1 fst 151 food freezing food science and technology 151 food freezing (basic concepts) lecture notes...

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1 FST 151 FOOD FREEZING FOOD SCIENCE AND TECHNOLOGY 151 Food Freezing (Basic concepts) Lecture Notes Prof. Vinod K. Jindal (Formerly Professor, Asian Institute of Technology) Visiting Professor Chemical Engineering Department Mahidol University Salaya, Nakornpathom Thailand

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FST 151FOOD FREEZINGFOOD SCIENCE AND TECHNOLOGY 151

Food Freezing (Basic concepts)Lecture Notes

Prof. Vinod K. Jindal(Formerly Professor, Asian Institute of Technology)Visiting ProfessorChemical Engineering DepartmentMahidol UniversitySalaya, NakornpathomThailand

2

Definitions

• Chilling: Temperatures between 15oCand slightly above freezing point.

• Freezing: From slightly below freezingpoint to -18oC.

3

Purpose

• Freezing stops / retards:- Growth of microorganisms- Rate of chemical reactions- Enzyme activity- Moisture loss, if properly packaged

4

Advantages of freezing:

• Preservation of color, flavor andnutritive value.

• Microorganisms do not grow andmultiply during frozen storage.

• Freezing kills some vegetative cells.Spores survive, and may grow whenthe food is thawed.

5

Disadvantages of freezing:

• Deterioration of texture depending on the nature of food and the freezing process

• Minor losses in nutritive value and quality

• Expensive preservation operation, andrequires energy even after the operation is complete

• Depending on the storage conditions, frozen foods may also lose water

6

The engineering aspects of the freezingprocess deal with the following:

• Computing the refrigeration requirements needed to accomplish the desired reductions in product temperature

• Determining the freezing time needed to reduce product temperature to desired levels

• Understanding the changes occurring within the food product during frozen food storage.

7

• A matter can exist in three states or phases by changing the

temperature and/or pressure:

- Gas- Liquid- Solid

Some basic concepts related to freezing…

8

SOLID TO LIQUID: MELTING LIQUID TO SOLID: FREEZING

GAS TO LIQUID: CONDENSATION

LIQUID TO GAS: EVAPORATION

SOLID TO GAS: SUBLIMATION GAS TO SOLID: DEPOSITION

PHASE TRANSISTIONS

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PHASE CHANGES

Solid Liquid Gas

Melting Boiling

Freezing Condensation

Deposition

Sublimation

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PHASE CHANGES

•Freezing point (FP) – the temperature where liquids change into solids

•Melting point (MP) – the temperature where solids change into liquids

•Boiling point (BP) - the temperature where liquids change into gases

11

HEAT TRANSFER

•Exothermic – heat is removed from the system

•Endothermic – heat is added to the system

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SPECIFIC HEAT CAPACITY •A substance’s resistance to temperature change when heat is added or removed. Symbol is c.•Measured in J/g•⁰C• Joules are a measurement of energy• 4.184 J = 1 calorie 1000 calorie = 1 k calorie•Specific heat is a physical property.

•High specific heat requires more heat to change the temperature.•Water has a very high specific heat. • cwater = 4.184 J/g•⁰C• cice = 2.05 J/g•⁰C• cwater vapor = 2.08 J/g• C⁰

15

PHASE CHANGE – HEAT CHANGE

•Heat of vaporization (Hvap)– the amount of heat required to change 1 g of a substance from liquid to gas or gas to liquid.

• q = mHvap

•Heat of fusion (Hf) – the amount of heat required to change 1 g of a substance from liquid to solid or solid to liquid

• q = mHf

•Hvap for water = 2260 J/g

•Hf for water = 334 J/g

•Does it take more energy to boil 100 g of water or freeze 100 g of water?

