Latent heat and phase change: what happens when ice meets water

Drop ice into water and the answer is usually not a temperature at all: the mixture stops dead at 0 °C part-way through melting. This page works out how much ice melts, how much survives, whether the water freezes instead, and only then what temperature everything ends at.

Last updated 11 Nov 2026

Ice into water: how much melts

Three outcomes, and the one in the middle is the usual one

Mass of iceIn kilograms. May be 0, which turns this into a plain cooling problem with no phase change in it at all — a reasonable thing to try first, so you can see what the latent-heat terms add.

Temperature of the iceIn degrees Celsius, and it cannot be above 0 because ice above 0 °C is not ice. Set it to 0 for the clean textbook case: the ice is already at its melting point, so every joule that reaches it goes into melting rather than warming, and the sensible-heat term for the ice vanishes. Ice from a domestic freezer is nearer −18 °C, and warming it to 0 absorbs 2100 × 18 = 37.8 kJ per kilogram — about a ninth of what melting it then costs.

Mass of waterIn kilograms; 200 g is 0.2. For water, litres and kilograms are interchangeable to well within the precision of anything else here.

Temperature of the waterIn degrees Celsius, and it cannot be below 0 because this field is the liquid. All the heat the mixture has to spend comes from cooling this water to 0 °C, and if that is not enough the water starts to freeze and the page says so.

Specific latent heat of fusion of waterTwo sourced values, 0.6 per cent apart, from the same official document. NCERT’s own CIET module on change of state gives the latent heat of fusion of water as 3.33 × 10⁵ J kg⁻¹ in its text and uses 3.35 × 10⁵ and 335 J g⁻¹ in its own worked examples. That is not a misprint worth tidying away: it is how big the real uncertainty in a school constant is, and it is why this page asks rather than choosing. On the default inputs the two differ by 0.34 g of melted ice out of 56.5.

Latent heat of fusion, typed inLocked unless the last option is chosen above. In joules per kilogram. A table in J g⁻¹ needs multiplying by 1000, and one in cal g⁻¹ by 4186 — the old figure of 80 cal g⁻¹ for ice is 334 880 J kg⁻¹, which lands between the two NCERT values.

56.55gExample

100 g of ice at −10 °C dropped into 200 g of water at 25 °C, on NCERT’s 3.33 x 10^5 J/kg

Three budgets, checked in order, before any temperature is calculated

Q_available = m_wc_wT_w · Q_{warm ice} = m_ic_i|T_i| · Q_{melt all} = m_iL_f · melted = (Q_available − Q_{warm ice}) / L_f

Lf: the specific latent heat of fusion: the energy to melt a kilogram with no change of temperature at all. For water it is about 333 kJ per kilogram, and the number to hold on to is that this equals cooling a kilogram of water by 79.55 K — L_f/c_w. Melting is a far bigger transaction than any ordinary temperature change
the order: the three budgets must be compared in order, and this is what students skip. Can the water warm the ice to 0? If not, the water freezes. If so, is what is left enough to melt all of it? If not, the mixture stops at 0 °C part-way through. Only if there is still a surplus does anything have a temperature above zero
0 °C: the temperature a melting mixture sits at, and it sits there while the melting happens. Heat going in does not raise the temperature; it changes the state. This is why a drink with ice still in it is at 0 °C whatever the room is doing, and why ice water is a usable calibration point
ci vs cw: ice is 2100 J kg⁻¹ K⁻¹ and water 4186, so ice warms twice as fast per joule. Using water’s figure for the ice is a common slip and doubles the first budget. The two are different substances as far as this calculation is concerned
Lv: the latent heat of VAPORISATION, for completeness, because the same three-budget logic runs at the other end: NCERT gives it as 22.6 × 10⁵ J kg⁻¹ in its text and 2.256 × 10⁶ in an example. It is about 6.8 times the latent heat of fusion, and more than five times the heat needed to take water from freezing to boiling

