IMK114 Introduction to Food Physics Chapter 4: Thermal Properties PDF

IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties IMK 114 (Introduction to Food Physics) Chapter 4: Thermal Properties...

IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties IMK 114 (Introduction to Food Physics) Chapter 4: Thermal Properties 1 Chapter Outline Temperature Heat Capacity Thermal Conductivity Thermal diffusivity Enthalpy Heat transfer in food Caloric value of food Refer to the Thermal analysis handouts on e-learn 2 2 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Chapter learning outcome 1. Explain the thermal properties, principles and laws that are applicable in food processing. 2. Solve problems related to thermal properties of food. 3 3 Application of thermal properties in food industry To perform various heat transfer calculations involved in: ✓ designing storage and refrigeration equipment. ✓ estimating process time for refrigerating, freezing, heating, drying, microbial inactivation and various thermal processing of foods and beverages. ✓ optimize process control 4 4 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Temperature 0TH law of thermodynamics Thermal contact Thermal equilibrium Temperature scale Unit conversion of temperature 5 5 Temperature The 0th law of thermodynamics states that if two thermodynamic systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other 6 6 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties 0th law of thermodynamics Thermal contacts: Thermal Energy can be equilibrium: exchanged Two objects would between objects not exchange energy due to by heat or radiation temperature if they were placed difference in thermal contacts 7 7 c 25°C 25°C If object A and B are separately in thermal equilibrium with a third object C, then A and B are in thermal equilibrium with each other 0th law of thermodynamics Thermal equilibrium = Same temperature a b 8 8 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Temperature scale Why SI unit for Why Kelvin is temperature is referred as Kelvin? absolute temperature scale? What is How do we absolute zero obtain the temperature absolute zero temperature? 9 9 Gay-Lussac’s Law graph Figure 4.1. Pressure versus temperature (°C) 10 10 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Figure 4.2. Pressure versus temperature (K) 11 11 Temperature scale Why SI unit for Why Kelvin is temperature is referred as Kelvin? absolute temperature scale? What is How do we absolute zero obtain the temperature absolute zero temperature? 12 12 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Unit conversion of temperature Unit conversion of temperature 𝑇℃ = 𝑇𝐾 − 273.15 (4.1) 9 𝑇℉ = 5 𝑇℃ + 32℉ (4.2) 5 ∆𝑇℃ = ∆𝑇𝐾 = 9 ∆𝑇℉ (4.3) 𝑇℉ = 𝑇°𝑅 − 459.67 (4.4) ∆𝑇℉ = ∆𝑇°𝑅 (4.5) 13 13 Exercise 4.1 Consider the following pairs of material. Which pair shows that one material is twice as hot as the other? a) boiling water at 100°C and a glass of water at 50°C. b) boiling water at 100°C and frozen methane at -50°C. c) flames from hot stove at 233°C and an ice cube at -20°C. 14 14 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Heat Capacity Specific heat capacity Estimating specific heat capacity of food Frozen food Unfrozen food Various model Techniques for specific heat measurement 15 15 Heat capacity, C The heat capacity of a material is a thermal property that indicates the ability of the material to hold and store heat. The amount of energy needed to raise the temperature of the material by 1°C or 1 K. 𝑑𝑄 ∆𝑄 𝐶= = (4.5) 𝑑𝑇 ∆𝑇 Q = C∆𝑇 (4.6) Unit: J/°C or J/K 16 16 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Specific heat capacity, Cp Heat capacity per unit mass 𝑑𝑄 ∆𝑄 𝐶𝑝 = = (4.7) 𝑚𝑑𝑇 𝑚∆𝑇 Q = m𝐶𝑝 ∆𝑇 (4.8) Indication of how much energy will be required to heat or cool an object of a given mass by a given amount Unit: J/kg.°C or J/kg.K 17 17 Specific heat In unfrozen foods, specific heat change slightly as the temperature rises from capacity in 0°C to 20°C. Specific heat capacity Cp of water is unfrozen ~4186-4200 J/kg.