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CHAPTER 19
THERMAL PROPERTIES
PROBLEM SOLUTIONS
Heat Capacity
19.1 Estimate the energy required to raise the temperature of 2 kg of the following materials from
20 to 100°C (293 to 373 K): aluminum, steel, soda–lime glass, and high-density polyethylene. Solution
The energy, E, required to raise the temperature of a given mass of material, m, is the product
of the specific heat, the mass of material, and the temperature change, ∆T, as
pE c m T= ∆
The ∆T in this problem is equal to 100°C – 20°C = 80°C (= 80 K), while the mass is 2 kg, and the
specific heats are presented in Table 19.1. Thus,
5(aluminum) = (900 J/kg K)(2 kg)(80 K) 1.44 10 JE ⋅ = ×
4(steel) = (486 J/kg K)(2 kg)(80 K) 7.78 10 JE ⋅ = ×
5(glass) = (840 J/kg K)(2 kg)(80 K) 1.34 10 JE ⋅ = ×
5(HDPE) = (1850 J/kg K)(2 kg)(80 K) 2.96 10 JE ⋅ = ×
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19.2 To what temperature would 11 kg of a 1025 steel specimen at 25°C (298 K) be raised if 130
kJ of heat is supplied? Solution
We are asked to determine the temperature to which 11 kg of steel initially at 25°C would be
raised if 130 kJ of heat is supplied. This is accomplished by utilization of a modified form of Equation
19.1 as
p
QTmc∆
∆ =
in which ∆Q is the amount of heat supplied, m is the mass of the specimen, and cp is the specific heat. From Table 19.1, cp = 486 J/ kg K⋅ for steel.
Thus
3130 10 Joules 24 K
(11 kg)(486 J/kg K)T ×
∆ = =⋅
and
0 + = 298K +24K =322K (49 C)fT T T= ∆ °
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19.3 (a) Determine the room temperature heat capacities at constant pressure for the following
materials: aluminum, silver, tungsten, and 70Cu–30Zn brass. (b) How do these values
compare with one another? How do you explain this? Solution
(a) This problem asks that we determine the room-temperature heat capacities at constant pressure, Cp,
for aluminum, silver, tungsten, and 70Cu–30Zn brass. All we need do is multiply the cp values in
Table 19.1 by the atomic weights (values are found inside the front cover), taking into account the
conversion from grams to kilograms (for the atomic weights). Thus, for Al
(900 J/kg K)(1 kg/1000 g)(26.98 g/mol) = 24.3 J/mol KpC = ⋅ ⋅
For Ag
(235 J/kg K)(1 kg/1000 g)(107.87 g/mol) = 25.4 J/mol KpC = ⋅ ⋅
For W
(138 J/kg K)(1 kg/1000 g)(183.84 g/mol) = 25.4 J/mol KpC = ⋅ ⋅
For brass it is first necessary to compute the alloy atomic weight (Aave) using Equation 4.11a
as follows:
aveCu Zn
Cu Zn
100A C CA A
=+
=100
70 wt%63.55 g /mol
+30 wt%
65.41 g /mol
= 64.09 g/mol
Thus (375 J/kg K)(1 kg/1000 g)(64.09 g/mol) = 24.0 J/mol KpC = ⋅ ⋅
(b) These values of Cp are very close to one another because room temperature is considerably above
the Debye temperature for these metals; therefore, the values of Cp should be about equal to
3R [(3)(8.31 J/ mol K⋅ ) = 24.9 J/ mol K⋅ ], which is indeed the case for all four of these metals.
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19.4 For aluminum, the heat capacity at constant volume Cv at 30 K is 0.81 J/ mol K⋅ , and the
Debye temperature is 375 K. Estimate the specific heat (a) at 50 K and (b) at 425 K. Solution
(a) For aluminum, Cv at 50 K may be approximated by Equation 19.2, since this temperature is
significantly below the Debye temperature (375 K). The value of Cv at 30 K is given, and thus, we
may compute the constant A as
5 43 3
0.81 J/mol K 3.00 10 J/mol K(30 K)
vCA
T−⋅
= = = × ⋅
Therefore, at 50 K
3 5 4 3 3.00 10 J/mol K (50 K) 3.75 J/mol.K( )vC AT −= = × ⋅ =
and
= (3.75 J/mol K)(1 mol/26.98 g)(1000 g/kg) = 139 J/kg Kvc ⋅ ⋅
(b) Since 425 K is above the Debye temperature, a good approximation for Cv is
3vC R=
= (3)(8.31 J/mol K) = 24.9 J/mol K⋅ ⋅
And, converting this to specific heat
= (24.9 J/mol K)(1 mol/26.98 g)(1000 g/kg) = 923 J/kg Kvc ⋅ ⋅
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19.5 The constant A in Equation 19.2 is 4 3D12 /5Rπ θ , where R is the gas constant and θD is the
Debye temperature (K). Estimate θD for copper, given that the specific heat is 0.78 J/ kg K⋅ at
10 K. Solution
For copper, we want to compute the Debye temperature, θD, given the expression for A in
Equation 19.2 and the heat capacity at 10 K. First of all, let us determine the magnitude of A, as
3 vCA
T=
3
(0.78 J/mol K)(1 kg/1000 g)(63.55 g/mol)= (10 K)
⋅
5 44.96 10 J/mol K−= × ⋅
As stipulated in the problem statement
4
3D
12 5
RA πθ
=
Or, solving for θD
1/ 34
D12
5R
Aπ
θ⎛ ⎞
= ⎜ ⎟⎝ ⎠
1/ 34
5 4
(12)( ) (8.31 J/mol K) 340 K(5) 4.96 10 J/mol K( )
π−
⎡ ⎤⋅= =⎢ ⎥× ⋅⎣ ⎦
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19.6 (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) The reason that Cv rises with increasing temperature at temperatures near 0 K is 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.
