How much ice forms in ice cream?

Initial freezing point marks the temperature at which ice begins to form. It does not give the amount of ice present at −12 °C or −18 °C.

Ice fraction depends on method used to estimate it. Tchigeov uses water content and initial freezing point. Chen also uses effective molecular weight. Goff and Hartel recalculate concentration of liquid serum as water freezes. Schwartzberg accounts for bound water. Differential scanning calorimetry, or DSC, measures thermal response in an experiment.

The previous article explains how sugars and salts lower the freezing point. It also explains how liquid serum becomes more concentrated as water freezes. Here, the aim is to calculate how much water has become ice at a given temperature.

Define what ice fraction means

Three related quantities need clear names before methods can be compared.

Initial freezing point, $T_f$, is the temperature at which ice can first coexist with the liquid mix. It is measured in degrees Celsius.

Fraction of original water frozen compares ice mass with the water initially present:

$$X_{i,water}=\frac{m_{ice}}{m_{water,0}}$$

Ice mass fraction of the product compares ice mass with the complete mix:

$$X_{i,product}=\frac{m_{ice}}{m_{mix}}$$

For a 1000 g formulation containing 620 g water, 400 g of ice means:

$$X_{i,water}=\frac{400}{620}=0.645$$

but:

$$X_{i,product}=\frac{400}{1000}=0.400$$

Both values are correct because they use different denominators. Every result below states which denominator it uses.

Tchigeov method

Tchigeov proposed an empirical equation for estimating ice mass fraction in frozen foods. Fikiin later reviewed this equation and related methods. ASHRAE also includes it in its guidance on food thermal properties:

$$X_i= \frac{1.105W} {1+\dfrac{0.7138}{\ln\left(T_f-T+1\right)}}$$

where:

• $X_i$ is ice mass fraction of the complete product;
• $W$ is initial water mass fraction of the product;
• $T_f$ is initial freezing point in degrees Celsius;
• $T$ is product temperature, with $T<T_f$.

The equation needs three values. Once $W$ and $T_f$ are known, it does not use ingredient composition.

Tchigeov calculation

Consider a 1000 g mix with:

• water fraction $W=0.620$;
• calculated initial freezing point $T_f=-2.892\ ^\circ\mathrm{C}$;
• target temperature $T=-12\ ^\circ\mathrm{C}$.

First, calculate logarithm argument:

$$T_f-T+1=-2.892-(-12)+1=10.108$$

Then:

$$X_i= \frac{1.105\times0.620} {1+0.7138/\ln(10.108)} =0.524$$

Tchigeov estimates that ice accounts for 52.4% of product mass. Relative to original water:

$$X_{i,water}=\frac{0.524}{0.620}=0.844$$

The model estimates that 84.4% of original water has frozen.

Fricke and Becker compared several models of food thermal properties with experimental data from published studies. Chen’s equation gave the best ice fraction results in their comparison, followed closely by Tchigeov. This supports the use of Tchigeov as a general correlation for frozen foods. It does not validate it as an equilibrium model for ice cream.

Information retained by Tchigeov

Tchigeov can be summarized as:

$$X_i=f(W,T_f,T)$$

Detailed formulation is used only to calculate initial freezing point. Sucrose, dextrose, lactose, milk salts, and other solutes are not separate inputs after that step.

Two formulations with the same $W$ and $T_f$ therefore produce the same Tchigeov curve.

This result comes from model structure. It does not show that both products have identical freezing curves. Tchigeov retains initial freezing point but not solute composition that produced it.

Liquid water changes during freezing

Serum composition changes during cooling. If 100 g of water freezes, sugars and salts remain in the liquid phase while the mass of liquid water falls by 100 g. Their concentration rises, so the equilibrium freezing point changes.

Tchigeov has no variable for the changing liquid water mass. It calculates ice fraction directly from the initial $W$ and $T_f$. A model that follows freeze concentration must instead solve for the amount of water that remains liquid:

1. Guess liquid water.
2. Recalculate serum concentration and FPD.
3. Compare FPD with |T|, then adjust liquid water.

The Goff and Hartel method calculates freezing point depression, or FPD, from sucrose equivalents, milk salts, and available water. Repeating this calculation for each candidate liquid water mass accounts for freeze concentration.

Question	Tchigeov	Goff and Hartel
Inputs at target temperature	$W$, initial $T_f$, $T$	formulation, $W_{liquid}$, $T$
Quantity solved	ice fraction directly	remaining liquid water
Serum concentration updated	No	Yes
Solute contributions retained	No, compressed into $T_f$	Partly, through SE and separate salt terms
Numerical root solve	No	Yes
Main strength	compact general correlation	accounts for freeze concentration
Main limitation	composition disappears after $T_f$	empirical ingredient grouping and FPD relationships remain

The Goff and Hartel method retains more composition information than Tchigeov, but it does not represent every ingredient separately. Sugars are grouped as sucrose equivalents. Formulations with the same $SE$, water, and salt terms still produce the same calculated curve.

Goff and Hartel calculation

Goff and Hartel describe an ice cream calculation based on sucrose equivalents. Carbohydrate sources are converted to an equivalent sucrose mass. Milk salts contribute a separate term.