16

PHASE CHANGE PROBLEM

• How much energy is required to boil 250 g of water that is at 100⁰C?• q = mHvap

• q = heat energy

• m = mass in grams

• Hvap = heat of vaporization of water

q = mHvap

q = 250g x 2260 J/gq = 565,000 J

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THREE STEP PROBLEM

•How much energy is released when 500 g of liquid water at 25⁰C is cooled to -15⁰C?•First calculate 25⁰C to 0⁰C

•Next, calculate freezing

•Next, calculate 0⁰C to -15 ⁰C

•Finally, add them together

q = mc∆Tq = 500g x 4.184 J/g•⁰C x 25⁰C q = 52,300 J

q = mc∆Tq = 500g x 2.03 J/g•⁰C x 15⁰C q = 15,225 J

q = mHf

q = 500g x 334J/gq = 167,000J

52,300J + 167,000J + 15,225 = 234,525 J

18

PHASE CHANGE DIAGRAM

•Things to notice:•Pressure and temperature both affect the phase of matter.•All three phases of matter exist at the triple point

Melting/Freezing

Boiling/Condensating

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In the phase diagram for pure water, three lines indicate the phase transition between solid, liquid and gas. All three lines meet at the triple point where all three phases are in equilibrium. If the pressure is lowered, we note that the boiling point will be lowered and the melting point raised (very slightly).

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Now look at the following diagram indicating the phase transitions for pure water and for water with some solute dissolved in it (not to scale).

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Freezing Point Depression

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Phase Change: Freezing/Melting

•Liquid cools at

constant ra

te

Tem

pera

tu

re

Time

•Liquid starts

freezing

•More & more

liquid freezes

FP

23

Actual FreezingT

em

pera

tur

e

Time

•Rounded: T not

uniform throughout

FP

24

Freezing & Freezing PointFreezing & Freezing Point

• A pure liquid will freeze when enough internal energy is removed from the system to the surroundings, this is usually initiated by a decrease in the surrounding’s temperature (an exothermicexothermic process).

• The exact temperature at which the solid phase is in equilibrium with the liquid phase is referred to as the “Freezing or Melting Point” and if the pressure is 1 atm (760 mmHg) then that temperature is called the “normal Freezing/Melting Point”.

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Freezing Point Depression in Solutions

The freezing point of pure water is 0°C, but that melting point can be depressed by the adding of a solvent such as a salt. A solution typically has a measurably lower melting point than the pure solvent.

A 10% salt solution may lower the melting point to -6°C (20°F) and a 20% salt solution to to -16°C (2°F).

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GGEENNEERRAAL PL PRROOPPEERRTTIIEESS O OFF S SOOLLUUTTIIOONNSS

1. A solution is a homogeneous mixture of two or more components.

2. The dissolved solute is molecular or ionic in size.

3. The solute remains uniformly distributed throughout the solution and will not settle out through time.

4. The solute can be separated from the solvent by physical methods.

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Colligative Properties of Solutions

• Colligative properties of solutions are properties that depend upon the concentration of solute molecules or ions, but not upon the identity of the solute.

• Colligative properties include freezing point depression, boiling point elevation, vapor pressure lowering, and osmotic pressure.

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Colligative Properties

• Solution properties differ from those of pure solvent

• Proportional to molality (concentration) of solute– Vapor pressure reduction

– Freezing point depression

– Boiling point elevation

– Osmotic pressure

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Colligative Property

• Magnitude of freezing point depression:

0 f,solventf,solutionf TTΔT

soluteKΔT ff

Depends on concentration of solute (not identity)

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Freezing Point Depression

Relationship for freezing point depression is:

solventfor constant depressionpoint freezing K

soltuion of ionconcentrat molal

solvent of depressionpoint freezingT :where

KT

f

f

ff

m

m

31

Freezing point depression

•Kf: “molal freezing point constant”

specific to solvent

• Units: °C

molal

•What is molal?

f,solventf,solutionff TTsoluteKΔT

32

Molality

•Concentration in molality, m:

solvent

solute

m

n

nt (kg)mass solve

te (mol)moles solumolality,m

•Independent of solution volume (V varies with T)

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Freezing point depression

• Substitution gives:

gsolvent, k

solutefff m

nKsoluteKΔT

molsolute, g/

solute, gsolute MM

mn • and:

gsolvent, kolsolute,g/m

solute,gff m MM

mK ΔT

• Using:

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Kf and Kb for some solvents

• Freezing point is lower• Boiling point is higher

Water H2O 0.0 100 1.858 0.521

Acetic acid HC2H3O2 16.6 118.5 3.59 3.08

Benzene C6H6 5.455 80.2 5.065 2.61

Camphor C10H16O 179.5 ... 40 ...

