A white-hot cube of a
silica fiber insulation
material, which, only
seconds after having been
removed from a hot furnace, can
be held by its edges with the bare
hands. Initially, the heat transfer
from the surface is relatively
rapid; however, the thermal
conductivity of this material is so
small that heat conduction from
the interior [maximum
temperature approximately
1250C (2300F)] is extremely slow.
This material was developed
especially for the tiles that cover
the Space Shuttle Orbiters and
protect and insulate them during
their fiery reentry into the
atmosphere. Other attractive
features of this high-temperature
reusable surface insulation (HRSI)
include low density and a low
coefficient of thermal expansion.
Chapter 17: Thermal Properties
CHAPTER 17:
THERMAL PROPERTIES
17.1 Introduction
Thermal property: Response of materials to the
application of heat
ISSUES TO ADDRESS...
• How does a material respond to heat?
• How do we define and measure...
--heat capacity
--coefficient of thermal expansion
--thermal conductivity
--thermal shock resistance
• How do ceramics, metals, and polymers rank?
1
17.2 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)
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.
2
Vibrational Heat Capacity
Generation of lattice waves in a crystal by atomic vibrations.
The phonon versus photon
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Heat Capacity vs T
• Atomic view:
--Energy is stored as
atomic vibrations.
--As T goes up, energy
of atomic vibration goes
up too
--increases with temperature
--reaches a limiting value of 3R
Cv = constant = ~3R
The temperature
dependence of the heat
capacity at constant
volume.
qD = Debye temperature
qD = ħnmax/k
qD < Troom
HEAT CAPACITY: COMPARISON
• Why is cp significantly
larger for polymers?
Selected values from Table 19.1, Callister 6e.
4
17.3 THERMAL EXPANSION
• Materials change size when heating.
L final  L initial
 (Tfinal  Tinitial )
L initial
coefficient of
thermal expansion (1/K)
Linit
Tinit
Lfinal
Tfinal
• Atomic view: Mean bond length increases with T.
Adapted from Fig. 19.3(a), Callister
6e. (Fig. 19.3(a) adapted from R.M.
Rose, L.A. Shepard, and J. Wulff,
The Structure and Properties of
Materials, Vol. 4, Electronic
Properties, John Wiley and Sons,
Inc., 1966.)
5
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Potential energy versus interatomic
distance. Interatomic separation increases
with rising temperature. With heating, the
interatomic separation increases from r0 to
r1 to r2, and so on.
For a symmetric potential energyversus-interatomic distance curve,
there is no increase in interatomic
separation with rising temperature
(i.e., r1 r2 r3).
THERMAL EXPANSION: COMPARISON
• Q: Why does 
generally decrease
with increasing
bond energy?
Selected values from Table 19.1, Callister 6e.
For thermal expansion of fractional volume
For isotropic materials v = ~3l
6
17.4 THERMAL CONDUCTIVITY
• General: The ability of a material to transfer heat.
• Quantitative:
temperature
dT
gradient
q  k
heat flux
dx
(J/m2-s)
thermal conductivity (J/m-K-s)
• Atomic view: Atomic vibrations in hotter region carry
energy (vibrations) to cooler regions.
7
THERMAL CONDUCTIVITY
for pure metals
Selected values from Table 19.1, Callister 6e.
8
Thermal conductivity versus composition for copper–zinc alloys.
Impurities decrease thermal conductivity (scattering centers in solid solutions)
Dependence of thermal conductivity on temperature for ceramics
Nonmetallic materials
Thermal insulators
Phonons for thermal conduction
Phonon scattering by imperfections
At higher T, radiant heat transfer
Porosity: increasing pore volume
reduces thermal conductivity also
gaseous convection ineffective
17.5 Thermal Stresses
REVIEW OF ELASTIC PROPERTIES
• Modulus of Elasticity, E:
(also known as Young's modulus)
• Hooke's Law:
s=Ee
E: [GPa] or [psi]
s=Ee
s = stress
E = modulus of elasticity
e= displacement
10
17.5 THERMAL STRESSES
• Occurs due to:
--uneven heating/cooling
--mismatch in thermal expansion.
• Example:
--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 -172 MPa?
L
T
L
L room
 e thermal  (T  Troom )
100GPa
20 x 10-6 /C
s  E(e thermal )  E(T  Troom )
-172MPa
Answer: 106C
20C
9
THERMAL SHOCK RESISTANCE
• Occurs due to: uneven heating/cooling.
• Ex: Assume top thin layer is rapidly cooled from T1 to T2:
rapid quench
tries to contract during cooling T 2
doesn’t want to contract
T1
Temperature difference that
can be produced by cooling:
quench rate
(T1  T2 ) 
k
s
Tension develops at surface
s  E(T1  T2 )
Critical temperature difference
for fracture (set s = sf)
sf
(T1  T2 ) fracture 
E
set equal
sf k
• Result: (quench rate ) for fracture 
E
s k
• Large thermal shock resistance when f is large.
E
10
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THERMAL PROTECTION SYSTEM
• Application:
Space Shuttle Orbiter
Fig. 23.0, Callister 5e. (Fig. 23.0 courtesy the
National Aeronautics and Space Administration.
• Silica tiles (400-1260C):
--large scale application
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.
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
Systems, Sunnyvale, CA.)
11
Low expansion alloys
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SUMMARY
• A material responds to heat by:
--increased vibrational energy
--redistribution of this energy to achieve thermal equil.
• 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 sfk/E.
12
Heat Capacity
To what temperature would 10 lbm of a brass specimen at 25°C (77°F) be raised if 65 Btu of heat is
supplied?
Solution
We are asked to determine the temperature to which 10 lbm of brass initially at 25°C would be raised if
65 Btu of heat is supplied. This is accomplished by utilization of a modified form of Equation 17.1 as
C=dQ/dT
c=(1/m) dQ/dT
T =
Q
m cp
in which Q is the amount of heat supplied, m is the mass of the specimen, and cp is the specific heat.
From Table 17.1, cp = 375 J/kg-K for brass, which in Customary U.S. units is just
cp
2.39  10 4 Btu /lb - F 
m