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Thermal Expansion
19.7 An aluminum wire 10 m long is cooled from 38 to −1°C (311 to 272 K). How much change in
length will it experience? Solution
In order to determine the change in length of the aluminum wire, we must employ a
rearranged form of Equation 19.3b and using the value of α1 taken from Table 19.1 [23.6 × 10−6
(°C)−1] as
0 0 1 0 ( )l fl l T l T Tα α∆ = ∆ = −
6 1 (10 m) 23.6 10 ( C) ( 1 C 38 C)− −⎡ ⎤= × ° − ° − °⎣ ⎦
3= 9.2 10 m = 9.2 mm−− × −
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19.8 A 0.1 m rod of a metal elongates 0.2 mm on heating from 20 to 100°C (293K to 373 K).
Determine the value of the linear coefficient of thermal expansion for this material. Solution
The linear coefficient of thermal expansion for this material may be determined using a
rearranged form of Equation 19.3b as
3
0 0 0
0.2 10 m ( ) (0.1 m)(100 C 20 C)l
f
l ll T l T T
α−∆ ∆ ×
= = =∆ − ° − °
6 125.0 10 ( C)− −= × °
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19.9 Briefly explain thermal expansion using the potential energy-versus-interatomic spacing
curve.
The phenomenon of thermal expansion using the potential energy-versus-interatomic spacing
curve is explained in Section 19.3.
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19.10 Compute the density for nickel at 500°C, given that its room-temperature density is
8.902 g/cm3. Assume that the volume coefficient of thermal expansion, αv, is equal to 3αl. Solution
In this problem we are asked to determine the density of nickel at 500°C. Let us use as the
basis for this determination 1 cm3 of material at 20°C, which has a mass of 8.902 g; it is assumed that
this mass will remain constant upon heating to 500°C. Let us compute the volume expansion of this
cubic centimeter of nickel as it is heated to 500°C. A volume expansion expression is given in
Equation 19.4—viz.,
0
vV T
Vα
∆= ∆
or
0 vV V Tα∆ = ∆
Also, αv = 3αl, as stipulated in the problem. The value of αl given in Table 19.1 for nickel is
13.3 × 10−6 (°C)−1. Therefore, the volume, V, of this specimen of Ni at 500°C is just
( ) ( )0 0 0 1 1 3v lV V V V T V Tα α= + ∆ = + ∆ = + ∆
3 6 1 1 cm 1 (3) 13.3 10 ( C) (500 C 20 C)( ) − −⎡ ⎤= + × ° ° − °⎣ ⎦
31.01915 cm=
Thus, the density is just the 8.902 g divided by this new volume—i.e.,
3
3
8.902 g 8.735 g/cm1.01915 cm
ρ = =
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19.11 When a metal is heated its density decreases. There are two sources that give rise to this
diminishment of ρ: (1) the thermal expansion of the solid, and (2) the formation of vacancies
(Section 4.2). Consider a specimen of copper at room temperature (20°C) that has a density of
8.940 g/cm3. (a) Determine its density upon heating to 1000°C when only thermal expansion
is considered. (b) Repeat the calculation when the introduction of vacancies is taken into
account. Assume that the energy of vacancy formation is 0.90 eV/atom, and that the volume
coefficient of thermal expansion, αv is equal to 3αl. Solution
(a) In this portion of the problem we are asked to determine the density of copper at 1000°C on the
basis of thermal expansion considerations. The basis for this determination will be 1 cm3 of material at
20°C; this volume of copper has a mass of 8.940 g, which mass is assumed to remain constant upon
heating to the 1000°C. Let us first compute the volume expansion of this cubic centimeter of copper as
it is heated to 1000°C. According to Equation 19.4 volume expansion is equal to
0
vV T
Vα
∆= ∆
where αv, the volume coefficient of thermal expansion, as stipulated in the problem statement, is equal
to 3αl. The value of αl given in Table 19.1 for copper is 17.0 × 10−6 (°C)−1. Therefore, the volume of
this specimen of Cu at 1000°C (V) is equal to
0 0 0 0 (1 )v vV V V V V T V Tα α+= + ∆ = ∆ = + ∆
0 0 01 3 1 3 ( )( )l l fV T V T Tα α⎡ ⎤⎢ ⎥⎣ ⎦
= + ∆ = + −
3 6 1 (1 cm ) 1 (3) 17.0 10 ( C) (1000 C 20 C)− −⎡ ⎤= + × ° ° − °⎣ ⎦
