Fisica para ingenieria y ciencias Bauer Vol I - solucioario capitulo 17

Chapter 17: Temperature
711
In-Class Exercises
17.1. c 17.2. a 17.3. e 17.4. c 17.5. d
Multiple Choice
17.1. a 17.2. a 17.3. c 17.4. b 17.5. d 17.6. d 17.7. a 17.8. d 17.9. c
Questions
17.10. Yes, this will still work. The lid will heat up more quickly if it is in direct contact with the warm water; the
lid will expand more than the container, making it easier to open (not as easy as with a glass jar, but easier
than before heating in the warm water).
17.11. As stated in Section 17.1, heat is defined as the transfer of a type of energy in the presence of a temperature
gradient, and this energy is the random motion of the atoms and molecules that make up the material
under study. In the conventional definition heat flows from high temperature to low temperature,
indicating that the material with the higher temperature has more thermal energy. It is conceivable for
heat to be defined in such a way as to represent the direction opposite to this energy transfer across a
temperature gradient, in which case heat would flow from lower temperature to higher temperature.
In addition to defining heat flow, the methods of measuring temperature involving thermal expansion
(Section 17.4) also depend upon temperature difference. Therefore it is entirely logical to define a
temperature scale in such a way that the temperature difference between a cold material and a hot material
is negative as well as it is to define a scale in such a way that the difference is positive; what is important is
the magnitude of the difference. In fact, the original scale devised by Celsius had 0 as the boiling point of
water and 100 as the freezing point. As thermodynamics became better understood, it was clear that there
was a unique point on any temperature scale, which we now call absolute zero. To define a scale, as we now
do, such that systems hotter than this have positive, rather than negative, temperatures (or even to label
this point 0) is a matter of convenience.
17.12. Although the corona is very hot, it is not dense, so that the actual energy contained within the corona is
small. This explains why a spaceship flying in the corona will not be burned up.
17.13. Different metals have different melting points and different heat capacities. Thus, one metal may liquefy
easily while the other remains solid, making welding very difficult.
17.14. The volume of each object changes by the same amount during an identical ΔT. This implies that ΔV/ΔT is
the same for both objects. Therefore:
1
1 1 1 1 2 1 2
2 1
2 2 2 2 2 1 2
2
1
2 2 .
2
V VV
V V V V V V
T T T
β
β β β β β β β β
β
Δ ΔΔ
= = = =  =  =  =  =
Δ Δ Δ
17.15. The only difference between these two temperature scales is where the zero point is. The units are of
identical size. Hence, a temperature difference on the Kelvin scale is numerically equal to a temperature
difference on the Celsius scale. The coefficient of linear expansion is used only in equations involving
temperature differences, so it will take on the same value for either 1
K−
or 1
C .−
°
17.16. (a) The system is not in equilibrium, so 0 d i .T T T> >

Bauer/Westfall: University Physics, 1E
712
(b) A typical hot day is: 0 40 C (104 F)T = ° ° The ice temperature is close to its melting point: i 0 CT = °
The drink temperature is somewhere in between, d40 C 0 CT° > > ° but hopefully closer to the ice
temperature than the air temperature, such as dT 10 C.≈ °
17.17. Rankine temperatures, R ,T differ from Fahrenheit temperatures, F ,T only in being measured from absolute
zero. 0 K 273.15 C 459.67 F.= − ° = − ° Thus, R F F R459.67 R 459.67 F.T T T T= + °  = − ° Rankine
temperatures differ from Kelvin temperatures, K ,T only in being measured in Fahrenheit-degree
increments. Thus,
R K K R
9 5
.
5 9
T T T T=  =
Finally, in terms of degrees Celsius, C :T
R C C R
9 5
( 273.15 C) 273.15 C.
5 9
T T T T= + °  = − °
17.18. For such a two-level system, ordinary postive absolute temperatures correspond to the normal situation in
which the lower energy level is more popluated than the higher. The lower the temperature, the more
dominat is the lower level, as expected: the limit T → 0+
corresponds to a “ground state” in which all
components of the system are in the lower level. The higher the temperature, the more components are
excited into the upper level, but the lower level is always more populated; the limit T→+∞ corresponds to a
limit in which both levels are equally populated.
Negative absolute temperatures correspond to a “population inversion,” in which the higher energy level
is more populated than the lower. The limit T → 0−
describes the limiting situation in which the entire
system is in the higher level. Population inversions are real: the lasing medium of a laser, for example,
must be driven (“pumped”) into a population inversion for the laser to operate. The total energy of the
system is higher at negative temperature than positive; negative absolute temperatures are not “colder than
absolute zero,” they are “hotter than infinity,” in this context. It may be noted that the time dependence of a
quantum state with energy E is given by the complex function:
exp cos sin
Et Et Et
i i
     
− = −     
       
,
where t is time, ħ is Planck’s constant divided by 2 ,π and i the imaginary unit. Comparison of this with the
temperature-dependent population factor (Maxwell-Boltzmann distribution) suggests that in a quantum
context, (inverse) temperature can be interpreted as imaginary time!
17.19. Material 1 expands and contracts more readily than does material 2.
(a) The strip will bend toward material 1, since material 1 will contract more than material 2 does.
(b) The strip will bend toward material 2, since material 1 will expand more than material 2 does.
17.20. The metal lid has a larger coefficient of thermal expansion than the glass. Hot food is placed in a hot glass
jar and then a hot metal lid is screwed on top. As the glass and lid cools, the lid shrinks by a larger
percentage than does the glass, making a tight seal.
17.21. Both will expand to the same outer radius. A cavity in a material will expand in the same manner as if it
was filled with the same material.
Problems
17.22. Use C F
5
( 32 F)
9
T T= − ° and K C( 273.15 C)T T= + °

713
(a) 19 F:− ° C
5
( 19 F 32 F) 28 C;
9
T = − ° − ° = − ° K ( 28 C 273.15) 245 KT = − ° + =
(b) 98.6 F:° C
5
(98.6 F 32.0 F) 37.0 C;
9
T = ° − ° = ° K (37 C 273.15 C) 310. KT = ° + ° =
(c) C
5
52 F: (52 F 32 F) 11 C;
9
T° = ° − ° = ° K (11 C 273.15 C) 284 KT = ° + ° =
17.23. The temperature 21.8 CCT = − ° is three times its equivalent value F 7.3 F.T = − ° To check this, recall the
conversion formula: C F
5
( 32 F).
9
T T= − ° C F3T T= F F
5
( 32 F) 3
9
T T − ° = F
5 27 5
( 32 F)
9 9
T
− 
 − ° =  
 
