Food Physics Thermodynamics PDF

Food Physics Prof. Enda Cummins Email: [email protected] Extn:7476 3. Thermodynamics 3.1 SPECIFIC HEAT • a measure of the amount of energy required to raise unit mass (1 kg) by unit temperature rise (1oC) •in batch heating or cooling, amount of energy required is Q = mass x average specific heat...

Food Physics Prof. Enda Cummins Email: [email protected] Extn:7476 3. Thermodynamics 3.1 SPECIFIC HEAT • a measure of the amount of energy required to raise unit mass (1 kg) by unit temperature rise (1oC) •in batch heating or cooling, amount of energy required is Q = mass x average specific heat x temp change = MCΔT (J) In a continuous process, rate of heat transfer is Q/t = mass flow rate x specific heat x temp Change 3.1.1 GASES AND VAPOURS •Two sets of conditions - Specific heat Cv at constant volume is amount of heat required to raise unit mass by unit temp rise under constant volume - Specific heat Cp at constant pressure is amount of heat required to raise unit mass by unit temp rise at constant pressure How can it be found •Calculation from composition Water: 4180 J/kg/k Protein: 1900 J/kg/k Fat: 1900J/kg/k Carbohydrate: 1220 J/kg/k Specific heat = sum(Xi. Cp,i) Where: Xi = fraction of a given component Cp,i = specific heat of that component Crude approximation Water rich foods: 3700 J/kg. oC Oil based foods: 2100 J/kg. oC Frozen foods: 2000 J/kg. oC Based on the fact that ice has a specific heat of 2110 J/kg. oC 3.1.2 DETERMINATION OF SPECIFIC HEAT 3.1.2.1 Method of Mixtures • Specimen of known mass and temp is dropped into a fluid of known mass and temp contained in a copper/aluminium container called a calorimeter • Readings required: ms, mc and mf are the masses of the specimen, calorimeter and fluid. Initial temp of specimen and fluid are Ts and Tf and the final temp of the mixture is Tm. •Then the heat lost by the specimen is mscs(Ts-Tm), heat gained by the fluid is mfcf(Tm-Tf) and heat gained by calorimeter is mccc(Tm-Tc) Therefore mscs(Ts-Tm) = mfcf(Tm-Tf) + mccc(Tm-Tc) and cs  T  m  T f mc c f  mc cc  Ts  Tm ms • Greatest source of experimental error come from heat loss to surroundings, especially if specific heat is low • Solution: insulate calorimeter Method of mixtures • Insulated copper beaker, weigh beaker •Add known weight of hot water, weigh again, measure tempetature •Add a known weight of food at a known temperature •Allow to come to equilibrium, measure temperature Insulation • Change in thermal energy of water + change in thermal energy of beaker = change in thermal energy of the food 3.1.2.2 Method of Cooling • At a given temp, rate of loss of heat from 2 liquids contained in calorimeters is equal • Newtons law of cooling: rate of heat loss from a body is directly proportional to excess temp of the body above that of its surroundings If m1 and m2 are the masses of sample and control liquid (often water), c1 and c2 are their respective specific heats, (DT/dt)1 and (DT/dt)2 are their respective cooling rates, and mc and cc are the mass and specific heat of the calorimeter, then: Rate of heat loss from sample = Rate of heat loss from reference material  dT   dT  m1c1  mc cc    m2c2  mc cc    dt 1  dt  2 • Since specific heat is only unknown • Rate of heat loss depends surface area of sample, temp of surroundings and nature of calorimeter surface • cooling rate at particular temp determined from gradient of tangent to the curve • Alternative method: Use a comparison calorimeter 3.1.2.3 Electrical methods • Insert an electrical element in a calorimeter containing the test fluid and measure temp rise by the application of a known voltage and fixed current for a fixed time • Heat loss is eliminated by cooling the liquid to below room temp and applying electrical energy until the temp reaches a value above room temp such that the temp differences above and below room temp are the same VIt = mcccΔT +m1c1ΔT V=voltage (V), I=current (A), t=times (s), ΔT= temp diff (oC), m1 and mc= masses of the liquid and calorimeter respectively, c1 and cc= specific heat of liquid and calorimeter respectively (J/kg/K) • Continuous flow calorimeters have been devised: liquid passes through a tube, containing electrical element at constant flow rate m’ (kg/s) with the application of steady voltage V (V) and current I (A) m’CLΔT = VI Flow rate can be measured and only unknown is the specific heat of liquid CL 3.2 LATENT HEAT • Previously - Sensible heat changes • Many stages in food processing encounter phase change = energy changes • Solid, liquid and gass • E.g. water 3.2 LATENT HEAT • Most usual form phase diagrams (Fig. 8.3) take is pressure vs temp • Where solid (S), liquid (L) and vapour (V) in equilibrium is triple point of water • Pressure and triple point temp are 4.6 and 0.01 oC • Line AT - pressure temp conditions where solid and liquid are in equilibrium. ie. melting point line • Line TB - vapour and liquid are in thermal equilibrium. It predicts how boiling point of liquid varies with pressure • Line TC - solid and vapour are in equilibrium. Thus if water vapour pressure is maintained below 4.6 and water is frozen, when energy is supplied solid will go directly to vapour. Process is known as sublimation and is how water is removed by freeze drying • Looking at CO2. Triple point conditions are -57 oC and 5.4 atm. So solid CO2 will sublime directly at atmospheric pressure. 