31.04998 cm=
Thus, the density is just the 8.940 g divided by this new volume—i.e.,
3
3
8.940 g 8.514 g/cm1.04998 cm
ρ = =
(b) Now we are asked to compute the density at 1000°C taking into consideration the creation of
vacancies which will further lower the density. To begin, this determination requires that we calculate
the number of vacancies using Equation 4.1. But it first becomes necessary to compute the number of
Cu atoms per cubic centimeter (NCu) at 1000°C using Equation 4.2. Thus,
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A CuCu
Cu
N
NA
ρ=
23 36.022 10 atoms/mol 8.514 g/cm
63.55 g/mol
( )( )×=
22 38.07 10 atoms/cm= ×
Now, from Equation 4.1, the total number of vacancies, Nv, is computed as
Cu exp vv
QN N
kT⎛ ⎞= −⎜ ⎟⎝ ⎠
22 35
0.90 eV/atom 8.07 10 atoms/cm exp8.62 10 eV/K (1000 273 K)( )( ) −
⎡ ⎤= × −⎢ ⎥× +⎣ ⎦
19 3 2.212 10 vacancies/cm= ×
We now want to determine the number of vacancies per unit cell, which is possible if the unit cell
volume is multiplied by Nv. The unit cell volume (VC) may be calculated using Equation 3.5 taking n =
4 inasmuch as Cu has the FCC crystal structure. Thus, from a rearranged form of Equation 3.5
Cu
Cu A
C
nAV
Nρ=
3 23
(4 atoms/unit cell)(63.55 g/mol) 8.514 g/cm 6.022 10 atoms/mol( )( )=
×
23 3 4.958 10 cm /unit cell−= ×
Now, the number of vacancies per unit cell, nv, is just
v v Cn N V=
19 3 23 3 2.212 10 vacancies/cm 4.958 10 cm /unit cell( )( )−= × ×
= 0.001097 vacancies/unit cell
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What this means is that instead of there being 4.0000 atoms per unit cell, there are only
4.0000 – 0.001097 = 3.998903 atoms per unit cell. And, finally, the density may be computed using
Equation 3.5 taking n = 3.998903; thus
Cu
CuA C
nAV N
ρ =
23 3 23
(3.998903 atoms/unit cell)(63.55 g/mol) 4.958 10 cm /unit cell 6.022 10 atoms/mol( )( )−=
× ×
38.512 g/cm=
Thus, the influence of the vacancies is almost insignificant—their presence reduces the density by only
0.002 g/cm3 (from 8.514 g/cm3 to 8.512 g/cm3).
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19.12 The difference between the specific heats at constant pressure and volume is described by the
expression 2
0vp v
v Tc c
αβ
− = (19.10)
where αv is the volume coefficient of thermal expansion, v0 is the specific volume (i.e.,
volume per unit mass, or the reciprocal of density), β is the compressibility, and T is the
absolute temperature. Compute the values of cv at room temperature (293 K) for copper and
nickel using the data in Table 19.1, assuming that αv = 3αl and given that the values of β for
Cu and Ni are 8.35 × 10−12 and 5.51 × 10−12 (Pa)−1, respectively. Solution
This problem asks that we calculate the values of cv for copper and nickel at room temperature
using Equation 19.10, the data in Table 19.1, given that αv = 3αl, and also values of the
compressibility. From Equation 19.10
2
0vv p
v Tc c
αβ
= −
And, from Table 19.1 and the problem statement
cp(Cu) = 386 J/ kg K⋅
cp(Ni) = 443 J/ kg K⋅
αv(Cu) = (3)[17.0 × 10−6 (°C)−1] = 5.10 × 10−5 (°C)−1
αv(Ni) = (3)[(13.3 × 10−6 (°C)−1] = 3.99 × 10−5 (°C)−1
β(Cu) = 8.35 × 10−12 (Pa)−1
β(Ni) = 5.51 × 10−12 (Pa)−1
The specific volume is just the reciprocal of the density; thus, in units of m3/kg
3
4 30 3
1 1 1000 g 1 m(Cu) 1.119 10 m /kgkg 100 cm8.94 g/cm
vρ
−⎛ ⎞ ⎛ ⎞⎛ ⎞= = = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠
34 3
0 3
1 1000 g 1 m(Ni) 1.124 10 m /kgkg 100 cm8.90 g/cm
v −⎛ ⎞ ⎛ ⎞⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠
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Therefore, for copper 2
0(Cu) (Cu)(Cu) (Cu)
(Cu)v
v pv T
c cα
β= −
25 1 4 3
12 2 1
5.10 10 ( C) 1.119 10 m /kg (293 K) 386 J/kg K
8.35 10 (N/m )( )− − −
− −
⎡ ⎤× ° ×⎣ ⎦= ⋅ −×
= 376 J/ kg K⋅
And, also for nickel
25 1 4 3
12 2 1
3.99 10 ( C) 1.124 10 m /kg (293 K)(Ni) 443 J/kg K
5.51 10 (N/m )( )
vc− − −⎡ ⎤
⎢ ⎥⎣ ⎦− −
× ° ×= ⋅ −
×
= 433 J/ kg K⋅