F
5( 32 F)
7.2727 F
22
T
− °
 = = − ° and C
5
( 7.3 F 32 F) 21.8 C
9
T = − ° − ° = − °
17.24. Using C F
5
( 32 F)
9
T T= − ° and F 134 F,T = ° it is found that 134 F 56.67 C.° = ° Thus,
56.67 C 47 C 9.67 C.TΔ = ° − ° = ° Rounding to two significant figures gives 9.7 C.TΔ = °
17.25. C F
5 5
( 32 ) ( 129 32 ) 89.4F C
9
F
9
FT T= − = − − = −° ° ° °
17.26. To compare temperature, it is necessary to use the absolute temperature scale, which is Kelvin. Since 0 F°
corresponds to 255 K , the temperature that is twice is warm as 0 F° is 511 K, which is 460. F.°
17.27. (a) K C 273.15 79 273.C 15 194C KCT T= + = − =° °+°
(b) F C
9 9
32 ( 79 ) 32 110 F
5 5
C C CT T= + = − + = −° ° °°
17.28. In the present-day Celsius scale, 77.0 F° corresponds to 25.0 C° , so that in the original Celsius scale,
room temperature is 100. C 25.0 C 75.0 C.° − ° = °
17.29. F K F K
9 9
( 273.15 K) 32. If , then ( 273.15 K) 32
5 5
T T T T T T T= − + = = = − + 
9 9 4
(273.15 K) 32 459.7,
5 5 5
T T T− = −  =
5(459.67)
so 574.59 574.59 K 574.59 F
4
T = =  = °
17.30. At higher temperatures, the mass of copper remains constant, but its volume increases. Hence, the density
is expected to decrease. The volume expansion is 0V V TβΔ = Δ . For copper, 5 1
5.1 10 K .β − −
= ⋅ Taking
room temperature to be 293 K, (ie. 20 C° ), 1356 K 293 K 1063 K.TΔ = − = Density is
mass divided by volume. Thus, 0 0(293 K) / ,M Vρ ρ= = and mp 0(1356 K) /( ).M V Vρ ρ= = + Δ
mp 0 0
5 1
0 0 0 0
/( ) 1 1 1
0.949
/ 1 ( / ) 1 1 (5.1 10 K )(1063 K)
M V V V
M V V V V V T
ρ
ρ β − −
+ Δ
= = = = = =
+ Δ + Δ + Δ + ⋅
At 1356 KT = , copper is only 94.9% of the density of copper at 293 K.T =
17.31. 3
density at 20.0 C 7800.0 kg/m ,ρ = ° =
0
density at 100.0 C ,
M
V V
ρ′ = ° =
+ Δ
0 volume at 20.0 CV = °
0 ;V V TβΔ = Δ
3
3
5 1
0 0 0
7800.0 kg/m
= 7780 kg/m
(1 ) 1 1 (3.6 10 K )(80.0 K)
M M
V V T V T T
ρ
ρ
β β β − −
′ = = = =
+ Δ + Δ + Δ + ⋅
17.32. The cubes will expand when they are heated and will have a total length of 201.00 mm when
al br al br201.00 mm or 1.00 mm.L L L L+ = Δ + Δ = So, al al br br 1.00 mm,L T L TΔ + Δ =α α and:

714
6 1 6 1
1.00 mm
240 K.
(100.00 mm)(22 10 K 19 10 K )
T − − − −
Δ = =
⋅ + ⋅
17.33. The piston ring expands when it is heated, and the inner diameter must increase by 0.10 cm, so,
br 6 1br
10.10 cm 10.00 cm
0.1 cm and 526.316 K 250 C
(10.00 cm)(19 10 )
L T Tα −−
−
Δ = Δ = = ≈ °
⋅ Κ
Note that this process cannot be reversed to remove a piston ring. Once seated, the piston ring is in
thermal contact with the piston and cannot be heated without heating and expanding the piston.
17.34. To calculate the dimension change due to heat, the Kelvin scale should be used. 100.0 F 311.0 K° = , and
200.0 F 366.0 K° = . Therefore the temperature change is 55.0 K.TΔ =
(a) The volume change = 6 1 3 3
3 (3)(22 10 K )(4 /3)(10.0 cm) (55.0 K) 15 cmV Tα π− −
Δ = ⋅ =
(b) The radius change = 6 1
(22 10 K )(10.0 cm)(55.0 K) 0.012 cmR Tα − −
Δ = ⋅ =
Note that the radius change could be found from 3 2
(4 /3) , ( ) 4 ( ).V R dV R dRπ π= = Therefore,
2
/(4 ) 0.012 cm.dR dV Rπ= =
17.35. The cross-sectional area has no relevance-this depends on linear expansion, which is governed
by f (1 )L L Tα= + Δ , where in this case f 5.2000 m, 60. CL T= Δ = ° and 6
13 10α −
= ⋅ per degree Celsius
from Table 17.2. This gives
( )( )
f
6
5.2000 m
5.195947 m
(1 ) 1 13 10 K 60. K
L
L
Tα −
= = =
+ Δ + ⋅
at 10 C.− ° Thus there
will be 4.1 mm between adjacent rails.
17.36. The track will be free of any built-in tension as long as the thermal expansion is less than the 10.0 mm gap.
The expansion of the track at temperature T is: steel ( 20.0 C).L L TαΔ = − ° At the maximum allowable
temperature, gap .L dΔ = Therefore, steel max gap( 20.0 C) ,L T dα − ° = so,
( )( )
2
gap
max 6
steel
1.00 10 m
20.0 C 20.0 C 51 C.
13 10 K 25.0 m
d
T
Lα
−
−
⋅
= + ° = + ° = °
⋅
For most places in a temperate climate, this is enough for secure operation of the tracks, although larger
gaps (on the order of 13 mm) may also be used.
17.37. Suppose Tf is the temperature at which the two screws will touch. Use the equation 0 f 0( )L L T TαΔ = − to
find the increase in length for both brass and the aluminum screws at this temperature and set
Brass Aluminum 1.00 mmL LΔ + Δ = (1)
Now:
Brass Brass f(20.0 cm)( )( 22.0 C)L TαΔ = − ° (2)
( )( )Aluminum Aluminum f(30.0 cm) 22.0 CL TαΔ = − ° (3)
Substituting equations (2) and (3) into equation (1) yields:
Brass f Aluminum f(20.0 cm)( )( 22.0 C) (30.0 cm)( )( 22.0 C) 0.100 cm.T Tα α− ° + − ° =
Given 6
Brass 18.9 10 / Cα −
= ⋅ ° and 6
Aluminum 23.0 10 / C:α −
= ⋅ °
6 6
f f
1 1
(20.0 cm)(18.9 10 C )( 22.0 C) (30.0 cm)(23.0 10 C )( 22.0 C) 0.100 cmT T− −−−
⋅ ° − ° + ⋅ ° − ° =
f 6 6
0.100 cm
22.0 C 115 C
(20.0 cm)(18.9 10 / C) (30.0 cm)(23.0 10 / C)
T − −
= + ° = °
⋅ ° + ⋅ °

715
17.38. THINK: From the change in volume, the total volume can be determined from the equation of volume
expansion. It is necessary to find this volume V0 and the radius R of the sphere that can hold it.
SKETCH:
RESEARCH: 2 3
0 cylinder sphere
4
, = ,
3
dV V dT V r D V Rβ π π= =
SIMPLIFY:
2
2
0 0
1/3
2 2 2 2
3 3
0
4 3 3 3
,
3 4 4 4
r D
dV V dT r D V
dT
r D r D r D r D
V R R R
dT dT dT dT
π
β π
β
π π
π
β π β β β
= =  =
 
= = = =  =  
 
CALCULATE:
3 2 2
6 3
0 4 1
1/3
3 2 2
3
4 1
(0.10 10 m) (1.0 10 m)
1.745 10 m
(1.8 10 K )(1.0 K)
3(0.10 10 m) (1.0 10 m)
7.469 10 m
4(1.8 10 K )(1.0 K)
V
R
π − −
−
− −
− −
−
− −
⋅ ⋅
= = ⋅
⋅
 ⋅ ⋅
= = ⋅ 
 ⋅ 
ROUND: Two significant figures: 6 3 3
0 1.7 10 m , 7.5 10 m.V R− −
= ⋅ = ⋅
DOUBLE-CHECK: The radius at 7.5 mm for the sphere is considerably larger than the radius of 0.1 mm
for the capillary. This is as expected since the capillary is much longer than the radius of the base.
17.39. THINK: Assume that at 37 C° , the pool just begins to overflow. The volume expansion equation can
yield the depth of the pool. Let 1.0 cm = 0.010 m.d =
SKETCH:
RESEARCH: 2 3
0 ( ), ,V S S d V S= − = 0V V TβΔ = Δ
SIMPLIFY:
3 2 2 3 3 2 3 2
0 0 0
2 3 2 2
( ) ( ) ( )
1
0 ( ) 0 (1 ) 0 S
V V T V V V T S S S d S S d T S S S d S S d T
T
S d S T S d T S d d T S T d T S T d
T
β β β β
β
β β β β β β
β
Δ = Δ  − = Δ  − − = − Δ  − + = − Δ
 + Δ
 − Δ + Δ =  + Δ − Δ =  + Δ − Δ =  =  
Δ 
CALCULATE:
6 -1
6 -1
1 (207 10 K )(16 K)
(0.010 m) 3.029 m
(207 10 K )(16 K)
S
−
−
 + ⋅
= = 
⋅ 
ROUND: Two significant figures: S = 3.0 m.
DOUBLE-CHECK: 3.0 m is a realistic depth for a pool.

