Why ice cream never freezes completely

Two ice creams are stored overnight in the same freezer at −18 °C.

The first is hard enough to bend a spoon.

The second can be scooped almost immediately.

Their temperature is identical, yet their texture is completely different.

Why?

The usual answer is that one recipe contains more sugar. That is directionally correct, but incomplete. The amount of sugar matters, the type of sugar matters, the quantity of available water matters, and the salts naturally present in dairy ingredients matter as well.

All of these variables influence one central property:

Freezing point depression determines how much water can remain liquid at a given temperature.

Ice cream does not cross a single temperature and suddenly become solid. It freezes progressively. As temperature falls, part of water crystallizes while the rest remains trapped in an increasingly concentrated liquid phase.

Relationship between temperature and frozen water is called freezing curve.

Understanding that curve connects formulation decisions to practical outcomes:

• hardness at storage temperature;
• scoopability;
• resistance to heat shock;
• melting behaviour;
• flavour release;
• ice-crystal growth;
• perceived creaminess.

This article develops that relationship step by step. We begin with simplest possible system: sugar dissolved in water. We then move toward a complete ice cream mix containing several carbohydrates and milk salts. Finally, we compare calculation strategies and examine why they do not always predict same freezing curve.

Start with an experiment

Consider three solutions, each prepared with:

• 1000 g of water;
• 100 g of sugar.

Only sugar differs:

1. sucrose;
2. glucose;
3. fructose.

All three contain same mass of sugar. Should they freeze at same temperature?

At first glance, that seems reasonable. If mass concentration is identical, perhaps effect should also be identical.

It is not.

Glucose and fructose solutions experience approximately twice as much freezing point depression as sucrose solution.

Reason is not that glucose is intrinsically more powerful than sucrose. Freezing point depression depends primarily on number of dissolved particles, not directly on their mass.

Sucrose has molar mass of approximately:

$$M_{\mathrm{sucrose}} = 342.30\ \mathrm{g\,mol^{-1}}$$

Glucose and fructose have molar masses close to:

$$M_{\mathrm{glucose}} \approx M_{\mathrm{fructose}} \approx 180.16\ \mathrm{g\,mol^{-1}}$$

For 100 g of each sugar:

$$n_{\mathrm{sucrose}} = \frac{100}{342.30} = 0.292\ \mathrm{mol}$$ $$n_{\mathrm{glucose}} = \frac{100}{180.16} = 0.555\ \mathrm{mol}$$

Glucose solution therefore contains almost twice as many dissolved molecules. More molecules disturb equilibrium between liquid water and ice more strongly. Consequently, glucose solution must be cooled further before ice and liquid water can coexist.

Equal mass or equal particle count?

Equal massEqual moles

Water mass

1000 g

Sugar mass

100 g

Sugar	Mass	Molar mass	Moles	Molality	ΔTf
Sucrose	100.0 g	342.30 g/mol	0.292	0.292 mol/kg	0.54°C
Glucose	100.0 g	180.16 g/mol	0.555	0.555 mol/kg	1.03°C
Fructose	100.0 g	180.16 g/mol	0.555	0.555 mol/kg	1.03°C

Keep sugar mass constant, then switch to constant molar amount. Distinction is essential: at equal mass, glucose produces larger effect than sucrose; at equal number of moles, their ideal colligative effect is approximately same.

The simplest freezing-point equation

For an ideal dilute solution, freezing point depression can be estimated with:

$$\Delta T_f = iK_fm$$

where:

• $\Delta T_f$ is freezing point depression in degrees Celsius;
• $i$ is van ’t Hoff factor;
• $K_f$ is cryoscopic constant of solvent;
• $m$ is molality of dissolved solute.

For water:

$$K_f \approx 1.86\ ^\circ\mathrm{C\,kg\,mol^{-1}}$$

Molality is number of moles of solute per kilogram of solvent:

$$m = \frac{n_{\mathrm{solute}}}{m_{\mathrm{water,kg}}}$$

For sugars such as sucrose, glucose and fructose, no dissociation into ions is normally assumed in this elementary calculation, so:

$$i \approx 1$$

For solution containing 100 g of sucrose in 1 kg of water:

$$m_{\mathrm{sucrose}} = \frac{100/342.30}{1} = 0.292\ \mathrm{mol\,kg^{-1}}$$

Therefore:

$$\Delta T_{f,\mathrm{sucrose}} = 1 \times 1.86 \times 0.292 \approx 0.54^\circ\mathrm{C}$$

Predicted initial freezing temperature is approximately:

$$T_f \approx -0.54^\circ\mathrm{C}$$

For 100 g of glucose:

$$m_{\mathrm{glucose}} = \frac{100/180.16}{1} = 0.555\ \mathrm{mol\,kg^{-1}}$$ $$\Delta T_{f,\mathrm{glucose}} = 1 \times 1.86 \times 0.555 \approx 1.03^\circ\mathrm{C}$$

Glucose solution therefore begins to freeze near:

$$T_f \approx -1.03^\circ\mathrm{C}$$

Ideal model captures first practical lesson:

At equal mass, smaller molecules generally produce larger freezing point depression because they provide more dissolved particles.