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19.13 To what temperature must a cylindrical rod of tungsten 10.000 mm in diameter and a plate of
316 stainless steel having a circular hole 9.988 mm in diameter have to be heated for the rod
to just fit into the hole? Assume that the initial temperature is 25°C. Solution
This problem asks for us to determine the temperature to which a cylindrical rod of tungsten
10.000 mm in diameter must be heated in order for it of just fit into a 9.988 mm diameter circular hole
in a plate of 316 stainless steel, assuming that the initial temperature is 25°C. This requires the use of
Equation 19.3a, which is applied to the diameters of both the rod and hole. That is
0
00
( )fl f
d dT T
dα
−= −
Solving this expression for df yields
0 0[1 ( )]f l fd d T Tα= + −
Now all we need do is to establish expressions for df (316 stainless) and df (W), set them equal to one
another, and solve for Tf. According to Table 19.1, αl (316 stainless) = 16.0 × 10−6 (°C)−1 and
αl (W) = 4.5 × 10−6 (°C)−1. Thus
(316 stainless) (W)f fd d=
6 1(9.988 mm)[1 16.0 10 ( C) ( 25 C)]fT− −+ × ° − °
6 1(10.000 mm)[1 4.5 10 ( C) ( 25 C)]fT− −= + × ° − °
Now solving for Tf gives Tf = 129.5°C
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Thermal Conductivity
19.14 (a) Calculate the heat flux through a sheet of steel 10 mm thick if the temperatures at the two
faces are 300 and 100°C (573 and 373 K); assume steady-state heat flow. (b) What is the heat
loss per hour if the area of the sheet is 0.25 m2? (c) What will be the heat loss per hour if
soda–lime glass instead of steel is used? (d) Calculate the heat loss per hour if steel is used
and the thickness is increased to 20 mm. Solution
(a) The steady-state heat flux through the plate may be computed using Equation 19.5; the thermal
conductivity for steel, found in Table 19.1, is 51.9 W/m K.⋅ Therefore,
Tq kx
∆= −
∆
3
(100 273 K) (300 273 K) (51.9 W/m K) 10 10 m−
⎡ ⎤+ − += − ⋅ ⎢ ⎥×⎣ ⎦
6 21.04 10 W/m= ×
(b) Let dQ/dt represent the total heat loss such that
dQ qAtdt
=
where A and t are the cross-sectional area and time, respectively. Thus,
6 2 2 1.04 10 J/s m 0.25 m (60 s/min)(60 min/h)( )( )dQdt
= × ⋅
8= 9.3 10 J/h×
(c) If soda-lime glass is used (k = 1.7 W/m K,⋅ Table 19.1),
dQ Tk A tdt x
∆= −
∆
23
200 K (1.7 J/s m K) 0.25 m (3600 s/h)10 10 m
( ) −
⎛ ⎞−= − ⋅ ⋅ ⎜ ⎟×⎝ ⎠
7= 3.06 10 J/h×
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(d) If the thickness of the steel is increased to 20 mm, then
23
200 K(51.9 W/m K) 0.25 m (3600 s/h) 20 10 m
( )dQ Tk A tdt x −
⎛ ⎞∆ −= − = − ⋅ ⎜ ⎟∆ ×⎝ ⎠
8= 4.7 10 J/h×
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19.15 (a) Would you expect Equation 19.7 to be valid for ceramic and polymeric materials? Why or why not? (b) Estimate the value for the Wiedemann–Franz constant L [in 2W/(K)Ω⋅ ] at room
temperature (293 K) for the following nonmetals: silicon (intrinsic), glass-ceramic
(Pyroceram), fused silica, polycarbonate, and polytetrafluoroethylene. Consult Tables B.7 and
B.9 in Appendix B. Solution
(a) Equation 19.7 is not valid for ceramic and polymeric materials since, in the development of this
expression, it is assumed that free electrons are responsible for both electrical and thermal conduction.
Such is the case for most metals. For ceramics and polymers, free electrons are the primary
contributors to the electrical conductivity. However, free electrons do not contribute significantly to
the thermal conductivity. For ceramics, thermal conduction is primarily by means of phonons; for
polymers, the energy transfer is made by chain vibrations, translations, and rotations.
(b) Estimated room-temperature values of L, in 2W/(K) ,Ω⋅ for the several materials are determined
below. Electrical conductivity values were determined by taking reciprocals of the electrical
resistivities given in Table B.9, Appendix B; thermal conductivities are taken from Table B.7 in the
same appendix. (Note: when a range of values is given in these tables, an average value is used in the
computation.)