716
17.40. THINK: The change in length of the rods is equal to the sum of the change of the length of each rod
separately.
SKETCH:
RESEARCH: -5 -1 -5 -1
Steel Al, 1.3 10 K , 2.2 10 KL L Tα α αΔ = Δ = ⋅ = ⋅ and Al StL L L= +
SIMPLIFY: Al Al Al Al Al St St St St St(1 ), (1 )L L L L T L L L L Tα α′ ′= + Δ = + Δ = + Δ = + Δ ,
( ) ( )Al St Al St Al Al St Steel( )L L L L L L L Tα α′ ′Δ = + − + = + Δ
CALCULATE:
( )-5 -1 -5 -1
2.0 m 1.0 m (2.0 m)(2.2 10 K ) (1.0 m)(1.3 10 K ) 178 K 3.0 m 0.01015 mL  Δ = + + ⋅ + ⋅ − = 
ROUND: Rounding the change in length to two significant figures, 1.0 cm.LΔ =
DOUBLE-CHECK: The increase of 1 cm is small but quite noticeable. This is a reasonable result.
17.41. THINK: Since the period of a pendulum is proportional to the square of length, I can compute the change
of length due to thermal contraction and then compute the new period. Let 0t be the period of the
pendulum at a temperature of 0T , and let 1t be the period of the pendulum at temperature 1T .The
following quantities are given. 0 25. C,0T = ° 1 20 C.0T = − °
SKETCH:
RESEARCH: 5 1
Brass 1 0 0 Brass1.9 10 K , (1 ), 2
L
L L L L T t
g
− −
= ⋅ = + Δ = + Δ =α α π
SIMPLIFY: 1/2
1 0 Brass/ (1 ) ,t t Tα= + Δ Time elapsed = ( ) 1/2
Brass
1
0
24 h 24 h (1 )
t
T
t
α
 
= + Δ 
 
1/2 1/20 0 Brass 01
0 1 1 Brass 0 Brass
(1 )
2 , 2 2 2 (1 ) (1 )
L L T LL
t t t T t T
g g g g
α
π π π π α α
+ Δ
= =  = = + Δ = + Δ
CALCULATE: 1 0 20.0 25.0 45.0 K,T T TΔ = − = − − = −
Time elapsed = 5 1
(24 h) 1 (1.9 10 K )( 45.0 K) 23.990 h 23 h and 59.384 min− −
+ ⋅ − = =
ROUND: Taking 24 hours to be precise, have three significant figures. Subtract a precise 23 hours from
our result, and report the minutes to three significant figures: 23 hours and 59.4 minutes.
DOUBLE-CHECK: At the colder temperature, the period 1t of the pendulum is decreased; it is only
0.999 times the pendulum at the warmer temperature. This is reasonable since we wouldn’t expect the time
change to be very significant.
17.42. THINK: Both the capillary tube and the mercury will expand as the temperature increases. I can compute
the height of the mercury at 70 C° for both the silica and quartz capillary. If the heights differ by more
than 5%, then the quartz thermometers must be scrapped. The following quantities are given:

717
3
01 cm 0.25 mm,,sV r= = 1 0 70 C 20 C 50 C 50 K,T T TΔ = = ° − ° = ° =− 6 1
Si 0.4 10 K ,a − −
= ⋅
6 1
quartz 12.3 10 ,Ka − −
= ⋅ and 6 1 6 1
Hg 181 10 C 181 10 K− − − −
= = ⋅ ° = ⋅β β .
SKETCH:
RESEARCH: Hg Hg s ,V V TβΔ = Δ 2C rπ= = circumference of capillary, 0 ,C C TαΔ = Δ volume of a
cylinder: 2
cyl ,V r hπ= change in volume of spherical reservoir: S S3 .V V TαΔ = Δ
SIMPLIFY: When the volume of Hg expands by HgVΔ , the excess Hg goes into the excess volume in the
sphere, and the remainder goes up the capillary, up to a height h.
cyl
cyl Hg S2
0 0
0
0
Hg S
2 2
0
Hg S S
2 2
0
,
(1 )
2 2 2
2 (1 )
(1 )
2
(1 )
3
(1 )
V
h V V V
r
C C C TC
r
r T
r T
V V
h
r T
V T V T
r T
π
α
π π π
π α
α
π
π α
β α
π α
= = Δ − Δ
+ Δ + Δ
= = =
+ Δ
= = + Δ
Δ − Δ
=
+ Δ
Δ − Δ
=
+ Δ
For quartz:
Hg S quartz S
quartz 2 2
0 quartz
S Hg quartz
2 2
0 quartz
3
(1 )
( 3 )
.
(1 )
V T V T
h
r T
V T
r T
β α
π α
β α
π α
Δ − Δ
=
+ Δ
Δ −
=
+ Δ
The fractional change is height is:
quartz
silica
2
Hg quartz silica
Hg silica quartz
1
( 3 ) (1 )
1 .
( 3 ) (1 )
h
f
h
T
T
β α α
β α α
= −
 − + Δ
= −  
− + Δ  
CALCULATE:
2
6 16 1 6 1
6 1 6 1 6 1
1 0.4 10 K (50 K)181 10 K 3(12.3 10 K )
1 0.1995
181 10 K 3(0.4 10 K ) 1 12.3 10 K (50 K)
f
− −− − − −
− − − − − −
  + ⋅⋅ − ⋅   = − = 
⋅ − ⋅  + ⋅   
ROUND: One significant figure: 20 %.f =
DOUBLE-CHECK: The quartz thermometers will give a maximum error of about 20% at 70 C.° They
will have to be scrapped.