Freezing point is not frozen-water fraction

Previous calculation predicts temperature at which freezing begins. It does not tell us how much water is frozen at −5, −10 or −18 °C.

This distinction is fundamental.

Initial freezing point answers:

At what temperature can first ice crystals appear?

Freezing curve answers:

At a given temperature, what fraction of available water has crystallized?

For ice cream formulation, second question is usually more useful.

Imagine a mix initially containing:

• 600 g of water;
• 150 g of dissolved sugars;
• 250 g of fat, proteins and other solids.

When first ice appears, nearly all 600 g of water is still liquid. As cooling continues, perhaps 100 g crystallizes. Dissolved sugars remain in liquid phase. They are now dissolved in approximately 500 g of water instead of 600 g.

If another 100 g freezes, same sugars become concentrated in only 400 g of liquid water. Molality therefore rises continuously:

$$m = \frac{n_{\mathrm{solute}}}{m_{\mathrm{unfrozen\ water}}}$$

As $m_{\mathrm{unfrozen\ water}}$ decreases, molality increases. Freezing point of remaining serum becomes lower:

$$\Delta T_f = iK_fm$$

This creates characteristic progressive freezing process:

$$\text{cooling} \rightarrow \text{ice formation} \rightarrow \text{serum concentration} \rightarrow \text{larger FPD} \rightarrow \text{more cooling required}$$

Process is self-limiting. Every increment of ice formation makes remaining water harder to freeze.

Follow freeze concentration

Temperature

-18°C

Water

600 g

Sucrose

100 g

Dextrose

50 g

Ice mass

541.1 g

Unfrozen water

58.9 g

Water frozen

90.2%

Serum sugars

254.8%

Molality

9.68 mol/kg

Residual

3.6e-15°C

Inspect curve

Temperature -18.00°CWater frozen, ideal: 90.2%

At every temperature, simulation exposes current ice mass, remaining liquid water, serum sugar concentration, molality and residual equilibrium error. Numerical feedback loop matters more than animation of ice crystals.

From one sugar to a real formulation

Ideal equation is easy to apply to a single sugar in water. Real ice cream mix is more complicated.

It may contain:

• sucrose;
• dextrose;
• fructose;
• lactose from milk;
• glucose syrup containing molecules of several sizes;
• salts from milk and whey;
• sodium chloride;
• alcohol or polyols;
• proteins;
• stabilizers;
• fat.

Not all components affect freezing in same way.

Small dissolved molecules have strong colligative effect because many particles are present per unit mass. Large molecules contribute fewer particles per unit mass. Insoluble fat does not participate directly in aqueous freezing equilibrium.

Proteins and hydrocolloids can strongly influence viscosity, water mobility and ice-crystal growth, but their direct molar contribution to freezing point depression is comparatively small because of high molecular mass.

Component	Direct contribution to FPD	Other important effects
Sucrose	High	Sweetness, solids
Dextrose	Very high per unit mass	Sweetness, softness
Fructose	Very high per unit mass	High sweetness
Lactose	Moderate	Crystallization risk
Glucose syrup	Depends on molecular-weight distribution	Body, viscosity, solids
Milk salts	Significant	Ionic contribution
Fat	Negligible direct colligative effect	Structure, lubrication, flavour
Proteins	Small direct effect	Emulsification, water binding, body
Stabilizers	Small direct effect	Viscosity, recrystallization control

Important phrase is direct contribution.

A stabilizer may transform sensory and storage behaviour without materially changing calculated FPD. Conversely, a small quantity of low-molecular-mass solute may alter FPD significantly without adding much body.

FPD is important, but it is not a complete texture model.

