Chapter 19: Thermal Properties ISSUES TO ADDRESS... How do materials respond to the application of heat ? How do we define and measure... -- heat capacity? -- thermal expansion? -- thermal conductivity? -- thermal shock resistance? How do the thermal properties of ceramics, metals, and polymers differ? Chapter 19 - 1 Heat Capacity The ability of a material to absorb heat Quantitatively: The energy required to produce a unit rise in temperature for one mole of a material. heat capacity (J/mol-K) dQ C dT energy input (J/mol) temperature change (K) Two ways to measure heat capacity: Cp : Heat capacity at constant pressure.

Cv : Heat capacity at constant volume. Cp usually > Cv J Btu Heat capacity has units of mol K lb mol F Chapter 19 - 2 Dependence of Heat Capacity on Temperature Heat capacity... -- increases with temperature -- for solids it reaches a limiting value of 3R R = gas constant 3R = 8.31 J/mol-K Cv = constant Cv 0 0

qD From atomic perspective: T (K) Adapted from Fig. 19.2, Callister & Rethwisch 8e. Debye temperature (usually less than T room ) -- Energy is stored as atomic vibrations. -- As temperature increases, the average energy of atomic vibrations increases. Chapter 19 - 3 Atomic Vibrations Atomic vibrations are in the form of lattice waves or phonons Adapted from Fig. 19.1, Callister & Rethwisch 8e. Chapter 19 - 4 Specific Heat: Comparison increasing cp Material Polymers

Polypropylene Polyethylene Polystyrene Teflon cp (J/kg-K) at room T 1925 cp (specific heat): (J/kg-K) 1850 Cp (heat capacity): (J/mol-K) 1170 1050 Ceramics Magnesia (MgO) Alumina (Al2O3) Glass 940 775 840 Metals Aluminum Steel Tungsten Gold 900

486 138 128 Why is cp significantly larger for polymers? Selected values from Table 19.1, Callister & Rethwisch 8e. Chapter 19 - 5 Thermal Expansion Materials change size when temperature is changed initial final Tinitial Tfinal Tfinal > Tinitial l final l initial l (Tfinal Tinitial ) l initial linear coefficient of thermal expansion (1/K or 1/C)

Chapter 19 - 6 Atomic Perspective: Thermal Expansion Asymmetric curve: -- increase temperature, -- increase in interatomic separation -- thermal expansion Symmetric curve: -- increase temperature, -- no increase in interatomic separation -- no thermal expansion Adapted from Fig. 19.3, Callister & Rethwisch 8e. Chapter 19 - 7 Coefficient of Thermal Expansion: Comparison Material increasing Polymers Polypropylene

Polyethylene Polystyrene Teflon Metals Aluminum Steel Tungsten Gold Ceramics Magnesia (MgO) Alumina (Al2O3) Soda-lime glass Silica (cryst. SiO2) (10-6/C) at room T 145-180 106-198 90-150 126-216 23.6 12 4.5 14.2 13.5 7.6 9 0.4 Polymers have larger

values because of weak secondary bonds Q: Why does generally decrease with increasing bond energy? Selected values from Table 19.1, Callister & Rethwisch 8e. Chapter 19 - 8 Thermal Expansion: Example Ex: A copper wire 15 m long is cooled from 40 to -9C. How much change in length will it experience? Answer: For Cu 6 1 16.5 x 10 ( C) rearranging Equation 19.3b 0 T [

16.5 x 10 6 (1/ C)](15 m)[ 40 C ( 9 C)] 0.012 m 12 mm Chapter 19 - 9 Thermal Conductivity The ability of a material to transport heat. Fouriers Law heat flux (J/m2-s) dT q k dx temperature gradient thermal conductivity (J/m-K-s) T2 T1 x1 heat flux x2

T2 > T1 Atomic perspective: Atomic vibrations and free electrons in hotter regions transport energy to cooler regions. Chapter 19 -10 increasing k Thermal Conductivity: Comparison Material k (W/m-K) Metals Aluminum 247 Steel 52 Tungsten 178 Gold 315 Ceramics Magnesia (MgO) 38 Alumina (Al2O3) 39 Soda-lime glass 1.7 Silica (cryst. SiO2) 1.4