718
17.43. THINK: Assuming the brass and steel rods, L = 1.0 m each, do not sag, they will increase in length by
B Sand ,L LΔ Δ respectively. The rods will touch when their combined extensions equals the separation,
d = 5.0 mm. The linear expansion coefficients of the rods are 6 6
B S19 10 / C and 13 10 / Cα α− −
= ⋅ ° = ⋅ ° . The
initial temperature of the rods is i 25 C.T = °
SKETCH:
RESEARCH: The brass rod will increase by B BL L TαΔ = Δ . The steel rod will increase by S SL L TαΔ = Δ .
The final temperature will be iT TΔ + . The rods touch when B S .L L dΔ + Δ =
SIMPLIFY: B S B S( ) .d L L L Tα α= Δ + Δ = + Δ f i
B S B S
Thus, .
( ) ( )
d d
T T T
L Lα α α α
Δ =  = +
+ +
CALCULATE:
3
f 6
5.0 10 m
25 C 181.25 C
(1.0 m)(19 13) 10 / C
T
−
−
⋅
= + ° = °
+ ⋅ °
ROUND: Two significant figures: f 180 CT = ° .
DOUBLE-CHECK: Given the long length, d, for the total expansion, such a high temperature is not
unreasonable.
17.44. THINK: As the pendulum is heated, each bar increases in length. The steel bars, S 50.0 cmL = and
6
S 13 10 / C,α −
= ⋅ ° will increase in length such that the bob will move twice this distance from the pivot.
The lead bars, 6
Pb 29 10 / C,α −
= ⋅ ° will increase in length such that it will reduce the distance from the bob
to the pivot. Determine the length, Pb ,L L= of each of the two lead bars.
SKETCH:
RESEARCH: The change in length of the steel rods is S S SL L TαΔ = Δ , while that of the lead rods is
Pb Pb PbL L TαΔ = Δ .For the pendulum length , h, to remain unchanged, S Pb2 .L LΔ = Δ
SIMPLIFY: S Pb S S Pb Pb2 2L L L T L Tα αΔ = Δ  Δ = Δ . S
Pb S
Pb
2
Thus, L L
α
α
=
CALCULATE:
6
Pb 6
2(13 10 / C)
(50.0 cm) 44.83 cm
29 10 / C
L
−
−
⋅ °
= =
⋅ °
ROUND: The values in Table 17.2 are given to two significant figures. This results in a final answer of
Pb 45 cm.L =

719
DOUBLE-CHECK: The value of S2α is about 10% less than Pbα (26 and 29, respectively), and this means
the lead bars being 10% shorter is reasonable.
17.45. THINK: Since brass has a higher linear expansion coefficient than steel, 6 -1
B 19 10 Kα −
= ⋅ and
6 -1
S 13 10 Kα −
= ⋅ , the brass will become larger in length. After being heated up, 20. K,TΔ = the strip will
arc. It is vital to focus on the radius of the midline of each part of the strip. Since each material has a
thickness, 0.50 mmδ = , each will be an arc of different radius, so the radius must be considered. The
actual arc length of each strip will be its length after being heated up and each will share the same angle.
The end of the strip lowers by 3.0 mm.yΔ =
SKETCH:
RESEARCH: From the geometry of the system, θΔ = −(1 cos ).y r Also, B( /2)r Lδ θ ′+ = and
δ θ ′− = S( /2) ,r L because the arc length equals radius times angle. When the strips are heated up, their
lengths are increased by B BL L TαΔ = Δ and S S .L L TαΔ = Δ This means the final lengths of the strips are
B B(1 )L L Tα′ = + Δ and S S(1 ).L L Tα′ = + Δ
SIMPLIFY: Determine the radius, r, first:
B B
SS
1( /2) /2
.
( /2) /2 1
L Tr r
r r TL
αδ θ δ
δ θ δ α
′ + Δ+ +
=  =
− − + Δ′
Thus,
B
S
1( 1)
, where .
2 2 2 ( 1) 1
Tx
r r x r x
x T
αδ δ δ
α
+ Δ+ 
+ = −  = = 
− + Δ 
Now that the radius is known, the angle, θ, can be determined:
1
(1 cos ) cos 1 cos 1 .
y y
y r
r r
θ θ θ −Δ Δ 
Δ = −  = −  = − 
 
Next consider the difference in B S ,L L′ ′−
B S B S B S B S(1 ) (1 ) ( ) .L L L T L T L T L T L Tα α α α α α′ ′− = + Δ − + Δ = Δ − Δ = − Δ
However, B S ( /2) ( /2) .L L r rδ θ δ θ δθ′ ′− = + − − = Therefore, B S
B S
( ) .
( )
L T L
T
δθ
δθ α α
α α
= − Δ  =
− Δ
Further algebraic simplification leads to
δ
α α δ
−  Δ −  
= −   − Δ +   
1
B S
2 1
cos 1 .
( ) 1
y x
L
T x

720
CALCULATE:
6 1
6 1
1 (19 10 K )(20. K)
1.000119969
1 (13 10 K )(20. K)
x
− −
− −
+ ⋅
= =
+ ⋅
3 3
1
6 1 3
0.50 10 m 2(3.0 10 m) 1.000119969 1
cos 1 0.1581m
1.000119969 1(19 13) 10 K (20. K) 0.50 10 m
L
− −
−
− − −
   ⋅ ⋅ −   = − =     +− ⋅ ⋅     
ROUND: Two significant figures: 0.16 mL =
DOUBLE-CHECK: This length is a plausible length for a bimetallic strip of metal that deflects 3 mm
when heated by 20 K.
17.46. THINK: Since the bulk modulus, 160 GPa,B = is the pressure per fractional change in volume, a change
in temperature of 1.0 CTΔ = ° will cause a change in volume, and thus a change in pressure. The linear
expansion coefficient of steel is 5
1.3 10 / C.−
⋅ °
SKETCH: A sketch is not needed to solve this problem.
RESEARCH: The bulk modulus is given by ( )/ / .B P V V= Δ Δ The change in volume is given by
3 .V V TαΔ = Δ
SIMPLIFY: 3 3 , 3
/
V P V
V V T T B P B B T
V V V V
α α α
Δ Δ Δ 
Δ = Δ  = Δ =  Δ = = Δ Δ  
CALCULATE: 5
3(160 GPa)(1.3 10 / C)(1 C) 6.24 MPaP −
Δ = ⋅ ° ° =
ROUND: Two significant figures: 6.2 MPa.PΔ =
DOUBLE-CHECK: For comparison, atmospheric pressure is 0.10 MPa. The problem mentioned it could
produce very large pressures, so this answer seems reasonable.
17.47. THINK: When the horseshoe is put in the tank, 10.0 cmr = , the water rises by 0.25 cm.h = The
horseshoe, i 293 KT = (room temperature, 20. C° ) and f 700. K,T = will increase its volume. When it is
put back in water, it will raise the water level by h′ . The difference in water weights is .h h h′Δ = − The
linear expansion coefficient of the horseshoe is 6 1
11 10 K− −
= ⋅α .
SKETCH:
RESEARCH: When water rises by h or h′ , the volumes displaced are 2
V r hπ= and 2
V r hπ′ ′= . The
volume of the heated horseshoe is (1 3 )V V Tα′ = + Δ . The initial volume of the horseshoe, 0 ,V is the same
as the volume of water it displaced before it was heated, 2
.r hπ The volume of displaced heated water is
equal to the volume of the heated horseshoe.
SIMPLIFY: 2 2
(1 3 ) (1 3 ) (1 3 )V V T r h r h T h h T′ ′ ′= + Δ  = + Δ  = + Δα π π α α
The change in water height is 3 .h h h h Tα′Δ = − = Δ
CALCULATE: 6 1
3(0.25 cm)(11 10 K )(700. K 293 K) 0.003358 cmh − −
Δ = ⋅ − =
ROUND: The least precise value given in the question has two significant figures. Therefore the final
answer should be rounded to 2
3.4 10 mm.h −
Δ = ⋅
DOUBLE-CHECK: The change in the water height is small which seems reasonable since the change in
volume is small and the cross sectional area 4 2
(3 10 mm )⋅ of the tank of water is relatively large.