Additive colligative calculation

For a mixture of ideal, non-electrolyte solutes, total molality can be estimated by summing moles of each dissolved compound:

$$m_{\mathrm{total}} = \frac{\sum_j n_j}{m_{\mathrm{water,kg}}}$$

with:

$$n_j = \frac{m_j}{M_j}$$

Ideal freezing point depression becomes:

$$\Delta T_f = K_f \frac{\sum_j m_j/M_j}{m_{\mathrm{water,kg}}}$$

For a mixture containing sucrose and dextrose:

$$\Delta T_f = K_f\left( \frac{m_{\mathrm{sucrose}}/M_{\mathrm{sucrose}} + m_{\mathrm{dextrose}}/M_{\mathrm{dextrose}}}{m_{\mathrm{water,kg}}} \right)$$

Suppose 40 g of sucrose are replaced by 40 g of dextrose. Sugar mass remains constant, number of dissolved molecules increases, and therefore:

$$\Delta T_f \uparrow$$

At fixed storage temperature:

$$\text{unfrozen water} \uparrow$$

and usually:

$$\text{hardness} \downarrow$$

This is physical basis behind many practical sugar-balancing strategies. Locking total sugar mass, total sweetness, or total solids produces different substitutions; these constraints should not be conflated.

Replace sucrose with dextrose

Start with 150 g sucrose in 620 g water. Choose what must remain constant, then replace part of the sucrose.

Constraint

Keep sugar mass constantKeep sweetness constant

Sucrose replaced

50%

Quantity	Original	After substitution
Sucrose	150.0 g	75.0 g
Dextrose	0.0 g	75.0 g
Total sugar mass	150.0 g	150.0 g
Sweetness equivalent	150.0	127.5
Dissolved amount	0.438 mol	0.635 mol
Ideal initial FPD	1.31°C	1.91°C

The table keeps each constraint explicit. Replacing sugar gram for gram preserves solids, but not sweetness. Replacing it at equal sweetness may require a different total sugar mass. In both cases, the number of dissolved molecules, and therefore predicted FPD, changes.

Why ideal equation is not enough

Cryoscopic equation assumes a dilute ideal solution. An ice cream serum becomes neither dilute nor ideal during freezing.

As ice forms:

• solute concentrations increase sharply;
• interactions between molecules become more important;
• effective behaviour of glucose syrups depends on molecular distribution;
• salts dissociate into ions;
• some water may be associated with proteins or other solids;
• liquid phase can approach a highly concentrated, viscous state.

Equation $\Delta T_f = iK_fm$ remains an excellent conceptual model. It explains why particle count matters, predicts direction of ingredient substitutions, and provides useful estimates at modest concentrations.

But extrapolating it directly across complete freezing range can produce an unrealistic curve.

This leads to practical question:

How can we estimate FPD for a complete ice cream formulation without explicitly modelling every molecule?

One influential answer is sucrose-equivalent method described by Goff and Hartel.

The Goff–Hartel sucrose-equivalent approach

Method converts major carbohydrate sources in formulation into equivalent amount of sucrose. Resulting value is sucrose equivalent, or $SE$.

A published form of relationship is:

$$\begin{aligned} SE ={}& 0.545\,MSNF + 0.765\,WS + S \\ &+ 0.2\,(10DE\ CSS) + 0.6\,(36DE\ CSS) \\ &+ 0.8\,(42DE\ CSS) + 1.2\,(62DE\ CSS) \\ &+ 1.8\,HFCS + 1.9\,F \end{aligned}$$

where, depending on formulation convention:

• $MSNF$ represents milk solids-not-fat;
• $WS$ represents whey solids;
• $S$ represents sucrose;
• $CSS$ represents corn-syrup solids of specified dextrose equivalent;
• $HFCS$ represents high-fructose corn syrup;
• $F$ represents fructose or another pure monosaccharide such as dextrose.

Coefficients express approximate freezing-point contribution of each ingredient relative to sucrose. Sucrose has coefficient 1. Fructose has coefficient near 1.9. Low-DE glucose syrup has lower coefficient because average molecular mass is higher. Higher-DE syrup contains greater proportion of smaller saccharides and therefore has larger effect.

Method does not abandon molecular reasoning. It packages it into practical ingredient coefficients.

According to University of Guelph calculation procedure, $SE$ is first expressed in g per 100 g of mix, then converted to equivalent sucrose concentration in available water:

$$C_{SE} = 100\frac{SE}{W}$$

where $W$ is water content on same mass basis. Sugar contribution can then be approximated by polynomial used in published computational work:

$$FPD_{\mathrm{sugars}} = 0.00009\,C_{SE}^2 + 0.0612\,C_{SE}$$

Definition and basis of $SE$ must remain consistent. Dividing by water is essential during freeze concentration because available liquid water changes.

