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Callister Materials Science 7th Edition Chpater 19 Solutions Manual
41
Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful. 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 p E cm 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 J E = × 4 (steel) = (486 J/kg K)(2 kg)(80 K) 7.78 10 J E = × 5 (glass) = (840 J/kg K)(2 kg)(80 K) 1.34 10 J E = × 5 (HDPE) = (1850 J/kg K)(2 kg)(80 K) 2.96 10 J E = ×
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Page 1: Solution Manual Chap 19 Callister 7e Materials

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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 ⋅ = ×

Page 2: Solution Manual Chap 19 Callister 7e Materials

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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= ∆ °

Page 3: Solution Manual Chap 19 Callister 7e Materials

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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.

Page 4: Solution Manual Chap 19 Callister 7e Materials

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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 ⋅ ⋅

Page 5: Solution Manual Chap 19 Callister 7e Materials

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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

θ⎛ ⎞

= ⎜ ⎟⎝ ⎠

1/ 34

5 4

(12)( ) (8.31 J/mol K) 340 K(5) 4.96 10 J/mol K( )

π−

⎡ ⎤⋅= =⎢ ⎥× ⋅⎣ ⎦

Page 6: Solution Manual Chap 19 Callister 7e Materials

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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.

Page 7: Solution Manual Chap 19 Callister 7e Materials

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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−− × −

Page 8: Solution Manual Chap 19 Callister 7e Materials

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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.

Page 10: Solution Manual Chap 19 Callister 7e Materials

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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

∆= ∆

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 = =

Page 11: Solution Manual Chap 19 Callister 7e Materials

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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

∆= ∆

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).

Page 14: Solution Manual Chap 19 Callister 7e Materials

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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

−⎛ ⎞ ⎛ ⎞⎛ ⎞= = = ×⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

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⋅

Page 16: Solution Manual Chap 19 Callister 7e Materials

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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

−= −

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

Page 17: Solution Manual Chap 19 Callister 7e Materials

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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

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

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.

Page 41: Solution Manual Chap 19 Callister 7e Materials

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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


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