721
17.48. THINK: Since the period of a pendulum is proportional to the length, an increase in temperature will
increase the length and, hence, the period. If the pendulum makes n oscillations in one week when at
20.0 C° , it will take longer to go through n oscillations when the period is greater, making the week
appear longer (the clock will run slow). The initial period of the pendulum is i 1.000 s,T = and then
increases while the temperature increases to30.0 C° . Use 6
Al 22 10 / C.α −
= ⋅ °
SKETCH:
RESEARCH: The period of the pendulum is given by 2 /T L gπ= . The length of pendulum after being
heated is Al(1 )L L Tα′ = + Δ . The number of oscillations that the pendulum makes over the period of time, t,
is / .n t T=
SIMPLIFY:
(a) Period after temperature change:
Al Al i Al2 / 2 (1 )/ 2 / 1 1 .T L g L T g L g T T T′ ′= = + Δ = + Δ = + Δπ π α π α α
(b) Number of oscillations in one week at 20 C is / .n t T° = The amount of time for n oscillations at
30 C is .t nT′ ′° = The difference in time between the pendulum at 30 C° and the pendulum at 20 C° is:
( ) ( / 1).t n T T t T T′ ′Δ = − = −
CALCULATE:
(a) 6
(1.000 s) 1 (22 10 / C)(30.0 C 20.0 C) 1.00011 sT −
′ = + ⋅ ° ° − ° =
(b)
7 days1.0011 s 24 hr 3600 s
1 week 1 66.52 s
1.000 s 1 week 1 day 1 hr
t
 
Δ = − ⋅ ⋅ ⋅ = 
 
ROUND: The answers should be rounded to two significant figures.
(a) 1.0 sT′ =
(b) 67 stΔ =
DOUBLE-CHECK: Losing 67 seconds over a full week is a reasonable amount for a temperature change
of 10. C.°
17.49. THINK: Use the subscript 1 to refer to the thin arm and the subscript 2 to refer to the thick arm. After the
temperature change (using a room temperature of 20. C° ), 1 380 KTΔ = and 2 180 KTΔ = , for the upper
and lower arms, respectively, each length will increase by a different amount. Since the ends are fixed in
position, the device overall will begin to angle downward so the tip is pointing below its original position.
The initial length and linear expansion coefficient of each arm is 1800 μmL = and 6 -1
3.2 10 K−
⋅ .The
separation of the electrical contacts is 45 μm.h =

722
SKETCH:
RESEARCH: The upper and lower arms each increase in length by 1 1L L TαΔ = Δ and 2 2 ,L L TαΔ = Δ
respectively. The change in height of the tip is 1( )siny L L θΔ = + Δ . The difference in the extended length is
1 2L LΔ − Δ and this length is equal to sinh θ .
SIMPLIFY: Determine the angle: 1 2 1 2sin sin ( )/ .h L L L L hθ θ= Δ − Δ  = Δ − Δ The change in tip height:
1 1 21( )sin ( )( )/ .y L L L L L L hθΔ = + Δ = + Δ Δ − Δ Thus, 2
1 1 2(1 )( )/ .y L T T T hα αΔ = + Δ Δ − Δ
CALCULATE:
( )6 1 2 6 1
(3.2 10 K )(1800μm) 1 (3.2 10 K )(380 K) (380 K 180 K)/ 45μm 46.14 μmy − − − −
 Δ = ⋅ + ⋅ − = 
ROUND: Two significant figures: 46 μmyΔ = downwards
DOUBLE-CHECK: The change in the tip height is of the same order of magnitude as the separation
between the contact points, so the result is reasonable.
17.50. THINK: The tip is located at the midpoint of the beam which is also midway between the contact points,
which are separated by a distance d. The silicon beam, 6 1
Si 3.2 10 K ,α − −
= ⋅ makes an angle of
0.10 radθ = from the horizontal when it is at a temperature of 20 C° . As the beam heats up to 500 C,° its
length will increase, but since the tip must remain in the same position horizontally, the angle the beam
makes must also increase, which in turn causes motion of the tip.
SKETCH:
RESEARCH: Before heating, the length, ,L of the beam is given by /(2cos )L d θ= . After heating, the
beam increases in length by SiL L TαΔ = Δ . Even after the length increases, the tip does not moves
horizontally, so that /(2cos ).L L d φ+ Δ = The tip moves vertically by an amount
Si( )sin sin (1 )sin sin .h L L L L Tφ θ α φ θ= + Δ − = + Δ −  
SIMPLIFY: The initial length of the beam is /(2cos )L d θ= . Therefore, the length of the beam after
heating is given by: Si Si(1 ) (1 )/(2cos ) .L L L T d Tα α θ+ Δ = + Δ = + Δ Since the new angle of the beam after
heating is /(2cos ),L L d φ+ Δ = this means that:
1
Si
Si Si
cos cos
(1 ) cos cos .
2cos 2cos 1 1
d d
T
T T
θ θ
α φ φ
θ φ α α
−
 
+ Δ =  =  =  
+ Δ + Δ 
The change in height of the tip is then:

723
Si Si
1
Si
Si
sin
(1 )sin sin (1 ) tan
2 cos
cos
sin cos
1
(1 ) tan .
2 cos
d
h L T T
Td
T
φ
α φ θ α θ
θ
θ
α
α θ
θ
−
 
= + Δ − = + Δ −    
 
   
   
+ Δ    = + Δ − 
 
 
 
CALCULATE:
( )
1
6 1
6 1
cos(0.10 rad)
sin cos
1 (3.2 10 K )(480 K)1800 μm
1 (3.2 10 K )(480 K) tan(0.10 rad)
2 cos(0.10 rad)
12.993μm
h
−
− −
− −
   
    + ⋅    = + ⋅ − 
 
 
 
 
=
ROUND: The answer should be rounded to two significant figures: 13 μm,h = upwards.
DOUBLE-CHECK: This value is the same order of magnitude as the beam width, meaning that it has a
great sensitivity, which would be desired for such a device. This is sensible.
17.51. THINK: For simplicity, define 1.00016,a = 5
4.52 10b −
= ⋅ and 6
5.68 10 .c −
= ⋅ In part (a), a derivative can
be used to determine the properties of the water. The volume, V, as a function of temperature, T, is given
by 2
V a bT cT= − + when the temperature is in the range [0.00 C, 50.0 C).° ° In part (b), evaluate β when
20.0 C.T = °
RESEARCH: The general function to evaluate the change in volume is .V V TΔ = Δβ The differences can
be approximated as differentials, i.e. / / .Y X dy dxΔ Δ ≈
SIMPLIFY: 2
( ) 2
dV d
a bT cT b cT
dT dT
= − + = − + . Since ,V V TβΔ = Δ it follows that:
2
1 1 2V dV b cT
V T V dT a bT cT
β
Δ − +   
= ≈ =   
Δ − +   
CALCULATE:
(a)
5 6
5 6 2
4.52 10 11.36 10
( )
1.00016 4.52 10 5.68 10
T
T
T T
β
− −
− −
− ⋅ + ⋅
=
− ⋅ + ⋅
(b)
5 6
5 6 2
4
4.52 10 (11.36 10 )(20.0 C)
(20 C)
1.00016 (4.52 10 )(20.0 C) (5.68 10 )(20.0 C)
1.8172 10 / C
β
− −
− −
−
− ⋅ + ⋅ °
° =
− ⋅ ° + ⋅ °
= ⋅ °
ROUND:
(a) Not necessary.
(b) Round to three significant figures: 4
( 20.0 C) 1.82 10 / CTβ −
= ° = ⋅ °
DOUBLE-CHECK: The value for β for water at 20.0 C° from Table 17.3 is 4
2.07 10 / C.−
⋅ ° Since the
calculated value is close, this is a reasonable result.
17.52. THINK: Since copper has a higher linear expansion coefficient than steel ( 6 1
C 17 10 Kα − −
= ⋅ and
6 1
S 13 10 Kα − −
= ⋅ ), the copper will become shorter in length than the steel. After a change in temperature
of 5.0 KTΔ = − , the strip will arc. Since each material has a thickness of 1.0 mmδ = each will be an arc
with a different radius, so the radius to the midpoint of each strip must be considered. The actual arc