Worked example

100 g of ice at −10 °C dropped into 200 g of water at 25 °C, on NCERT's 3.33 x 10^5 J/kg
BUDGET ONE: what the water can give. Cooling 0.200 kg of water from 25 °C to 0 °C releases 0.200 × 4186 × 25 = 20 930 J. That is every joule available, because the water cannot go below 0 °C without freezing
BUDGET TWO: warming the ice to its melting point. 0.100 × 2100 × 10 = 2100 J, on ice's own specific heat of 2100 and not water's 4186. The water can afford it, so the ice does reach 0 °C and the mixture does not freeze. Surplus: 20 930 − 2100 = 18 830 J
BUDGET THREE: melting it all would cost 0.100 × 333 000 = 33 300 J, and only 18 830 J are left. So not all of it melts, and THIS is the decision the question turns on. Everything stops at 0 °C
HOW MUCH MELTS. 18 830 / 333 000 = 0.056 55 kg = 56.55 g, leaving 43.45 g of ice still floating. The final temperature is exactly 0 °C, and it would still be exactly 0 °C if the water had started at 20 °C or at 30 — only the melted mass would move. A question answered with a temperature here has answered the wrong question
AND WHAT IT WOULD TAKE TO MELT THE LOT. The water would have to supply 2100 + 33 300 = 35 400 J, which at 25 °C needs 35 400 / (4186 × 25) = 0.338 kg of water rather than 0.200. On NCERT's other figure of 3.35 × 10⁵ the melted mass becomes 56.21 g instead of 56.55 — a 0.6 per cent change from a 0.6 per cent change in a constant that one official document prints two ways

The question is usually how much melted, not what temperature it reached

The thing that makes ice-and-water questions hard is that the answer is often not a temperature. Put enough ice into warm water and the mixture settles at exactly 0 °C with ice still in it, and stays there: more heat arriving melts more ice instead of raising the temperature. The quantity that actually varies is the melted mass, which is why this page leads with that and treats the final temperature as a secondary row. A student who reaches for one m·c·ΔT equation gets a confident number that is simply not what happened.

So the work is three budgets, compared in order. First, how much heat the warm water can release before it reaches 0 °C. Second, how much of that is spent warming the ice from wherever it started up to 0 °C — on ICE’S specific heat of 2100, not water’s 4186. Third, whether the surplus covers melting all the ice, at 333 kJ per kilogram. If the first budget cannot even cover the second, heat runs the other way and the water starts to freeze. If the surplus covers the third, everything is water and the mixture has a real temperature above zero. In between — which is most of the time — it stops at 0 °C part-way through.

The number worth remembering is 79.55. Melting a kilogram of ice takes as much heat as cooling a kilogram of water by 79.55 K, because that is L_f / c_w = 333 000 / 4186. That single ratio explains why ice is such an effective coolant, why a bag of ice outlasts a bottle of cold water many times over, and why the warming term for freezer ice is almost an afterthought: taking ice from −18 °C up to 0 costs 37.8 kJ per kilogram against the 333 kJ that melting it then costs, about a ninth. Students routinely spend most of their working on the term that matters least.

Where this goes wrong, specifically. Using water’s 4186 for the ice doubles the second budget. Forgetting the ice was below zero at all removes it. Using the latent heat with a temperature difference attached — mLΔT instead of mL — is the error that marks itself, because latent heat has no ΔT: the whole point is that the temperature does not change. And the one that is hardest to spot is answering with a temperature when the mixture ended inside the phase change: the arithmetic runs, the number looks reasonable, and it describes a state that cannot exist, with water at −4 °C sitting next to ice that has not finished melting.

Two sourced values, from one document, 0.6 per cent apart. NCERT’s own CIET module on change of state gives the latent heat of fusion of water as 3.33 × 10⁵ J kg⁻¹ in its text and then uses 3.35 × 10⁵ and 335 J g⁻¹ in its own worked examples; the same module gives the latent heat of vaporisation as 22.6 × 10⁵ in the text and 2.256 × 10⁶ in an example, 0.18 per cent apart. The old imperial-era figure of 80 cal g⁻¹ converts to 334 880, which sits between them. This page offers the choice instead of picking for you, and prints what the choice is worth: 0.34 g of melted ice out of 56.5 on its own default inputs.

Frequently asked questions

Why does the mixture stay at 0 °C while the ice is melting?

Because the energy arriving is going into breaking the bonds that hold the solid together rather than into the motion that temperature measures. A melting mixture is the textbook case of a system absorbing heat with no temperature change at all, and it is stable in a useful way: as long as any ice remains, the mixture is at 0 °C whatever the room is doing, which is exactly why ice water is a usable calibration point for a thermometer. The temperature only starts rising again once the last of the ice has gone.

How do I know whether all the ice melts without solving for a temperature first?

Compare two numbers. The heat the water can release before it hits 0 °C is m_wc_wT_w. The heat the ice needs to warm to 0 and then melt completely is m_ic_i|T_i| + m_iL_f. If the first is larger, all the ice melts and there is a final temperature above zero; if it is smaller but still bigger than the warming term alone, the mixture ends at 0 °C with ice left; if it is smaller than the warming term, the water begins to freeze. Doing it in that order means you never solve an equation whose assumptions have already failed.

Can water really end up freezing when I add ice to it?

