FST 151 FOOD FREEZING FOOD SCIENCE AND TECHNOLOGY 151 Food Freezing (Basic concepts) Lecture Notes

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

2. Definitions • Chilling: Temperatures between 15oC and slightly above freezing point. • Freezing: From slightly below freezing point 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 and nutritive value. • Microorganisms do not grow and multiply during frozen storage. • Freezing kills some vegetative cells. Spores survive, and may grow when the 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, and requires energy even after the operation is complete • Depending on the storage conditions, frozen foods may also lose water

6. The engineering aspects of the freezing • process 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. Some basic concepts related to freezing… A matter can exist in three states or phases by changing the temperature and/or pressure: - Gas - Liquid - Solid

8. PHASE TRANSISTIONS • SOLID TO LIQUID: MELTING • LIQUID TO SOLID:FREEZING • GAS TO LIQUID: CONDENSATION • LIQUID TO GAS:EVAPORATION • SOLID TO GAS: SUBLIMATION • GAS TO SOLID: DEPOSITION

9. PHASE CHANGES Sublimation Melting Boiling Solid Liquid Gas Freezing Condensation Deposition

10. 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

12. 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•⁰C

13. 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?

14. PHASE CHANGE PROBLEM q = mHvap q = 250g x 2260 J/g q = 565,000 J • 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

15. THREE STEP PROBLEM q = mc∆T q = 500g x 4.184 J/g•⁰C x 25⁰C q = 52,300 J q = mHf q = 500g x 334J/g q = 167,000J q = mc∆T q = 500g x 2.03 J/g•⁰C x 15⁰C q = 15,225 J 52,300J + 167,000J + 15,225 = 234,525 J 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

16. PHASE CHANGE DIAGRAM Melting/Freezing Boiling/Condensating Things to notice: Pressure and temperature both affect the phase of matter. All three phases of matter exist at the triple point

17. 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).

18. 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).

19. Freezing Point Depression

20. Phase Change: Freezing/Melting • Liquid cools at constantrate Temperature • Liquid starts freezing • More & more liquid freezes FP Time

21. Actual Freezing • Rounded: T not uniform throughout Temperature FP Time

22. Freezing & 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 exothermic 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”.

23. 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).

24. GENERAL PROPERTIES OF SOLUTIONS 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.

25. 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.

26. 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

27. Colligative Property • Magnitude of freezing point depression: Depends on concentration of solute (not identity)

28. Freezing Point Depression • Relationship for freezing point depression is:

29. Freezing point depression • Kf: “molal freezing point constant” specific to solvent • Units: °C molal • What is molal?

30. Molality • Concentration in molality,m: • Independent of solution volume (V varies with T)

31. Freezing point depression • Using: • Substitution gives: • and:

32. Kf and Kb for some solvents • Freezing point is lower • Boiling point is higher

33. Freezing Point DepressionDTf = - kfm Q. 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: m = moles of solute / kg of solvent = 1.067 mol / 2.00 kg = 0.5335 mol/kg Last calculate the temperature change: DTf = - kfm = -(1.86 oC kg/mol) (0.5335 mol/kg) = 0.992 oC The freezing point of seawater is Tsolvent - DT = 0 oC - 0.992 oC = - 0.992 oC

34. MOLALITY • 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 solution. Q. Calculate the molality of a solution consisting of 25 g of KCl in 250.0 mL of pure water at 20oC? 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 m (molal) solution

35. Molality and Mole Fraction Molarity Two important concentration units are: % by mass of solute

36. Molality and Mole Fraction (contd) Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.

37. 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.

38. 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.

39. 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

40. Freezing Curve A 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.

41. Figure 1. Typical freezing curve of foods

42. 3-21 Typical Freezing Curve (food)

43. SUPERCOOLING A Removal of sensible heat Temperature Removal of latent heat Removal of sensible heat Cooling time

45. 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.

46. 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.

47. 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.

48. Time–temperature data during freezing.

49. 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.

50. 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