724
length will be the length after being cooled and each will share the same angle or curvature. The initial
length of each strip is 25 mm.L =
SKETCH:
RESEARCH: The lengths of each strip after cooling are C C(1 )L L Tα′ = + Δ and S S(1 )L L Tα′ = + Δ . These
lengths are the arc lengths of circles of radius ( )/2r δ− and ( )/2 ,r δ+ respectively, so:
C
2
r L
δ
θ
  ′− = 
 
and S .
2
r L
δ
θ
  ′+ = 
 
The deflection of the strip is given by (1 cos ).y r θΔ = −
SIMPLIFY:
(a) Determine the radius of curvature, :r
( )
( )
C C C
S SS
C
S
S C
S C
/2 (1 ) 1/2
(1 ) /2 1/2
11
/2 ( /2) , where
2 1 1
2 ( )
2 ( )
rL L T Tr
L T r TrL
Tx
r x r r x
x T
T
r
T
′ − + Δ + Δ−
= =  =
+ Δ + + Δ+′
+ Δ+ 
 − = +  = = 
− + Δ 
 + + Δ
 =  
− Δ 
δ θ α αδ
α δ αδ θ
αδ
δ δ
α
α αδ
α α
(b) To find the deflection, yΔ , I need ( ) ( )C S: /2 , /2 .r L r Lθ δ θ δ θ′ ′− = + = Thus,
C(1 )
/2
L T
r
α
θ
δ
+ Δ
=
−
or S(1 )
.
/2
L T
r
α
θ
δ
+ Δ
=
+
So, the deflection is C S(1 ) (1 )
(1 cos ) 1 cos 1 cos .
/2 /2
L T L T
y r r r
r r
α α
θ
δ δ
   + Δ + Δ   
Δ = − = − = −      
− +      
CALCULATE:
(a)
6 13
6 1
2 (13 17) 10 K ( 5.0 K)1.0 10 m
49.996 m
2 (13 17) 10 K ( 5.0 K)
r
− −−
− −
 + + ⋅ −⋅
= = 
 − ⋅ − 
(b)
( )3 6 1
3
25 10 m 1 17 10 K ( 5.0 K)
49.996 m 1 cos 0.00625 mm
49.996 m 1.0 10 m /2
y
− − −
−
  ⋅ + ⋅ −
  Δ = − =
  − ⋅
  
ROUND: The results should be rounded to two significant figures.
(a) 50. mr =
(b) 6.3 μmyΔ =
DOUBLE-CHECK: Since the expansion coefficients of each are close to each other and the change in
temperature was small, it is reasonable that the strip barely curves (it has a large radius of curvature and a
small dip).

725
Additional Problems
17.53. Each side of the cube has length 40 cml = and its initial volume before heating is 3
i .V l= The change in
temperature is 100. CTΔ = ° and linear expansion coefficient of copper is 6
Cu 17 10 / C.α −
= ⋅ °
3 6 3 3
i3 3 3(17 10 / C)(40. cm) (100. C) 326.4 cmV V T l Tα α −
Δ = Δ = Δ = ⋅ ° ° =
Thus, change in volume is 3
330 cm .VΔ =
17.54. The initial length of the pipe is 50.0 m,L = the change in temperature is 30.0 CTΔ = ° , and the change in
length is 2.85 cm.LΔ =
(a) 50.0285 m
1.90 10 /K
(50.0 m)(30.0 C)
L
L L T
L T
α α −Δ
Δ = Δ  = = = ⋅
Δ °
(b) This linear expansion coefficient matches that of brass.
17.55. When the aluminum container is filled with turpentine, the turpentine will have a volume of 5.00 gal.V =
Since the volume expansion coefficient of the turpentine, 4
turp 9.00 10 / C,β −
= ⋅ ° is much greater than that
of aluminum, assume that all of the volume gained by the turpentine spills out. The change in temperature
is 12.0 C.TΔ = ° The change in volume is given by:
4
turp
3.785 L
(9.00 10 / C)(5.00 gal)(12.0 C) 0.2044 L.
1 gal
V V Tβ −  
Δ = Δ = ⋅ ° ° = 
 
Thus, 0.204 L of turpentine spills out of the container.
17.56. The building has initial height of 600. m.L = The change in temperature is 45.0 C.TΔ = ° The linear
expansion coefficient of steel is 5
S 1.30 10 / C.α −
= ⋅ °
5
S (1.30 10 / C)(600. m)(45.0 C) 0.351 mL L Tα −
Δ = Δ = ⋅ ° ° =
Thus, the building grows by0.351 m.
17.57. The initial diameter of the rod at 20. C° is 1 ,D and after being cooled by a change in temperature of
77.0 K (20. C 273 K) 216 K,T  Δ = − ° + = −  it will have a diameter of 2 10.000 mm.D = The linear
expansion coefficient of aluminum is 6 1
Al 22 10 K .α − −
= ⋅
Al 1 2 1 1 Al
2
2 Al 1 1 1 6 1
Al
(1 )
10.000 mm
(1 ) 10.0478 mm
1 1 (22 10 K )( 216 K)
D D T D D D D T
D
D T D D D
T
α α
α
α − −
Δ = Δ  = + Δ = + Δ
= + Δ  =  = =
+ Δ + ⋅ −
Thus, the maximum diameter the aluminum rod can have at 20. C° is 1 10. mm.D =
17.58. After the gas is heated up, its final volume is f 213 L.V = The change in temperature is 63 F.TΔ = ° The
volume expansion coefficient of gas is 6 1
950 10 K .− −
⋅ Convert the change in temperature to Kelvin:
f C
5
9
T TΔ = Δ and C K
5
(63 F) 35 K.
9
T T TΔ = Δ  Δ = ° =
f
gas i f i i gas i 6 1
gas
213 L
, (1 ) 206.15 L
1 1 (950 10 K )(35 K)
V
V V T V V V V T V
T
β β
β − −
Δ = Δ = + Δ = + Δ  = = =
+ Δ + ⋅
Thus, the maximum amount of gasoline that should be put into the tank at 57 F° is 206.15 L. Rounding
this value is dangerous, since the tank would overflow or possibly explode if 210 L is added.
17.59. The initial volume of the mercury is 8.0 mL,V = the cross-sectional area of the tube is 2
1.0 mmA = and
the volume expansion coefficient of mercury is 6
Hg 181 10 / C.β −
= ⋅ ° Consider a change in temperature
of 1.0 C.TΔ = ° Since the cross-sectional area remains closely the same, .V A LΔ = Δ

726
6 3
Hg
Hg 2
(181 10 / C)(8.0 mL)(1.0 C) 1000 mm
1.5 mm
mL1.0 mm
V T
V V T A L L
A
−Δ  ⋅ ° °
Δ = Δ = Δ  Δ = = = 
 
