, and it spends a little of the reserve on the way in.
  
  $t = \frac{345 , 616 J}{3 W} = 115, 205 s = 32.0 h$
6. Now put a leak on it. Take a small box in a warm room at 3 watts, which is a joule every third of a second. Divide, and the reserve is 32 hours.
  
  $t_{- 18} = \frac{900 ( 334 + 2.09 \times 18 ) - 2 , 009}{3} = \frac{332 , 449}{3} = 30.8 h$
7. Run it again with ice from a domestic freezer at −18 °C instead of a chest freezer at −25. The reserve falls to 332 kJ and the box lasts 30.8 hours. Seven degrees are worth 1.2 hours, and the whole 25 degrees are worth 4.4 of the 32 — which in ice is 122 grams, an eighth of the load.
  
  $\frac{m _{i} c _{ice}}{c _{ice} ∣ T _{i} ∣ + L} = \frac{900 \times 2.09}{386.25} = 4.87 g per^{\circ} C \Rightarrow 25^{\circ} C \equiv 122 g$
Answer

345.6 kJ of reserve, which is 32 hours at a 3-watt leak, and 13.6% of it came from the freezer rather than from the ice. Colder ice is worth having and it is not worth much: dropping the freezer another seven degrees buys about as long as tipping in another 120 g would.

The insight above this one prices the same extra cold at 0.071 °C of drink per degree of freezer and calls it almost nothing, and both readings are right, because a drink and a cool box are asking different questions. A drink wants a temperature, and temperature is set by the melting, which happens at one fixed value however cold the ice started. A box wants a duration, and duration is set by joules, where every degree below zero is real and counts. The tool prints temperatures because the drink is the case people arrive with. The quantity behind all of it is the enthalpy the module carries and never displays — the whole contents measured against liquid water at 0 °C — and once you have it in joules the box, the picnic and the freezer failure are all the same division.

$Q_{reserve} = - H = m_{i} (L + c_{ice} ∣ T_{i} ∣) - m_{w} c T_{w}$

By the 42Math team · Updated 2026-08-26

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Freezer ice in a glass of water - 300 g of water at 20 °C with 50 g of ice from a −18 °C freezer settles at 4.5 °C and the ice all goes. 85.6% of the heat the water gives up is spent on melting.
The same ice, at 0 °C - The same 50 g of ice, taken at 0 °C instead of −18 °C, leaves the glass at 5.7 °C. Eighteen degrees of freezer cold bought 1.3 °C.
More ice than the glass can use - 150 g of freezer ice in the same glass lands on 0.0 °C with 91.7 g still floating. Anything above 67.6 g gives that same 0.0 °C.
Tea brought to 55 °C - 43.5 g of freezer ice takes 250 g of tea from 80 °C to 55.0 °C. Getting there with water already at 0 °C would take 114 g.
A cup packed in ice - 120 g of water at 4 °C packed in 900 g of ice at −25 °C freezes solid and ends at −2.3 °C. Freezing starts once the ice passes 806 g.