Follow one calculation from ingredients to FPD

Use the formulation developed below: 110 g sucrose, 40 g dextrose, 100 g milk solids-not-fat, and 620 g water.

1. Convert each source to sucrose equivalent.

$$SE = 110 + (1.9 \times 40) + (0.545 \times 100) = 240.5\ \mathrm{g}$$

The three visible contributions are 110 g from sucrose, 76 g from dextrose, and 54.5 g from lactose associated with milk solids-not-fat.

2. Express that equivalent amount relative to available water.

$$C_{SE} = 100\frac{240.5}{620} = 38.79$$

3. Calculate the sugar contribution.

$$FPD_{\mathrm{sugars}} = 0.00009(38.79)^2 + 0.0612(38.79) = 2.51^\circ\mathrm{C}$$

4. Calculate the separate milk-salt contribution.

$$FPD_{\mathrm{MSNF}} = 2.37\frac{100}{620} = 0.38^\circ\mathrm{C}$$

5. Add both contributions.

$$FPD_{\mathrm{total}} = 2.51 + 0.38 = 2.89^\circ\mathrm{C}$$

The estimated initial freezing point is therefore approximately $-2.89^\circ\mathrm{C}$. This sequence follows the same separation used in the FDS implementation: normalize sucrose equivalents by free water, calculate their FPD, then add the milk-salt term.

Contribution of milk salts

Sugars are not only dissolved particles in an ice cream mix. Milk ingredients introduce mineral salts that also depress freezing point.

Goff and Hartel give practical relationship for contribution associated with milk solids-not-fat:

$$FPD_{\mathrm{MSNF}} = 2.37\left(\frac{MSNF}{W}\right)$$

where $MSNF$ and $W$ use consistent units. Factor 2.37 represents empirical approximation based on average mineral composition and molecular behaviour of milk salts.

Total initial freezing point depression can then be estimated as:

$$FPD_{\mathrm{total}} = FPD_{\mathrm{sugars}} + FPD_{\mathrm{MSNF}}$$

and predicted initial freezing temperature is:

$$T_{f,0} = -FPD_{\mathrm{total}}$$

For formulations containing separately added sodium chloride, another explicit salt term may be introduced:

$$FPD_{\mathrm{NaCl}} = 0.581855\,C + 0.0034889\,C^2 + 0.0004314\,C^3$$

with:

$$C = 100\frac{m_{\mathrm{NaCl}}}{m_{\mathrm{water}}}$$

Complete expression becomes:

$$FPD_{\mathrm{total}} = FPD_{\mathrm{sugars}} + FPD_{\mathrm{MSNF}} + FPD_{\mathrm{NaCl}}$$

Separation is useful because milk salts and deliberately added sodium chloride need not be represented by same empirical term.

See what changes after half the water freezes

If 50% of the original 620 g of water crystallizes, only 310 g remain liquid. Ingredient masses have not changed, so $SE$ is still 240.5 g. Its concentration in the serum doubles:

$$C_{SE,50\%} = 100\frac{240.5}{310} = 77.58$$

The two contributions then become:

$$FPD_{\mathrm{sugars}} = 0.00009(77.58)^2 + 0.0612(77.58) = 5.29^\circ\mathrm{C}$$ $$FPD_{\mathrm{MSNF}} = 2.37\frac{100}{310} = 0.76^\circ\mathrm{C}$$

Therefore:

$$FPD_{\mathrm{total}} = 5.29 + 0.76 = 6.05^\circ\mathrm{C}$$

Halving the liquid water raises predicted FPD from 2.89 °C to 6.05 °C. This numerical comparison is the freeze-concentration feedback loop in concrete form.

Worked example

Consider simplified 1000 g ice cream mix:

Component	Mass
Water	620 g
Sucrose	110 g
Dextrose	40 g
Milk solids-not-fat	100 g
Fat and other non-colligative solids	130 g

For ideal molar calculation:

$$n_{\mathrm{sucrose}} = \frac{110}{342.30} = 0.321\ \mathrm{mol}$$ $$n_{\mathrm{dextrose}} = \frac{40}{180.16} = 0.222\ \mathrm{mol}$$