Carbon disulfide CS2 ... 46.3 ... 2.4

Cyclohexane C6H12 6.55 80.74 20 2.79

Ethanol C2H5OH ... 78.3 ... 1.07

Mpt(oC) Bpt(oC) Kf(°C/ m) Kb(°C/ m)Solvent Formula

35

Freezing Point DepressionFreezing Point Depression

TTff = - k = - kff mmQ. Estimate the freezing point of a 2.00 L sample of seawater (kf = 1.86 oC kg / mol), which

has the following composition:

0.458 mol of Na+ 0.052 mol of Mg2+ 0.010 mol Ca2+

0.010 mol K+ 0.533 mol Cl- 0.002 mol HCO3

-

0.001 mol Br- 0.001 mol neutral species.

Since colligative properties are dependent on the NUMBER of particles and not the character of the particles, you must first add up all the moles of solute in the solution.

Total moles = 1.067 moles of solute

Now calculate the molality of the solution:

mm = moles of solute / kg of solvent = 1.067 mol / 2.00 kg

= 0.5335 mol/kg

Last calculate the temperature change:

Tf = - kf m = -(1.86 oC kg/mol) (0.5335 mol/kg) = 0.992 oC

The freezing point of seawater is Tsolvent - T = 0 oC - 0.992 oC

= - 0.992 oC

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MOLALITYMOLALITY• Molality = moles of solute per kg of solvent

• m = nsolute / kg solvent

• If the concentration of a solution is given in terms of molality, it is referred to as a molal solutionmolal solution.

Q. Calculate the molality of a solution consisting of 25 g of KCl in Q. Calculate the molality of a solution consisting of 25 g of KCl in 250.0 mL of pure water at 20250.0 mL of pure water at 20ooC?C?

First calculate the mass in kilograms of solvent using the density of solvent:

250.0 mL of H2O (1 g/ 1 mL) = 250.0 g of H2O (1 kg / 1000 g) = 0.2500 kg of H2O

Next calculate the moles of solute using the molar mass:

25 g KCl (1 mol / 54.5 g) = 0.46 moles of solute

Lastly calculate the molality:

m = n / kg = 0.46 mol / 0.2500 kg = 1.8 1.8 mm (molal) solution (molal) solution

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Molality and Mole Fraction

• Two important concentration units are:

1. % by mass of solute

%100solution of mass

solute of mass = w/w%

2. Molarity

solution of Liters

solute of moles = M

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Molality and Mole Fraction (contd)

m moles of solute

kg of solvent

in dilute aqueous solutions molarity and

molality are nearly equal

• Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.

39

Food Freezing Theory

• During freezing, sensible heat is first removed to lower the temperature of a food to the freezing point. In fresh foods, heat produced by respiration is also removed. This is termed the heat load, and is important in determining the correct size of freezing equipment for a particular production rate.

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• A substantial amount of energy is therefore needed to remove latent heat, form ice crystals and hence to freeze foods.

• The latent heat of other components of the food (for example fats) must also be removed before they can solidify but in most foods these other components are present in smaller amounts and removal of a relatively small amount of heat is needed for crystallization to take place.

41

FOOD FREEZING

• Temperature lowered• Most water transformed into ice

crystals• Liquid phase concentrated• As volume of ice 10% larger than

volume of water, internal pressure in the food rised to 10 bar or more

42

Freezing CurveA typical freezing curve of a food consists of the following regions:

• The initial sensible heat removal section: To bring it to the freezing point. the temperature changes but

without change in phase.• Supercooling: In slow freezing, food temperature may

drop temporarily below the freezing point, without phase change.• Latent heat: When ice crystals form, they release heat

of fusion, and temperature increases to the freezing point.

• Final sensible heat removal: Frozen foods are kept at or below -18oC. Since the freezing points of most foods is above that, we need to cool the frozen food.

43

Figure 1. Typical freezing curve of foods

44

Typical Freezing Curve (food) Typical Freezing Curve (food) 3-21

45

SUPERCOOLING

Cooling time

A

Tem

pera

ture

Removal of latent heat

Removal of sensible heat

Removal of sensible heat

46

47

• During freezing the product temperature decreases gradually as the latent heat of fusion is removed from water within the product.

• In foods the equilibrium temperature for initial formation of ice crystals is lower than the equilibrium temperature for ice crystal formation in pure water.

• The magnitude of the depression in equilibrium freezing temperature is a function of product composition.

48

• After the formation of initial ice crystals in the food product, the removal of phase change energy occurs gradually over a range of decreasing product temperatures.