For intrinsic silicon 2141 W/m K = = = 1203 W/K
1 (293 K)2500 (Ω m)
kLTσ
⋅Ω⋅
⎡ ⎤⎢ ⎥⋅⎣ ⎦
For Pyroceram glass-ceramic 12 2
14
3.3 W/m K 2.3 10 W/K1 (293 K)
2 10 (Ω m)
L
−
⋅= = × Ω⋅
⎡ ⎤⎢ ⎥× ⋅⎣ ⎦
For fused silica 15 2
18
1.4 W/m K 4.8 10 W/K1 (293 K)
10 (Ω m)
L ⋅= = × Ω⋅
⎡ ⎤⎢ ⎥⋅⎣ ⎦
For polycarbonate 11 2
14
0.20 W/m K 1.4 10 W/K1 (293 K)
2 10 (Ω m)
L ⋅= = × Ω⋅
⎡ ⎤⎢ ⎥× ⋅⎣ ⎦
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For polytetrafluoroethylene
13 2
17
0.25 W/m K 8.5 10 W/K1 (293 K)
10 (Ω m)
L ⋅= = × Ω⋅
⎡ ⎤⎢ ⎥⋅⎣ ⎦
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19.16 Briefly explain why the thermal conductivities are higher for crystalline than noncrystalline
ceramics. Solution
Thermal conductivities are higher for crystalline than for noncrystalline ceramics because, for
noncrystalline, phonon scattering, and thus the resistance to heat transport, is much more effective due
to the highly disordered and irregular atomic structure.
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19.17 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.
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19.18 (a) Briefly explain why porosity decreases the thermal conductivity of ceramic and polymeric
materials, rendering them more thermally insulative. (b) Briefly explain how the degree of
crystallinity affects the thermal conductivity of polymeric materials and why. Solution
(a) Porosity decreases the thermal conductivity of ceramic and polymeric materials because the
thermal conductivity of a gas phase that occupies pore space is extremely small relative to that of the
solid material. Furthermore, contributions from gaseous convection are generally insignificant.
(b) Increasing the degree of crystallinity of a semicrystalline polymer enhances its thermal
conductivity; the vibrations, rotations, etc. of the molecular chains are more effective modes of thermal
transport when a crystalline structure prevails.
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19.19 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.
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19.20 For each of the following pairs of materials, decide which has the larger thermal conductivity.
Justify your choices.
(a) Pure copper; aluminum bronze (95 wt% Cu–5 wt% Al).
(b) Fused silica; quartz.
(c) Linear polyethylene; branched polyethylene.
(d) Random poly(styrene-butadiene) copolymer; alternating poly(styrene-butadiene)
copolymer. Solution
This question asks for us to decide, for each of several pairs of materials, which has the larger
thermal conductivity and why.
(a) Pure copper will have a larger conductivity than aluminum bronze because the impurity atoms in
the latter will lead to a greater degree of free electron scattering.
(b) Quartz will have a larger conductivity than fused silica because fused silica is noncrystalline
(whereas quartz is crystalline) and lattice vibrations are more effectively scattered in noncrystalline
materials.
(c) The linear polyethylene will have the larger conductivity than the branched polyethylene because
the former will have the higher degree of crystallinity. Linear polymers have higher degrees of
crystallinity than branched polymers. 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.
(d) The alternating poly(styrene-butadiene) copolymer will have a higher crystallinity than the random
copolymer; alternating copolymers crystallize more easily than random ones. The influence of
crystallinity on conductivity is explained in part (c).
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19.21 We might think of a porous material as being a composite wherein one of the phases is a pore
phase. Estimate upper and lower limits for the room-temperature thermal conductivity of a
magnesium oxide material having a volume fraction of 0.30 of pores that are filled with still
air. Solution
This problem asks that we treat a porous material as a composite wherein one of the phases is
a pore phase, and that we estimate upper and lower limits for the room-temperature thermal
conductivity of a magnesium oxide material having a 0.30 volume fraction of pores. The upper limit
of k (kupper) may be determined using Equation 16.1 with thermal conductivity substituted for the
elastic modulus, E. From Table 19.1, the value of k for MgO is 37.7 W/m K,⋅ while for still air in the
pore phase, k = 0.02 W/m K⋅ (Section 19.4). Thus
kupper = Vpkair + VMgOkMgO
= (0.30)(0.02 W/m K⋅ ) + (0.70)(37.7 W/m K⋅ ) = 26.4 W/m K⋅
For the lower limit we employ a modification of Equation 16.2 as
klower =
kairkMgOVpkMgO + VMgOkair
(0.02 W/m K)(37.7 W/m K) 0.067 W/m K
(0.30)(37.7 W/m K) (0.70)(0.02 W/m K)⋅ ⋅
= = ⋅⋅ + ⋅
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19.22 Nonsteady-state heat flow may be described by the following partial differential equation:
2
2TT TDt x
∂ ∂=
∂ ∂
where DT is the thermal diffusivity; this expression is the thermal equivalent of Fick’s second
law of diffusion (Equation 5.4b). The thermal diffusivity is defined according to
Tp
kDcρ
=
In this expression, k, ρ, and cp represent the thermal conductivity, the mass density, and the
specific heat at constant pressure, respectively.
(a) What are the SI units for DT?