°C food 18 18 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Specific heat capacity of frozen food For frozen foods, there is a large decrease in specific heat as the temperature decreases. Below the food’s freezing point, the sensible heat from temperature change and the latent heat from the fusion of water must be considered. Latent heat is not released at a constant temperature, but rather over a range of temperatures, so an apparent specific heat must be used to account for both the sensible and latent heat effects. 19 19 Specific heat capacity of selected food 20 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Specific heat capacity of unfrozen food There were models that had been used to estimate the specific heat of unfrozen food. The models and equations are: Heldman and Singh (1981) Siebel’s equation (1982) Chen’s model (1985) Choi and Okos model (1986) Mass average of the specific heats of the food components 21 21 Specific heat capacity for unfrozen food Choi and Okos (1986) presented a comprehensive model to predict specific heat based on composition and temperature. Refer to the attachment for the model equation. 𝑛 𝑐𝑝 = ෍ 𝑐𝑝𝑖 𝑋𝑖 where, (4.9) cp = specific heat capacity of the materials 𝑖=1 cpi = specific heat of ith component Xi = mass fraction of the ith component n = the total number of components in a food 22 22 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties 23 23 Measurement technique-Proximate analysis Water: moisture content Obtain the proximate analysis Fat: Soxhlet extraction Protein: Kjedahl method content of the food Fiber: Fiber analysis Carbohydrate: calculation Mineral: ash analysis Estimate the specific heat by using models 24 24 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Exercise 4.2 150.0 kg of lamb meat is to be cooled from 10°C to 0°C. By using mass average of the specific heats of the food components and Choi and Okos model, a) Estimate the specific heat capacity of the lamb b) Determine the amount of heat that must be removed from the lamb (Q=mcpΔT) The composition of lamb is given as follows: Water, xw= 0.7342 Fat, xf = 0.0525 Protein, xp = 0.2029 Ash, xa = 0.0106 State your answer until 4 significant figures 25 25 Specific heat of frozen food Schwartzberg (1981) developed an alternative method for determining the apparent specific heat of a food below the initial freezing point, as follows: (4.9) Where, 26 26 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties Specific heat of frozen food A slightly simpler apparent specific heat model, which is similar in form to that of Schwartzberg (1976), was developed by Chen (1985). Chen’s model is an expansion of Siebel’s equation (Siebel 1892) for specific heat and has the following form: (4.10) Where, 27 27 Measurement technique-Calorimetry Calorimetry: quantitative measurement of heat exchange Heat lost = heat gained Or Energy out of one part = energy goes into another part Calorimeter: the device used to measure heat transfer/heat exchange 28 28 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties O2 bomb calorimeter 29 Exercise 4.3: Unknown specific heat determined by the calorimetry A food technologist needs to determine the specific heat of a new type metal alloy. This metal alloy will be used to make the new baking tray. A 0.150 kg sample of the alloy is heated to 541°C. It is then placed in 0.400 kg of water at 10.0°C, which is contained in a 0.200 kg aluminum calorimeter cup. The final recorded temperature of the system is 30.5°C. Calculate the specific heat of the alloy. Given: 𝐽 𝑐𝑤 = 4186.℃ 𝑘𝑔 𝑘𝐽 𝑐𝑐 = 0.900.℃ (It is assumed that the air space between the 𝑘𝑔 insulating jacket and the cup insulates it well, so that the temperature does not change significantly. Thus, we do not need to know the mass of insulating jacket) 30 30 IMK114 : Introduction to Food Physics Chapter 4: Thermal Properties E4.3 ℎ𝑒𝑎𝑡 𝑙𝑜𝑠𝑡 𝑏𝑦 𝑎𝑙𝑙𝑜𝑦 = ℎ𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑤𝑎𝑡𝑒𝑟 + (ℎ𝑒𝑎𝑡 𝑔𝑎𝑖𝑛𝑒𝑑 𝑏𝑦 𝑐𝑎𝑙𝑜𝑟𝑖𝑚𝑒𝑡𝑒𝑟 𝑐𝑢𝑝) 𝑄 = 𝑚𝑐∆𝑡 𝑚𝑎 𝑐𝑎 ∆𝑇𝑎 = 𝑚𝑤 𝑐𝑤 ∆𝑇𝑤 + 𝑚𝑐 𝑐𝑐 ∆𝑇𝑐 𝑚𝑎 𝑐𝑎 ∆𝑇𝑎 = 𝑚𝑤 𝑐𝑤 ∆𝑇𝑤 + 𝑚𝑐 𝑐𝑐 ∆𝑇𝑐 𝐽 𝑐𝑤 = 4186.℃ 𝑘𝑔 𝐽 𝑐𝑐 = 900.℃ 𝑘𝑔 31 31