• The temperature–time relationship during phase change is a function of the percent water frozen at any time during the freezing process.

• The shape of the temperature–time curve during the freezing process will vary with product composition and with the location within the product structure.

49

• The gradual decrease in temperature with time will continue until reaching the eutectic temperatures for major product components. In practice, food products are not frozen to sufficiently low temperatures to reach these eutectic temperatures.

50

Time–temperature data during freezing.

51

• AS The food is cooled to below its freezing point f which, with the exception of pure water, is always below 0ºC . At point S the water remains liquid, although the temperature is below the freezing point. This phenomenon is known as supercooling and may be as much as 10ºC below the freezing point.

• SB The temperature rises rapidly to the freezing point as ice crystals begin to form and latent heat of crystallisation is released.

52

• BC Heat is removed from the food at the same rate as before, but it is latent heat being removed as ice forms and the temperature therefore remains almost constant. The freezing point is gradually depressed by the increase in solute concentration in the unfrozen liquor, and the temperature therefore falls slightly. It is during this stage that the major part of the ice is formed .

• CD One of the solutes becomes supersaturated and crystallizes out. The latent heat of crystallization is released

53

• DE Crystallization of water and solutes continues. The total time tf taken (the freezing plateau) is determined by the rate at which heat is removed.

• EF The temperature of the ice–water mixture falls to the temperature of the freezer. A proportion of the water remains unfrozen at the temperatures used in commercial freezing; the amount depends on the type and composition of the food and the temperature of storage. For example at a storage temperature of -20ºC the percentage of water frozen is 88% in lamb, 91% in fish and 93% in egg albumin.

54

55

INITIAL FREEZING POINT OF DIFFERENT FOODS (OC)

• Beef -1.1• Common fruits - 0. to – 2.7• Common vegetables - 0.8 to – 2.8• Eggs - 0.5• Milk - 0.5

56

Effects of Freezing

• The goal of freezing is to reduce thetemperature of the food below -18oC

• This results in the crystallization of partof the water and some of the solutes inthe food

• Water in the frozen state does not act asa solvent, can not enter into chemicalreactions, and is not available tomicroorganisms.

57

Effect of water

• Water has a major impact on the freezing behavior of foods. Most foods contain large amounts of water (fruits/vegetables: up to 90 wt% water, meats: 65 - 75% water).

• Freezing point, the heat capacity aboveand below freezing, and the latent heatof freezing are strongly affected by themoisture content.

58

Freezing point

• Pure water freezes at 00C under 1 atm.

• The freezing point of foods will be close to 00C depending on the amount of water present.

• In general, the higher the moisture content, the closer the freezing point to 00C.

59

Water binding

• The degree of binding of water by thefood is also important. Water that istightly bound will not freeze as readilyas "free water".

• In foods, there are inorganic and organic substances such as sugars, acids, salts, colloids, etc. dissolved in water.

60

Supercooling

• It is possible to reduce the temperature ofwater below 00C, and still have liquid water at 1 atm. This is called supercooling.

• If we cool a solution with no ice crystals, the temperature can be lowered below the

freezing point.

• Supercooling is an unstable and temporarystate. It occurs because there is no "nucleus" for the crystals to form around. Occurrence of supercooling depends on the rate of freezing.

61

Crystallization of Water

• Formation of a regularly organized solid phase from a solution.

• There are two steps in the process:

• Nucleation• Crystal growth

62

Nucleation

• When the first ice crystal forms, it starts nucleation and the solidification process. The nucleus occurs homogeneously or heterogeneously.

• In very pure water, nucleation is homogeneous,the "nucleus" is water molecules orienting as crystals. Homogeneous nucleation does not occur in foods.

• In heterogeneous nucleation, crystals form around foreign particles, surface films, container walls. Ice crystal formation is easier in this case.

63

Crystal growth

• The rate of crystal growth depends on:

- How fast water molecules get to the nucleus.

- How fast heat is removed from the system to favor orientation and ordering of the water molecule.

64

Crystal growth

• If a material is cooled rapidly, the rate ofnucleation is greater than the rate of

crystal growth, and many small crystalswill form.

• These crystals have no time to attractmore water molecules and grow because new crystal formation is faster, and it

consumes the available water.

65

Crystal growth

• If the rate of cooling is slow, the rate ofcrystal growth is faster than the formation

of new crystals, and there will be a smaller number of large crystals.