β
β
Thus, the 1 C° tick marks should be spaced about 1.5 mm apart.
17.60. The initial volume of gasoline is 14 gallons and the change in temperature is 27 F.TΔ = ° The volume
expansion coefficient of gas is −
⋅ °4
9.6 10 / C. Convert the temperature change from Fahrenheit to Celsius:
C F
5
.
9
T TΔ = Δ Thus
5
(27 F) 15 C.
9
TΔ = ° = °
Thus, β −
Δ = Δ = ⋅ ° ° =4
gas (9.6 10 / C)(14 gal)(15 C) 0.2016 gal.V V T So, 0.20 gallons of gasoline are lost.
17.61. The change in temperature is 37.8 CTΔ = ° . The initial length of the slabs is 12.0 m.L = The linear
expansion coefficient of concrete is 6
con 15 10 / C.α −
= ⋅ °
6
con (15 10 / C )(12.0 m)(37.8 C) 0.006804 mL L Tα −
Δ = Δ = ⋅ ° ° =
Since the slabs expand uniformly, each side will grow by /2LΔ . However, the slabs expand towards each
other, so each can grow by / 2.LΔ Thus, the gap must be 2( /2) 6.8 mm.L LΔ = Δ =
17.62. Since water and aluminum have similar volume expansion coefficients, both must be accounted for. The
water has a volume of 3
500. cmV = . Though the volume of the aluminum can is not known, it has a
capacity to carry a volume .V For simplicity assume that the amount of water that it can hold is the same
as the volume of the aluminum vessel after heating. The change in temperature is 30. C,TΔ = ° the volume
expansion coefficient of the water is 6
w 207 10 / C−
= ⋅ °β and the linear expansion coefficient of aluminum
is 6
Al 22 10 / C.−
= ⋅ °α The change in volume of the water is given by: w w .V V TβΔ = Δ The change in
volume of the aluminum vessel is given by: Al Al3 .V V TαΔ = Δ The difference in the change in volumes is
w Al w Al( 3 ).V V V V T β α′ = Δ − Δ = Δ −
3 6 6 3
(500. cm )(30. C)(207 10 / C 3(22 10 / C)) 2.115 cmV − −
′ = ° ⋅ ° − ⋅ ° =
Thus, about 3
2.1 cm of water spills out, since the volume change of the water is larger.
17.63. The volume expansion coefficient of kerosene is 6
k 990 10 / C.β −
= ⋅ ° If the volume increases by 1.0%,
then / 0.010.V VΔ =
k 6
k
1 0.010
10.1 C
990 10 / C
V
V V T T
V
β
β −
 Δ
Δ = Δ  Δ = = = ° 
⋅ ° 
Thus, the kerosene must be heated up by at least 10. C° in order for its volume to increase by 1.0%.
17.64. The radius of the holes is h 1.99 cmr = and the radius of the ball bearings is bb 2.00 cmr = . The linear
expansion coefficient of epoxy is 4
e 1.3 10 / C,α −
= ⋅ ° the cross-sectional area of the ball bearings
is 2
bb bbA rπ= and the cross-sectional area of the holes is 2
h h .A rπ= The epoxy is heated so that h bb= .A A
bb h e
2 2
bb
2 2 2
bb h
2 4
h
2 (1 2 )
(2.00 cm)1 1
(1.99 cm)
1 2 38.752 C
2 2(1.3 10 / C)
A A T A A T
r
r r
T T
r
α α
α
α −
Δ = Δ  = + Δ
− −
− = Δ  Δ = = = °
⋅ °
Thus, the epoxy needs to be heated up by about39 C.°
17.65. THINK: When the disk (mass ,M radius R and moment of inertia I ) is heated up from
i 20. CT = ° to f 100. C,T = ° its radius, and hence area, will increase but the mass will stay the same. This
allows us to determine the new moment of inertia and compare to the initial one.

727
SKETCH:
RESEARCH: Since the object is a disk, its moment of inertia, before and after heating, is:
21
2
I MR= and 21
2
I MR′ ′= , respectively.
The area of the disk is 2
A Rπ= and the area changes by 2A A TαΔ = Δ upon heating. The linear expansion
coefficient of the brass disk is 6
B 19 10 / C.α −
= ⋅ °
SIMPLIFY: Area after heating: 2 2
f i B B(1 2 ) (1 2 ).A A T R R Tα α′= + Δ  = + Δ The fractional change in
moment of inertia given by:
2 2
2 22 2
B
B2 2
2
1 1
(1 2 )2 2 2 .
1
2
MR MR R T RI I I R R
T
I I R R
MR
α
α
′ −′ ′ + Δ −Δ − −
= = = = = Δ
CALCULATE: 6
2(19 10 / C)(100. C 20. C) 0.00304
I
I
−Δ
= ⋅ ° ° − ° =
ROUND: Two significant figures: the moment of inertia changes by 0.30%.
DOUBLE-CHECK: From our experience we would not expect the moment of inertia of a disk to change
very dramatically for such a modest temperature change. A change of 0.30% is a reasonable result.
17.66. THINK: Initially, the brass sphere of diameter B 25.01 mmd = is too big to fit through the
hole, Al 25.00 mm,d = in the aluminum plate. As both are heated up, both will expand. Since aluminum
has a higher expansion coefficient, the hole will eventually become larger than the sphere.
SKETCH:
RESEARCH: The area of the hole and the cross-sectional area of the sphere increase with temperature
as Al Al Al2A A TαΔ = Δ and B B B2 ,A A TαΔ = Δ respectively, where the initial areas of the hole and sphere are
2
Al Al( /2)A dπ= and 2
B B( /2) ,A dπ= respectively. The sphere will fall into the hole when the final areas of
the two are equal: Al Al B B(1 2 ) (1 2 ).A T A Tα α+ Δ = + Δ The linear expansion coefficients of brass and
aluminum are 6
B 19 10 / Cα −
= ⋅ ° and 6
A 22 10 / C,α −
= ⋅ ° respectively. The initial temperature of two
objects is room temperature, i 20. C.T = °

728
SIMPLIFY:
2
B
B B(1 2 )
2
d
A Tπ α
 ′ = + Δ 
 
，
2
Al
Al Al(1 2 )
2
d
A Tπ α
 ′ = + Δ 
 
2 2
B Al B B Al Al
2 2
2 2 2 2 B Al
B Al Al Al B B 2 2
Al Al B B
2 2
B Al
f i2 2
Al Al B B
(1 2 ) (1 2 )
2 ( )
2( )
2( )
A A d T d T
d d
d d T d d T
d d
d d
T T
d d
α α
α α
α α
α α
′ ′=  + Δ = + Δ
−
− = Δ −  Δ =
−
−
= +
−
CALCULATE:
2 2
f 2 6 2 6
(25.01 mm) (25.00 mm)
20. C 134.04 C 20. C 154.04 C
2 (25.00 mm) (22 10 / C) (25.01 mm) (19 10 / C)
T − −
−
= + ° = ° + ° = °
 ⋅ ° − ⋅ ° 
ROUND: In the previous calculation, the quotient should be rounded to two significant figures, so the
final answer is f 150 C.T = °
DOUBLE-CHECK: Since the expansion coefficients of the two materials are close in value, such a high
temperature is expected.
17.67. THINK: The steel band has an initial diameter of i 4.4 mm,d = width 3.5 mm,w = and thickness
0.45 mm.t = As the band cools from i 70 CT = ° to f 37 CT = ° its diameter will decrease. Since the
circumference of the band is directly proportional to the diameter, both the circumference and the
diameter have the same relative change with the decrease in temperature. The tension in the band can be
found by considering the Young’s modulus of the steel band.
SKETCH:
RESEARCH: The change in the area of the band (i.e. area around the tooth), is S2 ,A A TαΔ = Δ where the
area is ( )
2
/2dπ . Young’s modulus is the ratio of the stress to the strain where the stress is the force per
unit area and the strain is the relative change in length,
i.e.
/
/
F wt
Y
L L
=
Δ
.
For steel, 10 2
20. 10 N/m .Y = ⋅ The length of the band is the circumference, so L dπ= . For this problem,
use TΔ in place of .TΔ The linear expansion coefficient of steel is 6
S 13 10 / C.α −
= ⋅ °
SIMPLIFY: The relative change in area is: S2 .
A
T
A
α
Δ
= Δ Since the length is proportional to the
diameter: .
L d
L d
Δ Δ
= Since ( )
2
2
S/2 , 2 .
A d L A
A d T
A d L A
π α
Δ Δ Δ Δ 
= =  = = Δ 
 