For formulation convention used here:

$$SE=S+1.9D+0.545MSNF+0.765WS$$

Here, $S$ is sucrose mass, $D$ is dextrose mass, and $WS$ is whey solids represented separately. The interactive example sets $WS=0$. Coefficients and ingredient bases must use the same formulation convention.

At initial freezing point, sucrose-equivalent concentration is calculated against all available water:

$$C_{SE,0}=100\frac{SE}{W_0}$$

As ice forms, dissolved material remains in serum. Available liquid water becomes:

$$W_{liquid}=W_0-m_{ice}$$

Sucrose-equivalent concentration must therefore be recalculated:

$$C_{SE}=100\frac{SE}{W_{liquid}}$$

The University of Guelph procedure obtains sugar FPD from a sucrose solution table. The interactive calculation uses the following polynomial fit with a zero intercept:

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

The milk salt contribution is recalculated using the same remaining water basis:

$$FPD_{milk\ salts}=2.37\frac{MSNF+WS}{W_{liquid}}$$

When sodium chloride is added separately, the calculation includes a concentration dependent salt term. The total calculated depression is:

$$FPD_{total}=FPD_{sugars}+FPD_{milk\ salts}+FPD_{NaCl}$$

Solve for liquid water

At the target temperature, the equilibrium calculation finds the amount of liquid water that satisfies:

$$FPD\left(formulation,W_{liquid}\right)=|T|$$

Once $W_{liquid}$ is found:

$$m_{ice}=W_0-W_{liquid}$$

Equilibrium calculation

Use 1000 g formulation containing:

Component	Mass
Water	620 g
Sucrose	110 g
Dextrose	40 g
MSNF	100 g
Other non-colligative solids	130 g

At −12 °C, bounded bisection gives:

$$W_{liquid}=168.20\ \mathrm{g}$$

Therefore:

$$m_{ice}=620-168.20=451.80\ \mathrm{g}$$

At that candidate liquid-water mass:

$$C_{SE}=142.99\ \mathrm{g\ sucrose\ equivalent}/100\ \mathrm{g\ liquid\ water}$$ $$FPD_{sugars}=10.591\ ^\circ\mathrm{C}$$ $$FPD_{milk\ salts}=1.409\ ^\circ\mathrm{C}$$

and:

$$FPD_{total}=12.000\ ^\circ\mathrm{C}$$

Residual is approximately $1.7\times10^{-8}\ ^\circ\mathrm{C}$ after 28 iterations. Residual describes numerical convergence, not measurement accuracy.

Ice and liquid water

Goff and Hartel calculation

Reset

Water

620 g

Sucrose

110 g

Dextrose

40 g

MSNF

100 g

Sodium chloride

0.00 g

Temperature

-12.00 °C

Initial freezing point-2.89°C

Liquid water168.2 g

Ice mass fraction of product45.2%

Fraction of original water frozen72.9%

Freezing point depression

Sucrose equivalent concentration142.988 g/100 g liquid water

Sugar FPD10.591 °C

Milk salt FPD1.409 °C

Sodium chloride FPD0.000 °C

Residual: 1.7e-8°C. Iterations: 28. Status: converged.

Change dextrose while keeping the temperature fixed. The module shows initial freezing point, liquid water, each FPD contribution, and both ice fraction measures. Reset restores the worked formulation.

Chen model

Chen derived ice content from freezing point depression using dry solids fraction $X_s$ and effective molecular weight $M_s$:

$$X_{ice}= \frac{X_sRT_0^2(T_f-T)} {M_sL_f(T_0)(T-T_0)(T_f-T_0)}$$

Here, $T_0$, $T_f$, and $T$ are absolute temperatures in kelvin. $R$ is the gas constant, and $L_f(T_0)$ is the latent heat of fusion at the freezing point of pure water. The calculation accepts Celsius as input and converts temperatures to kelvin before evaluating the equation.

Chen retains more physical detail than Tchigeov, but it still reduces formulation to $M_s$. Chen tested method against calorimetric data for meat, fish, and fruit juices. Tests did not include ice cream. Chen therefore provides independent comparison, not experimental reference for ice cream.

Schwartzberg model and bound water

Schwartzberg separates ice, free liquid water, and bound water that does not freeze. This gives the following low temperature limit:

$$X_{ice,max}<X_w$$

The bound water coefficient varies by food system and ingredient class. Current inputs include sugars, proteins, hydrocolloids, and salts, but they do not provide enough information to select a coefficient. The Schwartzberg curve therefore needs calibration before it can be plotted.

Experimental reference from DSC

Cogné and colleagues measured ice cream enthalpy with DSC. They used the latent heat contribution to estimate ice mass fraction on a product basis. Their standard composition contained 58.87% water, 26.08% carbohydrates, 5.75% proteins, and 9.30% lipids. The measured initial freezing point was about $-2.5\ ^\circ\mathrm{C}$. They compared physical models based on Raoult’s law, including Chen and Schwartzberg, with experimental data.