Yes, and the condition is just an inequality: enough ice, cold enough, against a small amount of water barely above freezing. Try 1 kg of ice at −30 °C with 50 g of water at 2 °C. The water can release only 419 J while warming that ice to 0 °C would take 63 kJ, so heat runs from the water to the ice: the water cools to 0, freezes, and the whole mass then settles below zero. Note which specific heat governs the last stage — it is 2100, because by then everything is ice.

Why is latent heat so much bigger than specific heat?

They are not really comparable quantities — one is per kelvin and one is not — so the fair comparison is a ratio with units that cancel. L_f/c_w = 79.55 K for melting and L_v/c_w = 540 K for boiling. In other words, melting a kilogram of ice is worth cooling a kilogram of water by eighty degrees, and boiling it is worth five hundred and forty — which is a temperature change water cannot even have as a liquid. That is why evaporation is such a powerful cooling mechanism, and why steam burns are far worse than hot-water burns at the same temperature.

Does this page handle boiling as well as melting?

Not as a calculation, no. The page is built around the ice-and-water mixture, which is where school questions and school mistakes concentrate, and its branches are about the fusion boundary at 0 °C. The same three-budget method works unchanged at 100 °C with L_v in place of L_f and the steam’s own specific heat beyond it — but the numbers are much larger and the partial-boil case is rarer in a sealed question, because the steam normally leaves. NCERT’s figures for vaporisation are 22.6 × 10⁵ and 2.256 × 10⁶ J kg⁻¹, which is the same 0.18 per cent disagreement in the same document.

Related calculators

References

NCERT / CIET, Physics-02 (Keph_201104): Thermal Properties of Matter, Unit 7 Module 15, “Change of State” (chapter 11, class XI), 2019, on the Central Institute of Educational Technology e-content server (read 11 November 2026). The source of every material constant this batch prints without hedging: ice 2100 J kg⁻¹ K⁻¹, water 4186 — written 4.2 × 10³ in a different worked example in the same module — copper 0.39 J g⁻¹ K⁻¹, latent heat of fusion 3.33 × 10⁵ J kg⁻¹ in the text against 3.35 × 10⁵ and 335 J g⁻¹ in its own examples, and latent heat of vaporisation 22.6 × 10⁵ in the text against 2.256 × 10⁶ in an example. Those are not misprints to tidy away: one official module gives the same quantity two ways, 0.6 per cent apart on fusion and 0.18 per cent apart on vaporisation. That is why this batch takes the constant from the reader instead of choosing one for them.
Specific Heat Capacity of Water: Temperature-Dependent Data and Calculator, The Engineering ToolBox (2004) (read 11 November 2026). The source of the 4181.6 J kg⁻¹ K⁻¹ option here and of the curve behind it: at saturation pressure the isobaric specific heat runs 4.2199 kJ kg⁻¹ K⁻¹ at 0.01 °C, 4.1955 at 10, 4.1844 at 20, 4.1816 at 25, a minimum of 4.1796 at 40, then back up through 4.1851 at 60, 4.1969 at 80 and 4.2157 at 100. Water’s specific heat is a shallow U with its floor near 40 °C, and the 0.96 per cent spread across the liquid range is wider than the gap between the two NCERT figures. The page states that it names no primary source of its own, which is why these figures sit beside the NCERT ones rather than replacing them.
The International System of Units (SI), 9th edition, BIPM, published May 2019; the version read for this batch is 4.01, revised June 2026 (read 11 November 2026). The source of the kelvin’s CURRENT definition: the kelvin is defined by fixing the Boltzmann constant at exactly k = 1.380 649 × 10⁻²³ kg m² s⁻² K⁻¹, which the brochure states as an exact relation. The degree Celsius is defined equal in magnitude to the kelvin, with T₀ = 273.15 K and t/°C = T/K − 273.15. Most school textbooks still define the kelvin as 1/273.16 of the triple point of water. The brochure records that as the PREVIOUS definition, and says the triple-point temperature must now be determined experimentally — it stood at 273.16 K with a relative standard uncertainty of 3.7 × 10⁻⁷ at the redefinition, and is a measured quantity from then on.
COULD NOT BE READ, and nothing was substituted for any of it. The NCERT textbook pages on ncert.nic.in, and four CIET modules in the same series as the ones cited here — Keph_201101 (temperature and its measurement), Keph_201102 (thermal expansion), Keph_201103 (specific heat capacity and calorimetry) and Keph_201105 (conduction and convection) — were all refused by the server’s robots policy on 11 November 2026. That is why this batch quotes no NCERT table number, and why the coefficients of expansion and the thermal conductivities are attributed to the reference actually read rather than to the textbook table you are more likely to be holding. If your table disagrees, use your table: these pages take the constant from you for exactly that reason.