Thermal Properties. Department of Materials Engineering. Heat Capacity. C = dq dt

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1 Thermal Properties Heat Capacity Thermal Expansion Thermal Conductivity Thermal Stresses 164 Heat Capacity General: The ability of a material to absorb heat. Quantitative: The energy required to increase the temperature of the material. heat capacity (J/mol-K) C = dq dt Two ways to measure heat capacity: -- Cp : Heat capacity at constant pressure. -- Cv : Heat capacity at constant volume. energy input (J/mol) temperature change (K) 165

2 Phonons Energy is stored as atomic vibrations. As T goes up, so does the average energy of atomic vibration. 166 Heat Capacity vs. Temperature 167

3 Heat Capacity: Comparison increasing cp material Polymers Polypropylene Polyethylene Polystyrene Teflon Ceramics Magnesia (MgO) Alumina (Al2O3) Glass Metals Aluminum Steel Tungsten Gold cp (J/kg-K) at room T cp: (J/kg-K) Cp: (J/mol-K) 168 Thermal Expansion Materials change size when heating. L final L initial L initial =α(t final T initial ) Linit coefficient of Lfinal thermal expansion (1/K) Atomic view: Mean bond length increases with T. Tinit Tfinal Increasing T T 1 T 5 T 1 T 3 169

4 Thermal Expansion: Comparison increasing α Material Polymers Polypropylene Polyethylene Polystyrene Teflon Metals Aluminum Steel Tungsten Gold Ceramics Magnesia (MgO) Alumina (Al2O3) Soda-lime glass Silica (cryst. SiO2) α (10-6 /K) at room T Thermal Conductivity General: The ability of a material to transfer heat. Heat is transported in solid materials by phonons and free electrons. Quantitative: heat flux (J/m 2 -s) q = k dt dx temperature gradient thermal conductivity (J/m-K-s) T1 x1 heat flux x2 T2 > T1 171

5 Thermal Conductivity: Comparison increasing k Material Metals Aluminum Steel Tungsten Gold Ceramics Magnesia (MgO) Alumina (Al2O3) Soda-lime glass Silica (cryst. SiO2) Polymers Polypropylene Polyethylene Polystyrene Teflon k (W/m-K) Energy Transfer By vibration of atoms and motion of electrons By vibration of atoms By vibration/ rotation of chain molecules 172 Thermal Stresses Stress in a body occurs due to: --uneven heating/cooling --mismatch in thermal expansion. T o T f : σ = Eα ( T T ) = Eα T l f T f > T o stress is compressive o l Tf < To stress is tensile 173

6 Example Example Problem 19.1, p. 666, Callister 6e. --A brass rod is stress-free at room temperature (20 C). --It is heated up, but prevented from lengthening. --At what T does the stress reach -172MPa? Troom Lroom L L T =ε thermal =α(t T room ) L room 100GPa 20 x 10-6 /C σ = E( ε thermal ) = Eα(T T room ) compressive σ keeps L = 0-172MPa 20 C Answer: 106 C 174 Thermal Shock Resistance (TSR) Ex: Assume top thin layer is rapidly cooled from T1 to T2: rapid quench σ tries to contract during cooling T2 Tension develops at surface doesn t want to contract T1 σ = Eα T Temperature difference that can be produced by cooling: Critical temperature difference for fracture (set s = s f ) quench rate (T 1 T 2 ) fracture = σ f (T 1 T 2 ) = k Eα set equal Result: (quench rate ) for fracture σ fk Eα σ f k Large thermal shock resistance when Eα is large. 175

7 Summary A material responds to heat by: --increased vibrational energy --redistribution of this energy to achieve thermal equilibrium. Heat capacity: --energy required to increase a unit mass by a unit T. --polymers have the largest values. Coefficient of thermal expansion: --the stress-free strain induced by heating by a unit T. --polymers have the largest values. Thermal conductivity: --the ability of a material to transfer heat. --metals have the largest values. Thermal shock resistance: --the ability of a material to be rapidly cooled and not crack. Maximize σ f k/eα. 176