Freezing & Freezing Equipment Notes

Last reviewed 1 Oct 2026 · 8 min read

Why freeze?

Freezing preserves food by lowering the temperature (to −18 °C or below) so that most water turns to ice, making water unavailable (lower $a_{w}$ ) and slowing microbial and enzyme activity, with minimal changes in flavour, colour and nutrients if done properly. Used for fruits, vegetables, meat, fish, ready meals, bakery and ice cream.

Principles

Freezing point depression

Dissolved solutes (sugars, salts) lower the freezing point of food below 0 °C: most fruits and vegetables begin to freeze at about −0.5 to −3 °C (initial freezing point), meat and fish at about −1 to −2 °C, juice with 12 % TSS at about −2 °C. As ice forms, the remaining solution becomes more concentrated and its freezing point falls further, so freezing is a gradual process and some water stays unfrozen even at −18 °C (about 10–15 % of the water in meat; more in sugary products), with ice-cream-like products remaining soft.

Temperature–time history

Stages of freezing a food:

Pre-cooling from the initial temperature to the freezing point.
Phase change (thermal arrest) — the temperature stays near the freezing point while the main part of the water freezes (the longest stage; most energy removed).
Tempering (sub-cooling) — further cooling to the final storage temperature (−18 °C or lower).

Super-cooling (the temperature falls below the freezing point before nucleation begins) can occur by a few degrees, then rises to the freezing point on nucleation.

Ice crystal formation

Nucleation (formation of small ice nuclei) and crystal growth.
Slow freezing: few nuclei, large crystals mostly outside the cells (extracellular); water leaves the cells (osmotic dehydration), cell walls are ruptured, giving drip loss on thawing, soft texture and flavour loss.
Quick freezing: many small nuclei, small crystals both inside and outside the cells; less cell damage and drip; better quality.
Definition: quick freezing — passing through the zone of maximum ice-crystal formation (about −1 to −5 °C) in about 30 minutes or less (the guide of an ice front advancing at 1–5 cm/h or more is also used).
Recrystallisation — temperature fluctuations during storage make small crystals melt and re-freeze into large ones, degrading texture (ice cream becomes gritty).

Other effects

Volume expansion (about 9 % for water to ice) — cracks in fruits (cherries) and packs.
Cell damage; enzyme activity; protein denaturation; lipid oxidation (rancidity); freezer burn (surface dehydration by sublimation — grey-white dry patches); colour changes (browning in fruits, discolouration of meat to brown).
Microbes are inhibited, not killed; some die from freezing and storage (vegetative cells, parasites partially), spores and toxins survive.

Freezing time — Plank's equation

Plank's equation (for a food of uniform shape freezing from an initial temperature at its freezing point) gives the time to freeze:

FormulaPlank's equation

t_{f} = \frac{ρ λ}{T _{f} - T _{a}} (\frac{P a}{h} + \frac{R a ^{2}}{k _{f}})

$ρ$ = density of the food (kg/m³), $λ$ = latent heat of freezing of the food (J/kg) (about $water fraction \times 334$ kJ/kg), $T_{f}$ = initial freezing point (°C), $T_{a}$ = freezing medium temperature (°C).
$a$ = thickness (slab) or diameter (cylinder, sphere) (m), $h$ = surface heat transfer coefficient (W/m²K), $k_{f}$ = thermal conductivity of the frozen food (W/mK).
Shape constants: infinite slab: $P = \frac{1}{2}$ , $R = \frac{1}{8}$ ; infinite cylinder: $P = \frac{1}{4}$ , $R = \frac{1}{16}$ ; sphere: $P = \frac{1}{6}$ , $R = \frac{1}{24}$ .

Interpretation: freezing time is proportional to the (thickness or diameter)² (at a large internal resistance), inversely proportional to the temperature difference, and falls with higher $h$ (faster air, liquid); it is an estimate (assumes a single freezing point and neglects pre-cooling and tempering).

Worked ExampleExample — freezing time of a fish slab

A fish block of thickness 0.04 m has $ρ = 1000$ kg/m³, effective latent heat $λ = 250$ kJ/kg (about 75 % water $\times$ 334 kJ/kg), initial freezing point $T_{f} = - 1.5$ °C, frozen conductivity $k_{f} = 1.6$ W/mK. The air-blast freezer is at $T_{a} = - 30$ °C with $h = 30$ W/m²K.

$Δ T = - 1.5 - (- 30) = 28.5$ K; $\frac{ρ λ}{Δ T} = \frac{1000 \times 2.5 \times 1 0 ^{5}}{28.5} = 8.77 \times 1 0^{6}$ .
$\frac{P a}{h} = \frac{0.5 \times 0.04}{30} = 6.67 \times 1 0^{- 4}$ ; $\frac{R a ^{2}}{k _{f}} = \frac{0.125 \times 0.0016}{1.6} = 1.25 \times 1 0^{- 4}$ ; sum $= 7.92 \times 1 0^{- 4}$ .
$t_{f} = 8.77 \times 1 0^{6} \times 7.92 \times 1 0^{- 4} = 6, 945$ s $\approx 1.9 h$ .

Doubling $h$ (to 60 W/m²K, e.g., by a faster air flow) gives $\frac{P a}{h} = 3.33 \times 1 0^{- 4}$ and a total of $4.58 \times 1 0^{- 4}$ , so $t_{f} = 4, 017$ s $\approx 1.1 h$ — a 42 % reduction.

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