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?
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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 
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•
Dependence of Heat Capacity on
Heat capacity... Temperature
-- increases with temperature
-- for solids it reaches a limiting value of 3R
R = gas constant 3R
= 8.31 J/mol-K
0
Cv = constant
0
qD
• From atomic perspective:
T (K)
Debye temperature
(usually less than T room )
-- Energy is stored as atomic vibrations.
-- As temperature increases, the average energy of
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atomic vibrations increases.
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Atomic Vibrations
Atomic vibrations are in the form of lattice waves or
phonons
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Specific Heat: Comparison
increasing cp
Material
• Polymers
Polypropylene
Polyethylene
Polystyrene
Teflon
cp (J/kg-K)
at room T
1925
1850
1170
1050
• Ceramics
Magnesia (MgO)
Alumina (Al2O3)
Glass
940
775
840
• Metals
Aluminum
Steel
Tungsten
Gold
900
486
138
128
cp (specific heat): (J/kg-K)
Cp (heat capacity): (J/mol-K)
• Why is cp significantly
larger for polymers?
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Thermal Expansion
Materials change size when temperature
is changed
 initial
 final
Tinitial
Tfinal
Tfinal > Tinitial
 final   initial
 α (Tfinal  Tinitial )
 initial
linear coefficient of
thermal expansion (1/K or 1/°C)
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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
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Coefficient of Thermal Expansion:
Comparison
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Material
increasing a
• Polymers
Polypropylene
Polyethylene
Polystyrene
Teflon
• Metals
Aluminum
Steel
Tungsten
Gold
• Ceramics
Magnesia (MgO)
Alumina (Al2O3)
Soda-lime glass
Silica (cryst. SiO2)
a (10 /C)
at room T
145-180
106-198
90-150
126-216
23.6
12
4.5
14.2
Polymers have larger
a values because of
weak secondary bonds
• Q: Why does a
generally decrease
with increasing
bond energy?
13.5
7.6
9
0.4
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Thermal Conductivity
The ability of a material to transport heat.
Fourier’s Law
heat flux
(J/m2-s)
dT
q  k
dx
temperature
gradient
thermal conductivity (J/m-K-s)
T2
T1
x1
heat flux
T2 > T1
x2
• Atomic perspective: Atomic vibrations and free electrons in
hotter regions transport energy to cooler regions.
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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
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Thermal Stresses
• Occur due to:
-- restrained thermal expansion/contraction
-- temperature gradients that lead to differential
dimensional changes
Thermal stress 
 Ea (T0 Tf )  Ea T

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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
  Ea (T1 T2 )
Critical temperature difference
for fracture (set  = f)
Temperature difference that
can be produced by cooling:
(T1  T2 ) 
quench rate
k

(T1 T2 ) f racture 
f
Ea
set equal
• (quench rate) f or f racture  Thermal
Shock Resistance ( TSR) 
• Large TSR when
f k
is large
Ea
f k
Ea
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Thermal Protection System
• Application:
Space Shuttle Orbiter
Re-entry T
Distribution
reinf C-C
silica tiles
(1650°C) (400-1260°C)
nylon felt, silicon rubber
coating (400°C)
• Silica tiles (400-1260C):
-- large scale application
-- microstructure:
~90% porosity!
Si fibers
bonded to one
another during
heat treatment.
100 mm
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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
Ea
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Thermal Expansion: Example
Ex: A copper wire 15 m long is cooled from
40 to -9°C. How much change in length will it
experience?
• Answer: For Cu
a  16.5 x 106 ( C)1
rearranging Equation 17.3b
  a   0 T  [
16.5 x 10 6 (1/ C)](15 m)[ 40C  ( 9C)]
  0.012 m  12 mm
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Example Problem
-- A brass rod is stress-free at room temperature (20°C).
-- 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  a (Tf T0 )
room
Step 2: Compress specimen back to original length
0




compress 

room
 thermal
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Example Problem (cont.)
0


The thermal stress can be directly
calculated as
  E(compress)
Noting that compress = -thermal and substituting gives
  E(thermal )  Ea
 (Tf T0 )  Ea (T0 Tf )
Rearranging and solving for Tf gives

20ºC
Tf  T0 
Answer: 106°C

100 GPa

Ea
-172 MPa (since in compression)
20 x 10-6/°C
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