3.2.1 Behaviour of ice and water during heating • Water frozen to -40 oC, energy supplied at constant rate to heat the ice. Temperature is plotted against time, following results (Fig. 8.4) • Temp rises steadily until it reaches B (melting point of water) • At this point latent heat energy is required to melt the ice plateau BC •Temp remains at 0 oC until all ice melted at C. Temp then rises steadily to boiling point D. • Slope CD for water heating is about half AB for the ice because of the different specific heat capacities • At point D, water will start to boil; latent heat energy is required to cause vaporisation. Temp remains constant until all water is converted to vapour. Point D represents a saturated liquid at its boiling point. • Point E represents a saturated vapour at its boiling point and anywhere between D and E represents the coexistence of vapour and liquid (wet vapour) • If the saturated vapour can be heated further, the timetemp relationship follows EF, here vapour is superheated. Lengths of plateau’s BC and DE are different reflecting different values for fusion and vaporisation SOLID LIQUID VAPOR (latent heat of fusion) (latent heat of vaporisation) 335 kJ/kg 2257 kJ/kg Energy absorbed Energy released 3.2.2 Behaviour of foods during freezing • When energy is removed from pure water, temp falls and at 0 oC ice will start to separate • Temp will remain constant as latent heat removed until all water is frozen; temp of ice will then fall • Foods – freeze –1 oC • Removal of concentrates solute and further depresses freezing point • As more ice crystallises, concentration of solutes in unfrozen material increases. Known as freeze concentration and occurs in solid and liquid foods. • Increase in concentration results in increase in reaction rates, despite reduction in temp • Therefore effective freezing time is defined as the time required to reduce the temp at the slowest cooling point from the ambient temp to -15 oC • Very dependent on the size and nature of product and method of freezing • Freezing times vary from < 1min (pea frozen in liquid nitrogen) to over 48 hours (for a large meat carcass) frozen in cold air 3.2.3 Energy Changes during Freezing Eg. amount of heat removed to reduce 200kg of apples from 25 oC to -20 oC is: Heat removed to reduce from 25 oC to -1 oC = mcΔT = 200 x 3.59 x26 =18668 kJ Latent heat of fusion Heat removed to freeze food = mλ = 200 x 2.815 x 102 = 56300 kJ Heat removed to reduce from -1 oC to -20 oC = mcΔT = 200 x 1.88 x 19 = 7144 kJ Therefore total heat = 18668 + 56300 + 7144 kJ = 82112 kJ Need to remove 82,112 kJ of energy to freeze Note: 68.6% latent heat 3.3 ENTHALPY •Enthalpy (H) is a thermodynamic property. Is the sum of the internal energy + pressure x volume H = U + pV Units: SI= joule, but also calorie The change in a food’s enthalpy can be used to estimate the energy that must be added or removed to effect a temperature change. Above the freezing point, enthalpy consists of sensible energy; below the freezing point, enthalpy consists of both sensible and latent energy. Enthalpy (H) ~ heat content (q) @ constant pressure DH = thermal (heat) energy change = q Physical process H2O (l) + energy  H2O (g) DHvap = 44 kJ/mole heat of vaporization What happens for the reverse process? DH = - 44 kJ/mole and exothermic 3.3 ENTHALPY •Changes in enthalpy is given by ΔH = ΔU + work + ΔpΔV But p is constant, ΔH = ΔU + work + pΔV Assume only work required is the pressure to change volume ΔH = ΔU - pΔV + pΔV ΔH ~ ΔU ΔH ~ change in heat q • For a process taking place at constant pressure, enthalpy change is equal to heat absorbed or evolved. • If enthalpy change is positive, heat is absorbed and reaction is endothermic • If negative, heat is evolved and reaction is exothermic • Because many reactions take place at constant pressure, enthalpy is often equated with heat content Eg. when methane is burned CH4 + 2O2 CO2 + 2H2O ΔH = -890 kj/mol Learning outcomes  Explain the basic science governing selected physical properties of food  Specific heat  Approximation for food  Analytical Methods of measurement  Latent heat  Phase diagrams  Enthalpy  Perform simple calculations involving physical properties  Energy Changes during Freezing