(b) Determine values of DT for aluminum, steel, aluminum oxide, soda–lime glass,
polystyrene, and nylon 6,6 using the data in Table 19.1. Density values are included in Table
B.1, Appendix B. Solution
(a) The units of DT are 2
3
k (J/s m K) m /skg/m (J/kg K)( )T
p
Dcρ⋅ ⋅
= =⋅
(b) The values of DT for the several materials are given below. (Note: values for k and cp are taken
from Table 19.1; density values are from Table B.1, Appendix B, and converted to units of kilograms
per meter cubed):
For aluminum 4 2
3 3
247 W/m K 1.0 10 m /s2.71 10 kg/m (900 J/kg K)( )T
p
kDcρ
−⋅= = = ×
× ⋅
For steel 5 2
3 3
51.9 W/m K 1.4 10 m /s7.9 10 kg/m (486 J/kg K)( )TD −⋅
= = ×× ⋅
For aluminum oxide 5 2
3 3
39 W/m K = = 1.26 10 m /s4 10 kg/m (775 J/kg K)( )TD −⋅
×× ⋅
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For soda-lime glass 7 2
3 3
1.7 W/m K 8.1 10 m /s2.5 10 kg/m (840 J/kg K)( )TD −⋅
= = ×× ⋅
For polystyrene 7 2
3 3
0.13 W/m K 1.06 10 m /s1.05 10 kg/m (1170 J/kg K)( )TD −⋅
= = ×× ⋅
For nylon 6,6 7 2
3 3
0.24 W/m K 1.3 10 m /s1.14 10 kg/m (1670 J/kg K)( )TD −⋅
= = ×× ⋅
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Thermal Stresses
19.23 Beginning with Equation 19.3, show that Equation 19.8 is valid. Solution
We want to show that Equation 19.8 is valid beginning with Equation 19.3. Upon
examination of Equation 19.3b,
0
ll T
lα∆
= ∆
it may be noted that the term on the left-hand side is the same expression as that for the definition of
engineering strain (Equation 6.2); that is
0
ll
∈∆
=
Furthermore, elastic stress and strain are related through Hooke's law, Equation 6.5:
Eσ ∈=
Making appropriate substitutions and algebraic manipulations gives
0
ll T
l Eσ∈ α∆
= = = ∆
Or, solving for σ
lE Tσ α= ∆
which is the form of Equation 19.8.
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19.24 (a) Briefly explain why thermal stresses may be introduced into a structure by rapid heating or
cooling. (b) For cooling, what is the nature of the surface stresses? (c) For heating, what is the
nature of the surface stresses? Solution
(a) Thermal stresses may be introduced into a structure by rapid heating or cooling because
temperature gradients will be established across the cross section due to more rapid temperature
changes at the surface than within the interior; thus, the surface will expand or contract at a different
rate than the interior and since this surface expansion or contraction will be restrained by the interior,
stresses will be introduced.
(b) For cooling, the surface stresses will be tensile in nature since the interior contracts to a lesser
degree than the cooler surface.
(c) For heating, the surface stresses will be compressive in nature since the interior expands to a lesser
degree than the hotter surface.
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19.25 (a) If a rod of 1025 steel 0.5 m long is heated from 20 to 80°C (293 to 353 K) while its ends
are maintained rigid, determine the type and magnitude of stress that develops. Assume that at
20°C the rod is stress free. (b) What will be the stress magnitude if a rod 1 m long is used?
(c) If the rod in part (a) is cooled from 20 to −10°C (293 K to 263 K), what type and
magnitude of stress will result? Solution
(a) We are asked to compute the magnitude of the stress within a 1025 steel rod that is heated while its
ends are maintained rigid. To do this we employ Equation 19.8, using a value of 207 GPa for the
modulus of elasticity of steel (Table 6.1), and a value of 12.0 × 10−6 (°C)−1 for αl (Table 19.1).
Therefore
0 ( )l fE T Tσ α= −
3 6 1 207 10 MPa 12.0 10 ( C) (20 C 80 C)( ) − −⎡ ⎤= × × ° ° − °⎣ ⎦
= –150 MPa
The stress will be compressive since its sign is negative.
(b) The stress will be the same [–150 MPa], since stress is independent of bar length.
(c) Upon cooling the indicated amount, the stress becomes
0 ( )l fE T Tσ α= −
[ ]3 6 1 (207 10 MPa) 12.0 10 ( C) (20 C ( 10 C)− −⎡ ⎤= × × ° ° − − °⎣ ⎦
= +74.5 MPa
This stress will be tensile since its sign is positive.