Determining the
force S
/
: ( / ) 2 .
/
F wt
F Y F Ywt L L F Ywt T
L L
α=  = Δ  = Δ
Δ
CALCULATE: 10 2 3 3 6
(20. 10 N/m )(3.5 10 m)(0.45 10 m) 2(13 10 / C)(33 C) 9226.8 NF − − −
= ⋅ ⋅ ⋅ ⋅ ° ° =
ROUND: The answer has to be rounded to two significant figures: 9.2 kN.F =
DOUBLE-CHECK: Since a tooth is very strong, this large tension that is created will be able to act on the
tooth without causing problems. The force must also be large in order to withstand the forces of biting
food. Therefore, this is a reasonable result.

729
17.68. THINK: To find the spacing between tick marks, I must consider how high the mercury, of initial
volume 3
i 8.63 cm ,V = rises in the tube of diameter 1.00 mmd = when the temperature increases
by 1.00 C.TΔ = ° I can assume that the cross-sectional area of the tube remains constant.
SKETCH:
RESEARCH: The cross-sectional area of the thermometer is ( )
2
/2A dπ= . The change in volume of the
mercury due to a temperature change is HgV V TβΔ = Δ . Since the expansion of the tube can be neglected,
.V A LΔ = Δ The coefficient of volumetric expansion for the mercury is 4
Hg 1.82 10 / C.β −
= ⋅ °
SIMPLIFY:
Hg Hg
Hg 2
2
V T V T
V V T A L L
A d
β β
β
π
Δ Δ
Δ = Δ = Δ  Δ = =
 
 
 
CALCULATE:
4 3
2
(1.82 10 / C)(8.63 cm )(1.00 C)
0.19998 cm
0.100 cm
2
L
π
−
⋅ ° °
Δ = =
 
 
 
ROUND: Three significant figures: 2.00 mmLΔ =
DOUBLE-CHECK: This length would indicate a thermometer to be 20 cm to measure from
0 C to 100 C .° ° This is a reasonable size for a thermometer, so this spacing size is logical.
17.69. THINK: The device has an initial volume of 3
i 0.0000500 m ,V = which will increase upon heating. Of
course, this volume change is proportional to the volume expansion coefficient of the material, β . A
change in temperature is proportional to a change in volume. This means that a temperature change rate
( 200. CTΔ = ° in T 3.00 seconds)tΔ = is also proportional to a volume change rate
3
( 0.000000100 mVΔ = in V 5.00 seconds).tΔ =
RESEARCH: The change in volume is iV V TβΔ = Δ . The maximum volume change rate is:
i
V Tmax
V T
V
t t
β
 Δ Δ
= 
Δ Δ 
.
SIMPLIFY: The value for β for the maximum volume change rate is when:
T
V i
1
.
tV
t T V
β
 Δ Δ
=   
Δ Δ  
CALCULATE:
3
6
3
(0.000000100 m )(3.00 s)
6.0000 10 / C
(5.00 s)(200. C)(0.0000500 m )
β −
= = ⋅ °
°
ROUND: Three significant figures: 6
6.00 10 / C.β −
= ⋅ °
DOUBLE-CHECK: This value has the same order of magnitude that as has been seen for many volume
expansion coefficients, so it is a reasonable answer.

730
17.70. THINK: The rod has a length of cross-sectional area of 1.0000 mL = and 4 2
5.00 10 m .A −
= ⋅ After an
increase in temperature from i 0 CT = ° to f 40. C,T = ° the rod will tend to expand. Since it cannot
expand between the two end points, it will experience stress. The stress can then be determined by using
Young’s modulus. 11 2
2.0 10 N/mY = ⋅ , 6
13 10 / C.α −
= ⋅ °
SKETCH:
RESEARCH: Young’s modulus (for steel is 10 2
20 10 N/m )Y = ⋅ is the ratio of the stress to the strain
where the stress is the force per unit area and the strain is the relative change in length,
i.e.
stress
/
Y
L L
=
Δ
Even though the rod does not actually extend because it is between to fixed points, we can then think of
the stress as preventing the expansion, which still depends on /L LΔ . The change in length of the rod is
S ,L L TαΔ = Δ where the linear expansion coefficient of steel is 6
S 13 10 / C.α −
= ⋅ °
SIMPLIFY: S
Sstress
L TL
Y Y Y T
L L
α
α
ΔΔ
= ⋅ = ⋅ = Δ
CALCULATE: 11 2 6 8
stress (2.0 10 N/m )(13 10 / C)(40. C 0 C) 1.04 10 Pa−
= ⋅ ⋅ ° ° − ° = ⋅
ROUND: Two significant figures: 8
stress 1.0 10 Pa= ⋅
DOUBLE-CHECK: Even though the rod only wants to increase by 0.52 mmLΔ = (which is small), the
rod is made of steel which is very strong material, so a large stress is reasonable.
17.71. THINK: The bugle is an open pipe of length 183.0 cm.L = The speed of sound in air is dependent on
temperature, as is the length of the bugle, so an increase in temperature from i 20.0 CT = ° to
f 41.0 CT = ° will cause both to change.
SKETCH:
RESEARCH: The fundamental frequency of an open pipe is 1 /2 ,f v L= where v is the speed of sound.
The speed of sound in air as a function of temperature is ( ) (331 0.6 ) m/s,v T T= + with T in units of C.°
The length of the tube increases by B ,L L TαΔ = Δ with a linear expansion coefficient for brass of
6
B 19 10 / C.α −
= ⋅ °
SIMPLIFY:
(a) If only the change in air temperature is considered, f
1
( )
.
2
v T
f
L
=
(b) If only the change in length of the bugle is considered, i
1
B
( )
.
2 (1 )
v T
f
L Tα
=
+ Δ
(c) If both effects are taken into account, f
1
B
( )
.
2 (1 )
v T
f
L Tα
=
+ Δ

731
CALCULATE:
(a)
( )
1
331 (0.6)(41.0) m/s
97.158 Hz
2(1.83 m)
f
+
= =
(b)
( )
( )1 6
331 (0.6)(20.0) m/s
93.678 Hz
2(1.83 m) 1 (19 10 / C)(41.0 C 20.0 C)
f −
+
= =
+ ⋅ ° ° − °
(c)
( )
( )1 6
331 (0.6)(41.0) m/s
97.120 Hz
2(1.83 m) 1 (19 10 / C)(41.0 C 20.0 C)
f −
+
= =
+ ⋅ ° ° − °
ROUND: Three significant figures:
(a) 1 97.2 Hzf =
(b) 1 93.7 Hzf =
(c) 1 97.1 Hzf =
DOUBLE-CHECK: If neither effect is considered, the fundamental frequency is actually 93.7 Hz . This is
essentially the same as the answer for (b) where only the change in length was considered. Since the
change in length is relatively small, it is reasonable that the change in sound speed would have a larger
effect as is demonstrated in the (a) and (c) answers.

Fisica para ingenieria y ciencias Bauer Vol I - solucioario capitulo 17

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Fisica para ingenieria y ciencias Bauer Vol I - solucioario capitulo 17