Polymers Polypropylene 0.12 Polyethylene 0.46-0.50 Polystyrene 0.13 Teflon 0.25 Energy Transfer Mechanism atomic vibrations and motion of free electrons atomic vibrations vibration/rotation of chain molecules Selected values from Table 19.1, Callister & Rethwisch 8e. Chapter 19 - 11 Thermal Stresses Occur due to: -- restrained thermal expansion/contraction -- temperature gradients that lead to differential dimensional changes

Thermal stress E (T0 Tf ) E T Chapter 19 -12 Example Problem -- A brass rod is stress-free at room temperature (20C). -- It is heated up, but prevented from lengthening. -- At what temperature does the stress reach -172 MPa? Solution: T0 Original conditions 0 Step 1: Assume unconstrained thermal expansion 0 Tf thermal (Tf T0 ) room

Step 2: Compress specimen back to original length 0 compress thermal room Chapter 19 - 13 Example Problem (cont.) 0 The thermal stress can be directly calculated as E(compress )

Noting that compress = -thermal and substituting gives E(thermal ) E (Tf T0 ) E (T0 Tf ) Rearranging and solving for Tf gives 20C -172 MPa (since in compression) Tf T0 E Answer: 106C 100 GPa 20 x 10-6/C Chapter 19 -14 Thermal Shock Resistance Occurs due to: nonuniform heating/cooling Ex: Assume top thin layer is rapidly cooled from T1 to T2 rapid quench

tries to contract during cooling T2 resists contraction T1 Tension develops at surface E (T1 T2 ) Critical temperature difference for fracture (set = f) Temperature difference that can be produced by cooling: quench rate (T1 T2 ) k (T1 T2 ) fracture f E

set equal (quench rate) for f k Thermal Shock Resistance (TSR) fracture E Large TSR when f k is large E Chapter 19 -15 Thermal Protection System Application: Re-entry T

Distribution Space Shuttle Orbiter Chapter-opening photograph, Chapter 23, Callister 5e (courtesy of the National Aeronautics and Space Administration.) Silica tiles (400-1260C): -- large scale application reinf C-C silica tiles (1650C) (400-1260C) nylon felt, silicon rubber coating (400C) Fig. 19.2W, Callister 6e. (Fig. 19.2W adapted from L.J. Korb, C.A. Morant, R.M. Calland, and C.S. Thatcher, "The Shuttle Orbiter Thermal Protection System", Ceramic Bulletin, No. 11, Nov. 1981, p. 1189.) -- microstructure: ~90% porosity! Si fibers bonded to one another during heat treatment. 100 mm

Fig. 19.3W, Callister 5e. (Fig. 19.3W courtesy the National Aeronautics and Space Administration.) Fig. 19.4W, Callister 5e. (Fig. 219.4W courtesy Lockheed Aerospace Ceramics Chapter 19 -16 Systems, Sunnyvale, CA.) THERMAL DIFFUSIVITY It measures the ability of a material to conduct thermal energy relative to its ability to store thermal energy.It has the SI unit of m/s. Thermal diffusivity is usually denoted by The formula is: = k/(c x rho ) where k is thermal conductivity (W/(mK)) rho is density (kg/m) c is specific heat capacity (J/(kgK)) Chapter 19 -17 Summary The thermal properties of materials include: Heat capacity: -- energy required to increase a mole of material by a unit T -- energy is stored as atomic vibrations Coefficient of thermal expansion: -- the size of a material changes with a change in temperature

-- polymers have the largest values Thermal conductivity: -- the ability of a material to transport heat -- metals have the largest values Thermal shock resistance: -- the ability of a material to be rapidly cooled and not fracture -- is proportional to f k E Chapter 19 -18 ANNOUNCEMENTS Reading: Core Problems: Self-help Problems: Chapter 19 -19