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19.26 A copper wire is stretched with a stress of 70 MPa at 20°C (293 K). If the length is held
constant, to what temperature must the wire be heated to reduce the stress to 35 MPa. Solution
We want to heat the copper wire in order to reduce the stress level from 70 MPa to 35 MPa; in
doing so, we reduce the stress in the wire by 70 MPa – 35 MPa = 35 MPa, which stress will be a
compressive one (i.e., σ = –35 MPa). Solving for Tf from Equation 19.8 [and using values for E and αl
of 110 GPa (Table 6.1) and 17.0 × 10−6 (°C)−1 (Table 19.1), respectively] yields
0 fl
T TEσα
= −
3 6 1
35 MPa 20 C (110 10 MPa)[17.0 10 ( C) ]− −
−= ° −
× × °
= 20°C + 19°C = 39°C (312 K)
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19.27 If a cylindrical rod of nickel 100.00 mm long and 8.000 mm in diameter is heated from 20°C
to 200°C while its ends are maintained rigid, determine its change in diameter. You may want
to consult Table 6.1. Solution
This problem asks for us to determine the change in diameter of a cylindrical nickel rod
100.00 mm long and 8.000 mm in diameter when it is heated from 20°C to 200°C while its ends are
maintained rigid. There will be two contributions to the diameter increase of the rod; the first is due to
thermal expansion (which will be denoted as ∆d1), while the second is from Poisson's lateral expansion
as a result of elastic deformation from stresses that are established from the inability of the rod to
elongate as it is heated (denoted as ∆d2). The magnitude of ∆d1 may be computed using a modified
form of Equation 19.3 as
1 0 0 ( )l fd d T Tα∆ = −
From Table 19.1 the value of αl for nickel is 13.3 × 10−6 (°C)−1. Thus,
6 1
1 (8.000 mm) 13.3 10 ( C) (200 C 20 C)d − −⎡ ⎤∆ = × ° ° − °⎣ ⎦
= 0.0192 mm
Now, ∆d2 is related to the transverse strain (∈x) according to a modified form of Equation 6.2
as
2
0x
dd
∈∆
=
Also, transverse strain and longitudinal strain (∈z) are related according to Equation 6.8:
x z∈ ν ∈= −
where ν is Poisson’s ratio. Substitution of this expression for ∈x into the first equation above leads to
2
0z
dd
ν ∈∆
= −
Furthermore, the longitudinal strain is related to the modulus of elasticity through Equation 6.5—i.e.,
z Eσ∈ =
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And, therefore, 2
0
dd E
σν∆
= −
Now, from Equation 19.8 stress is equal to
0( )l fE T Tσ α= −
which, when substituted into the preceding equation leads to
02
00
( )( )l f
l f
E T TdT T
d Eν α
να−∆
= − = − −
Solving for ∆d2 and realizing that, for nickel, ν = 0.31 (Table 6.1) yields
2 0 0( )l fd d T Tνα∆ = − −
6 1= (8.000 mm)(0.31) [13.3 10 ( C) ](20 C 200 C)− −− × ° ° − °
= 0.0059 mm
Finally, the total ∆d is just ∆d1 + ∆d2 = 0.0192 mm + 0.0059 mm = 0.0251 mm.
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19.28 The two ends of a cylindrical rod of 1025 steel 75.00 mm long and 10.000 mm in diameter are
maintained rigid. If the rod is initially at 25°C, to what temperature must it be cooled to have
a 0.008-mm reduction in diameter? Solution
This problem asks for us to determine to what temperature a cylindrical rod of 1025 steel
75.00 mm long and 10.000 mm in diameter must be cooled from 25°C in order to have a 0.008 mm
reduction in diameter if the rod ends are maintained rigid. There will be two contributions to the
diameter decrease of the rod; the first is due to thermal contraction (which will be denoted as ∆d1),
while the second is from Poisson's lateral contraction as a result of elastic deformation from stresses
that are established from the inability of the rod to contract as it is cooled (denoted as ∆d2). The
magnitude of ∆d1 may be computed using a modified form of Equation 19.3b as
1 0 0 ( )l fd d T Tα∆ = −
Now, ∆d2 is related to the transverse strain (∈x) according to a modified form of Equation 6.2
as 2
0x
dd
∈∆
=
Also, transverse strain and longitudinal strain (∈z) are related according to Equation 6.8:
x z∈ ν ∈= −
where ν is Poisson’s ratio. Substitution of this expression for ∈x into the first equation above leads to 2
0z
dd
ν ∈∆
= −
Furthermore, the longitudinal strain is related to the modulus of elasticity through Equation 6.5—i.e.,
z Eσ∈ =
And, therefore, 2
0
dd E
σν∆
= −
Now, from Equation 19.8 stress is equal to
0( )l fE T Tσ α= −
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which, when substituted into the preceding equation leads to
020
0
( ) ( )l fl f
E T TdT T
d Eν α
να−∆
= − = − −
And, solving for ∆d2 from this expression
2 0 0 ( )l fd d T Tνα∆ = − −
The total ∆d is just ∆d = ∆d1 + ∆d2, and
0 0 0 0 0 0 (1 )( )( ) ( )l f l f l fd d T T d T T d T Tα να α ν∆ = − + − = − +
The values of ν and αl for 1025 steel are 0.30 and 12.0 × 10−6 (°C)−1, respectively (Tables 6.1 and
19.1). Incorporating, into the above equation, these values, as well as those for ∆d, d0, and T0 cited in
the problem statement gives
6 1(0.008 mm) (10.000 mm) 12.0 10 ( C) 25 C (1 0.30)( )fT− −⎡ ⎤− = × ° − ° +⎣ ⎦
And, finally, solving the above expression for Tf yields Tf = – 26.3°C.
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19.29 What measures may be taken to reduce the likelihood of thermal shock of a ceramic piece? Solution
According to Equation 19.9, 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, σf and αl are most amenable to alteration,
usually be changing the composition and/or the microstructure.