Total sugar moles:

$$n_{\mathrm{total}} = 0.321 + 0.222 = 0.543\ \mathrm{mol}$$

Water mass:

$$m_{\mathrm{water}} = 0.620\ \mathrm{kg}$$

Ideal molality:

$$m = \frac{0.543}{0.620} = 0.876\ \mathrm{mol\,kg^{-1}}$$

Ideal sugar FPD:

$$FPD_{\mathrm{ideal,sugars}} = 1.86 \times 0.876 = 1.63^\circ\mathrm{C}$$

Milk-solids contribution is estimated as:

$$FPD_{\mathrm{MSNF}} = 2.37\left(\frac{100}{620}\right) = 0.38^\circ\mathrm{C}$$

A simple additive estimate gives:

$$FPD_{\mathrm{total}} \approx 1.63 + 0.38 = 2.01^\circ\mathrm{C}$$

Predicted initial freezing temperature is therefore approximately:

$$T_{f,0} \approx -2.01^\circ\mathrm{C}$$

This calculation is not yet a freezing curve. It provides only starting point. To estimate frozen fraction at lower temperatures, calculation must account for freeze concentration.

Building a freezing curve

At a temperature below initial freezing point, some water must be removed from liquid phase as ice until remaining serum has a freezing point equal to imposed temperature.

This equilibrium condition can be written conceptually as:

$$FPD\left(m_{\mathrm{unfrozen\ water}},\text{solute composition}\right) = |T|$$

Unknown is mass of unfrozen water. Once that value is found, ice mass follows from:

$$m_{\mathrm{ice}} = m_{\mathrm{freezable\ water}} - m_{\mathrm{unfrozen\ water}}$$

Fraction of water frozen is:

$$X_{\mathrm{ice,water}} = \frac{m_{\mathrm{ice}}}{m_{\mathrm{freezable\ water}}}$$

If result is expressed relative to complete mix:

$$X_{\mathrm{ice,mix}} = \frac{m_{\mathrm{ice}}}{m_{\mathrm{mix}}}$$

These percentages answer different questions and should be labelled clearly. Statement such as “ice fraction is 50%” is ambiguous unless denominator is specified.

An iterative solution

For complex empirical FPD function, most practical solution is numerical.

At each temperature:

1. begin with trial value for unfrozen water;
2. calculate concentration of sugars in that water;
3. recalculate sucrose equivalents or molality;
4. calculate corresponding FPD;
5. compare it with target temperature;
6. adjust unfrozen-water estimate;
7. repeat until error is acceptably small.

A bisection solver is appropriate when relationship is monotonic and valid interval is known. Newton’s method may converge more rapidly, but requires reliable derivative and can become unstable when initial estimate is poor. For educational simulator, robustness is more valuable than maximizing computational speed.

For example, when solving equilibrium at −12 °C, numerical trace could read:

Target temperature: −12.0 °C

Iteration 1
Unfrozen water: 310.0 g
Predicted FPD: 7.4 °C

Iteration 2
Unfrozen water: 205.0 g
Predicted FPD: 12.9 °C

Iteration 3
Unfrozen water: 222.5 g
Predicted FPD: 11.8 °C

...

Converged

Unfrozen water: 218.6 g
Ice mass: 401.4 g
Predicted FPD: 12.00 °C
Residual error: 0.001 °C

Every point on freezing curve is solution of an equilibrium problem. Curve is obtained by repeatedly solving same equilibrium at different temperatures.

Comparing different approaches

Several methods estimate freezing behaviour of an ice cream mix. They describe same phenomenon using different assumptions and therefore produce different predictions. Understanding assumptions is more valuable than memorizing equations.

PAC: freezing power

Simplest approach is empirical PAC (Pouvoir Anti-Congélant). Each ingredient receives coefficient representing relative freezing power compared with sucrose.

$$PAC = \sum_i m_i \times PAC_i$$

where $m_i$ is ingredient mass and $PAC_i$ is empirical coefficient.

Ingredient	Relative PAC
Sucrose	100
Dextrose	190
Fructose	190
Lactose	100
Glucose syrup	Depends on DE

Because dextrose has roughly half molecular weight of sucrose, same mass produces almost twice as many molecules in solution, leading to stronger freezing point depression.

PAC is useful for rapid formulation work. Its limitation is important: PAC predicts relative freezing power, not actual freezing curve. It cannot directly answer how much ice is present at −10 °C, how much unfrozen water remains at −18 °C, or what serum concentration becomes during freezing.

Ideal cryoscopy

Next level applies classical cryoscopic equation directly:

$$\Delta T_f = iK_fm$$

Unlike PAC, this approach is based on number of dissolved particles rather than empirical coefficients. It explains why replacing sucrose with dextrose softens ice cream even when total sugar mass remains unchanged.