• At temperature near the melting point, water molecules add to the mass of the

nucleus rather than forming new nuclei.

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Freezing point

• In foods, there are solutes, suspendedmatter, and cellular material in solution.

• Ice crystals "squeeze" other moleculesout. Solutes get concentrated. Thisreduces the freezing point.

• In most foods, there is a narrow rangeof temperature at which freezing occurs.

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Ice crystals

• The number and shape of ice crystals has amajor effect on the quality of frozen foods.

• Large crystals damage the tissue. Meat,poultry, fish, shellfish, fruit and vegetablecells contain jelly-like protoplasm. Large icecrystals puncture cells. After thawing, theycannot reach their former state.

• Small crystals do not injure the tissue asmuch. When thawed, they can be re-absorbedinto the protoplasm. Thaw-drip is minimized.

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• The location of ice crystals in food tissues depends on freezing rate, temperature, nature of the cells.

• Slow freezing (less than 10C/min) causes thecrystals to form exclusively in the extracellular spaces. This shrinks the cells, disrupts tissues and results in lower quality.

• Freezing starts at extracellular space. Inside the cell is a supercooled solution. Its water vapor pressure is higher than that of extracellular ice. This difference in vapor pressure causes water to migrate from inside the cell to extracellular space.

69

Re-crystallization

• Largest possible size, and no defects in thecrystal lattice is the thermodynamically stable form of a crystal.

• Small crystals try to coalesce into bigger ones.

• The number, size, shape and orientation ofcrystals change during storage. The rate ofre-crystallization depends on temperature.

70

Re-crystallization

• Lower temperature implies slower re-crystallization.

• Keep foods at as low a temperature aspossible. Fluctuations in temperaturehelp re-crystallization.

• Pure ice re-crystallizes at a significantrate at -70oC. In regular tissue food,the rate is very slow at -28oC.

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Eutectic

• Each solute has a solubility limit in water. As more water is removed by freezing, a point is reached where the solute is

saturated in the remaining solution at that temperature.

• Further freezing of water results incrystallization of solute together withwater. Such simultaneous crystallization is called a eutectic and the temperature is known as the eutectic point of the solute.

72

Examples

• For example, a NaCl solution in water has a eutectic point of -21oC. When this temperature is reached, water and salt crystallize together.

• Since the concentration of the remaining solution does not change after this point, temperature is constant until all water and salt crystallize.

• Sucrose solutions have a eutectic point of -14oC.

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Eutectic Point

• Temperature where there is no further concentration of solutes due to freezing, thus the solution freezes.• Temperature at which a crystals of individual

solute exists in equilibrium with the unfrozen liquor and ice.

• Difficult to determine individual eutectic points in the complex mixtures of solutes in foods so the term Final Eutectic Point is used

• This represents lowest Eutectic temperature of the solutes in the food.

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Eutectic temperatures

• Ice Cream -55oC• Meat -50 to -60oC• Bread -70oC

MAXIMUM ICE CRYSTALS FORMATION IS NOT POSSIBLE UNTIL THIS TEMPERATURE IS REACHED

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Effects of Freezing

• Volume change : Most substances shrinkwhen going from the liquid to the solid state.Water is anomalous : its volume increases

as it freezes.

• Conversion of water to ice causes about 9%volume increase. Volume change in food

during freezing depends on its composition. Concentrated sugar solutions do not expand when frozen, they may even shrink.

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Volume change

The change in volume depends on:

• Percent water. More water = larger expansion.• Air pockets. They absorb the growth of crystals.• Unfreezable water. If there are many solutes

present, water is bound, and will not freeze.• Temperature. Before the freezing point of water

is reached, its volume decreases. Maximumdensity is around 4oC. Below that temperature, it will expand.

77

Effects of volume change

• During freezing some parts of food contract, and some expand. This causes mechanical stresses. If these stresses are allowed enough time to dissipate, there will be mo major mechanical damage to the material.

• If the rate of freezing is very fast (cryogenic freezing) the material cracks.

78

Concentration Effects

• Freezing removes water from the food,and the solute concentration increases.

• Changes in electrolytes may causeirreversible changes in colloidal structures.

Milk proteins may coagulate. Changes in pH, ionic strength, viscosity, surface tension, redox potential, freezing point, etc.

may result.