= (375 J/kg - K) 
= 0.090 Btu/lb


1 J / kg - K


Thus
T =

65 Btu
= 72.2 F
(10 lb m )(0.090 Btu /lb m - F)
and

T f = T0 + T = 77F + 72.2 F = 149.2 F (65.1 C)
m - F
(a) Briefly explain why Cv rises with increasing temperature at temperatures near
0 K.
(b) Briefly explain why Cv becomes virtually independent of temperature at
temperatures far removed from 0 K.
Solution
(a) Cv rises with increasing temperature at temperatures near 0 K because, in this
temperature range, the allowed vibrational energy levels of the lattice waves are far
apart relative to the available thermal energy, and only a portion of the lattice
waves may be excited. As temperature increases, more of the lattice waves may be
excited by the available thermal energy, and, hence, the ability of the solid to
absorb energy (i.e., the magnitude of the heat capacity) increases.
(b) At temperatures far removed from 0 K, Cv becomes independent of
temperature because all of the lattice waves have been excited and the energy
required to produce an incremental temperature change is nearly constant.
Thermal Expansion
A copper wire 15 m (49.2 ft) long is cooled from 40 to –9°C (104 to 15°F). How much
change in length will it experience?
Solution
In order to determine the change in length of the copper wire, we must employ a
rearranged form of Equation 17.3b and using the value of l taken from Table 17.1 [17.0 
10-6 (°C)-1] as
l = l0 l T = l0 l (T f  T0 )


= (15 m) 17.0  10 6 (C)-1 (9C  40C)
= 1.25  10 -2 m = 12.5 mm (0.49 in.)
Briefly explain why metals are typically better thermal
conductors than ceramic materials.
Solution
Metals are typically better thermal conductors than are
ceramic materials because, for metals, most of the heat is
transported by free electrons (of which there are relatively
large numbers). In ceramic materials, the primary mode of
thermal conduction is via phonons, and phonons are more
easily scattered than are free electrons.
For some ceramic materials, why does the thermal
conductivity first decrease and then increase with
rising temperature?
Solution
For some ceramic materials, the thermal
conductivity first decreases with rising temperature
because the scattering of lattice vibrations increases
with temperature. At higher temperatures, the thermal
conductivity will increase for some ceramics that are
porous because radiant heat transfer across pores may
become important, which process increases with
rising temperature.
For each of the following pairs of materials, decide which has the larger
thermal conductivity. Justify your choices.
(a) Fused silica; polycrystalline silica.
(b) Atactic polypropylene (
Mw
Mw
= 106 g/mol); isotactic polypropylene (
= 5 × 105 g/mol).
Solution
(a) Polycrystalline silica will have a larger conductivity than fused silica
because fused silica is noncrystalline and lattice vibrations are more effectively scattered
in noncrystalline materials.
(b) The isotactic polypropylene will have a larger thermal conductivity than
the atactic polypropylene because isotactic polymers have a higher degree of
crystallinity. Since heat transfer is accomplished by molecular chain vibrations, and the
coordination of these vibrations increases with percent crystallinity, the higher the
crystallinity, the greater the thermal conductivity.
What measures may be taken to reduce the likelihood
of thermal shock of a ceramic piece?
Solution
According to Equation 17.9,
the thermal shock resistance of a ceramic piece may be enhanced by
increasing the fracture strength and thermal conductivity, and by
decreasing the elastic modulus and linear coefficient of thermal
expansion. Of these parameters, sf and l are most amenable to
alteration, usually be changing the composition and/or the
microstructure.