Unfortunately, ice cream rapidly leaves conditions under which equation is valid. As freezing progresses, serum becomes highly concentrated, interactions between solutes increase, viscosity rises dramatically, and solution is no longer dilute or ideal. Classical equation becomes progressively less accurate.

Goff–Hartel approach

Rather than explicitly calculating every dissolved molecule, Goff and Hartel use sucrose equivalents. Different sugars are converted into equivalent sucrose amounts with practical coefficients. Milk salts receive separate contribution.

This approach incorporates carbohydrate composition and milk salts while remaining computationally inexpensive. It still depends on empirical assumptions and average ingredient composition.

Compare predicted freezing curves

Temperature

-18°C

Water

620 g

Sucrose

110 g

Dextrose

40 g

MSNF

100 g

Added NaCl

0.0 g

FPD contribution at -18°C

The model first finds 119.2 g of unfrozen water. It then recomputes every dissolved contribution in that smaller liquid phase.

Sugars and lactose

16.01°C

Milk salts

1.99°C

Added NaCl

0.00°C

Total predicted FPD: 18.00°C

Bisection trace

First three and final two trials. Each step narrows the possible unfrozen-water interval.

Iteration	Trial water	Predicted FPD	Residual	Interval
1	310.00 g	6.054°C	1.2e+1°C	620.00 g
2	155.00 g	13.192°C	4.8e+0°C	310.00 g
3	77.50 g	30.716°C	1.3e+1°C	155.00 g
26	119.20 g	18.000°C	7.7e-7°C	0.00 g
27	119.20 g	18.000°C	7.4e-8°C	0.00 g

Ideal frozen water

90.9%

Goff–Hartel frozen water

80.8%

Goff–Hartel iterations

27

Solver residual

7.4e-8°C

Inspect curve

Temperature -18.00°CWater frozen, ideal: 90.9%Water frozen, Goff–Hartel: 80.8%

Ideal cryoscopyGoff–Hartel

Use temperature control to compare ideal and Goff–Hartel predictions at same formulation. Interface reports percentage of water frozen, bisection iterations, and residual error. Curves are predictions, not experimental measurements.

What freezing curve explains

Repeating equilibrium calculation from 0 °C to −25 °C produces complete freezing curve. As temperature falls:

• ice mass increases;
• unfrozen water decreases;
• serum concentration rises;
• freezing point depression becomes progressively larger.

Physical origin becomes apparent. Curve is not a property assigned independently to ice cream; it emerges from continuous concentration of unfrozen serum during freezing.

This explains why two ice creams stored at same −18 °C can have very different textures. Temperature alone does not specify phase composition. Formulation determines how much water has crystallized and how concentrated remaining serum has become.

Model limits

These calculations isolate direct colligative effects. They do not predict complete sensory texture.

• Proteins and stabilizers affect water mobility and viscosity.
• Fat destabilization and air structure influence mechanical resistance.
• Ice-crystal size depends on processing and thermal history.
• Real concentrated solutions deviate from ideality.
• Ingredient composition and milk-mineral content vary.

Freezing curve is a necessary part of hardness model, not sufficient model by itself.

Experimental work supports this distinction. Ice content and serum-phase viscosity both contribute to hardness, while processing conditions also change ice-crystal size, fat destabilization, and incorporated air. Two formulations with similar calculated FPD can therefore develop different mechanical properties.

References

• H. Douglas Goff, Freezing Point Depression of a Mix, University of Guelph.
• H. Douglas Goff and Richard W. Hartel, Ice Cream, Springer, 2013.
• Romain Bertin, FDS ice-cream calculation implementation, source code.
• A. Rinaldi et al., A Mathematical Modeling of Freezing Process in the Batch Production of Ice Cream, Foods 10(2), 2021.
• H. Douglas Goff, Low-temperature stability and the glassy state in frozen foods, Food Research International 25(4), 1992.
• C. Cogné, P. Laurent, J. Andrieu and J. Ferrand, Experimental Data and Modelling of Ice Cream Freezing, Chemical Engineering Research and Design 81(9), 2003.
• E. M. Drewett and R. W. Hartel, Ice crystallization in a scraped surface freezer, Journal of Food Engineering 78(3), 2007.
• K. Inoue et al., Modeling of the effect of freezer conditions on the hardness of ice cream using response surface methodology, Journal of Dairy Science 92(12), 2009.
• Q. Wang, G. Sala and E. Scholten, Functionality of sugars and sugar replacers in model frozen dessert systems, Current Research in Food Science, 2025.