Methods have different roles:

DSC measurement: experimental reference
Goff and Hartel: formulation model
Chen: independent thermodynamic comparison
Tchigeov: older empirical comparison
Schwartzberg: model that requires bound water calibration

Agreement between Goff and Hartel and Chen does not validate either model. A difference between them shows where their assumptions produce different results.

Ice fraction estimates

Goff and Hartel, Chen, and Tchigeov for the same formulation

Model	Information retained	Role
Goff and Hartel	SE, milk salts, liquid water	Formulation calculation
Chen	Dry solids, effective molecular weight, Tf	Independent comparison
Tchigeov	Initial water and Tf	Older empirical comparison
Schwartzberg	Bound water parameter	Not calculated without calibration

Inspect temperature

-18.00°C

Chen effective molecular weight, Ms

0.55 kg/mol

Goff and Hartel, product ice50.1%

Chen, product ice37.3%

Tchigeov, product ice54.5%

Goff and Hartel, initial water frozen80.8%

Goff minus Chen12.81 pp

Goff minus Tchigeov-4.43 pp

Goff and HartelChenTchigeov

Difference between estimates

Goff minus ChenGoff minus Tchigeov

Signed difference in percentage points of product mass. Zero means calculations agree. It does not show agreement with experiment.

Change Chen $M_s$ to see how the parameter affects the result. Every plotted line uses product mass as the denominator. A separate Goff and Hartel value reports the fraction of initial water frozen.

Difference panel reports:

$$\Delta_{GC}(T)=X_{ice}^{Goff}(T)-X_{ice}^{Chen}(T)$$

and:

$$\Delta_{GT}(T)=X_{ice}^{Goff}(T)-X_{ice}^{Tchigeov}(T)$$

Zero means calculations agree at that temperature. It does not show agreement with experiment.

Limits of experimental comparison

Cogné and colleagues compared ice fraction measured by DSC with Miles, Heldman, Chen, and simplified Raoult equations. For the ice cream they tested, the largest reported difference between physical correlations and experiment was about 7%. This result does not define an error bound for the Goff and Hartel method or the formulation used here.

The interactive chart does not include DSC points because the publication does not provide machine readable pairs of temperature and ice fraction. Extracting points from the figure would add uncertainty. The chart therefore compares model calculations only.

Choose method for purpose

Model	Best use	Information compressed or required	Status
Goff and Hartel	formulation calculations	sugars grouped as SE; average milk salt relationship	main calculation
Chen	independent thermodynamic check	effective molecular weight $M_s$ required	benchmark
Tchigeov	comparison with older method	formulation reduced to $W$ and $T_f$	comparison
Schwartzberg	bound water research	calibration for specific system required	not calculated
DSC	formulation-specific validation	measured enthalpy and experimental protocol required	reference

The Goff and Hartel result is a calculated estimate. Sucrose equivalent coefficients approximate ingredient classes. The MSNF term assumes average lactose and mineral composition. Glucose syrups need coefficients suited to their dextrose equivalent. Concentrated serum may also differ from the empirical sucrose relationship.

Equilibrium models do not describe the full product structure or processing history. Proteins, stabilizers, and fat affect water mobility. Processing affects the number and size of crystals, even when equilibrium ice mass is similar. The numerical residual reports the precision of the equation solver, not the uncertainty of the physical model.

Use Goff and Hartel for formulation calculations. Use Chen as an independent check and Tchigeov for comparison with an older empirical method. Use DSC when a specific formulation needs experimental validation.

Main limits

• Goff and Hartel groups sugars as sucrose equivalents. Empirical sucrose and milk salt relationships may differ from real serum.
• Chen depends on effective $M_s$ and $T_f$. Uncertainty in either input changes result.
• Tchigeov gives same curve to formulations with same $W$ and $T_f$, regardless of solute composition.
• Schwartzberg needs a bound water parameter for a specific product.
• DSC depends on calibration, scan rate, thermal history, baseline construction, and conversion of enthalpy to ice fraction.

Ice fraction does not describe texture

Ice fraction helps explain hardness and the amount of liquid serum at serving temperature. It does not describe crystal size, air structure, fat destabilization, viscosity, or lubrication in the mouth.

Next article explains what makes ice cream creamy.

References

• H. Douglas Goff, Freezing Point Depression of a Mix, University of Guelph.
• H. Douglas Goff and Richard W. Hartel, Ice Cream, Springer, 2013.
• K. A. Fikiin, Ice content prediction methods during food freezing: a survey of the Eastern European literature, Journal of Food Engineering 38(3), 1998.
• B. A. Fricke and B. R. Becker, Evaluation of thermophysical property models for foods, HVAC&R Research 7(4), 2001.
• C. S. Chen, Thermodynamic Analysis of the Freezing and Thawing of Foods: Ice Content and Mollier Diagram, Journal of Food Science 50(4), 1985.
• C. Cogné, J. Andrieu, P. Laurent, A. Besson, and J. Nocquet, Experimental data and modelling of thermal properties of ice creams, Journal of Food Engineering 58(4), 2003.
• ASHRAE, Thermal Properties of Foods, ASHRAE Handbook: Refrigeration.
• Romain Bertin, FDS ice-cream calculation implementation, source code.