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DESIGN PROBLEMS
Thermal Expansion
19.D1 Railroad tracks made of 1025 steel are to be laid during the time of year when the temperature
averages 10°C (283 K). If a joint space of 4.6 mm is allowed between the standard 11.9-m
long rails, what is the hottest possible temperature that can be tolerated without the
introduction of thermal stresses? Solution
For these railroad tracks, each end is allowed to expand one-half of the joint space distance, or
the track may expand a total of this distance (4.6 mm). Equation 19.3a is used to solve for Tf, where
the value αl for the 1025 steel [12.0 × 10−6 (°C)−1] is found in Table 19.1. Thus, solving for Tf from
Equation 19.3a leads to
00
fl
lT Tlα
∆= +
3
6 1
4.6 10 m 10 C12.0 10 ( C) (11.9 m)
−
− −
×= + °
⎡ ⎤× °⎣ ⎦
= 32.2°C + 10°C = 42.2°C (315.2 K)
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Thermal Stresses
19.D2 The ends of a cylindrical rod 6.4 mm in diameter and 250 mm long are mounted between rigid
supports. The rod is stress free at room temperature [20°C (293 K)]; and upon cooling to
−40°C (233 K); a maximum thermally induced tensile stress of 125 MPa is possible. Of which
of the following metals or alloys may the rod be fabricated: aluminum, copper, brass, 1025
steel, and tungsten? Why? Solution
This is really a materials selection problem in which we must decide for which of the five
metals listed, the stress in the rod will not exceed 125 MPa, when it is heated while its ends are
mounted in rigid supports. Upon examination of Equation 19.8, it may be noted that all we need do is
to compute the Eαl∆T product for each of the candidate materials, and then note for which of them the
stress is less than the stipulated maximum. [The value of ∆T is T0 – Tf = 20°C – (–40°C) = 60°C.]
These parameters and their product for each of the alloys are tabulated below. (Modulus of elasticity
values were taken from Table 6.1, while the αl values came from Table 19.1.)
Alloy αl (°C)−1 E (MPa) αlE∆T (MPa)
Aluminum 23.6 × 10−6 69 × 103 98
Copper 17.0 × 10−6 110 × 103 112
Brass 20.0 × 10−6 97 × 103 116
1025 Steel 12.0 × 10−6 207 × 103 149
Tungsten 4.5 × 10−6 407 × 103 110
Thus, aluminum, copper, brass, and tungsten are suitable candidates.
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19.D3 (a) What are the units for the thermal shock resistance parameter (TSR)? (b) Rank the
following ceramic materials according to their thermal shock resistance: glass-ceramic
(Pyroceram), partially stabilized zirconia, and borosilicate (Pyrex) glass. Appropriate data
may be found in Tables B.2, B.4, B.6, and B.7 of Appendix B. Solution
(a) This portion of the problem asks that we cite the units for the thermal shock resistance parameter
(TSR). From Equation 19.9
2
2 1
N/m (W/m K) W/m
N/m ( C)( )( )
f
l
kTSR
Eσ
α −
⋅= =
°
(Note: in reducing units in the above expression, we have assumed that units of temperature in K and
°C are equivalent)
(b) Now we are asked to rank glass-ceramic (Pyroceram), partially-stabilized zirconia, and borosilicate
(Pyrex) glass as to their thermal shock resistance. Thus, all we need do is calculate, for each, the value
of TSR using Equation 19.9. Values of E, σf, αl, and k are found, respectively, in Tables B.2, B.4, B.6,
and B.7, Appendix B. (Note: whenever a range for a property value in these tables is cited, the average
of the extremes is used.)
For the glass-ceramic
f
l
kTSR
Eσ
α=
3 6 1
(247 MPa)(3.3 W/m K) 1045 W/m120 10 MPa 6.5 10 ( C)( ) − −
⋅= =
⎡ ⎤× × °⎣ ⎦
For partially-stabilized zirconia
3 6 1
(1150 MPa)(2.7 W/m K) 1578 W/m205 10 MPa 9.6 10 ( C)( )TSR
− −
⋅= =
⎡ ⎤× × °⎣ ⎦
And, for borosilicate glass
3 6 1
(69 MPa)(1.4 W/m K) 418 W/m70 10 MPa 3.3 10 ( C)( )TSR
− −
⋅= =
⎡ ⎤× × °⎣ ⎦
Thus, these materials may be ranked according to their thermal shock resistance from the
greatest to the least as follows: partially-stabilized zirconia, glass-ceramic, and borosilicate glass.
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19.D4 Equation 19.9, for the thermal shock resistance of a material, is valid for relatively low rates
of heat transfer. When the rate is high, then, upon cooling of a body, the maximum
temperature change allowable without thermal shock, ∆Tf, is approximately
f
fl
TEσα
∆ =
where σf is the fracture strength. Using the data in Tables B.2, B.4, and B.6 (Appendix B),
determine ∆Tf for a glass-ceramic (Pyroceram), partially stabilized zirconia, and fused silica. Solution
We want to compute the maximum temperature change allowable without thermal shock for
these three ceramic materials, which temperature change is a function of the fracture strength, elastic
modulus, and linear coefficient of thermal expansion. These data and the ∆Tf's are tabulated below.
(Values for E, σf, and αl are taken from Tables B.2, B.4, B.6 in Appendix B.)
Material σf (MPa) E (MPa) αl (°C)−1 ∆Tf (°C)
Glass ceramic 247 120 × 103 6.5 × 10−6 317
Zirconia 1150 205 × 103 9.6 × 10−6 584
Fused silica 104 73 × 103 0.4 × 10−6 3562