Unformatted text preview: nts of thermal expansion for lead and quartz have been taken
from Table 12.1.
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18. REASONING Each section of concrete expands as the temperature increases by an amount
∆T. The amount of the expansion ∆L is proportional to the initial length of the section, as
indicated by Equation 12.2. Thus, to find the total expansion of the three sections, we can
apply this expression to the total length of concrete, which is L0 = 3(2.4 m). Since the two
gaps in the drawing are identical, each must have a minimum width that is one half the total
expansion. SOLUTION Using Equation 12.2 and taking the value for the coefficient of thermal
expansion for concrete from Table 12.1, we find
−1
∆L = α L0 ∆T = 12 × 10−6 ( C° ) 3 ( 2.4 m ) ( 32 C° ) 636 TEMPERATURE AND HEAT The minimum necessary gap width is one half this value or
1
2 12 × 10−6 ( C° )−1 3 ( 2.4 m ) ( 32 C° ) = 1.4 × 10−3 m 19. REASONING AND SOLUTION ∆L = αL0∆T gives for the expansion of the aluminum
∆LA = αALA∆T (1) and for the expansion of the brass
∆LB = αBLB∆T (2) The air gap will be closed when ∆LA + ∆LB = 1.3 × 10 −3 m . Thus, taking the coefficients of
thermal expansion for aluminum and brass from Table 12.1, adding Equations (1) and (2),
and solving for ∆T, we find that
∆T = ∆LA + ∆LB α A LA + α B LB = 1.3 × 10−3 m
= 21 C° 23 × 10−6 ( C° ) −1 (1.0 m ) + 19 × 10−6 ( C° )−1 ( 2.0 m ) The desired temperature is then
T = 28 °C + 21 C° = 49 °C
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20. REASONING The rod contracts upon cooling, and we can use Equation 12.2 to express the
change ∆T in temperature as
∆L
∆T =
(1)
α L0 where ∆L and L0 are the change in length and original length, respectively, and α is the
coefficient of linear expansion for brass. Originally, the rod was stretched by the 860N
block that hangs from the lower end of the rod. This change in length can be found from
Equation 10.17 as
F L0
(2)
∆L =
YA where F is the magnitude of the stretching force, Y is Young’s modulus for brass, and A is
the crosssectional area of the rod. Substituting Equation (2) into Equation (1) and noting
that L0 is algebraically eliminated from the final result, we have Chapter 12 Problems 637 F L0
∆T = F
YA
=
α L0 α Y A (Note: The length L0 that appears in Equation (1) is slightly larger than the length L0 that
appears in Equation (2). This difference is extremely small, however, so we assume that
they are the same. Thus, the two lengths appearing in the equation above can be
algebraically eliminated, as shown above.)
SOLUTION The stretching force F is the weight of the block (860 N). Young’s modulus Y
for brass can be obtained from Table 10.1, and the coefficient of linear expansion α can be
found in Table 12.1. Thus, the change in temperature of the brass rod is
F
860 N
=
= 39 C°
−1 −6 (
α Y A 19 ×10
C° ) ( 9.0 ×1010 N/m 2 )(1.3 ×10−5 m 2 ) ______________________________________________________________________________
∆T = 21. SSM WWW REASONING AND SOLUTION Recall that ω = 2π / T (Equation 10.6),
where ω is the angular frequency of the pendulum and T is the period. Using this fact and
Equation 10.16, we know that the period of the pendulum before the temperature rise is
given by T1 = 2π L 0 / g , where L0 is the length of the pendulum. After the temperature
has risen, the period becomes (using Equation 12.2), T2 = 2π [L 0 + α L0 ∆T ]/ g . Dividing these expressions, solving for T2 , and taking the coefficient of thermal expansion
of brass from Table 12.1, we find that T2 = T1 1 + α ∆T = (2.0000 s) 1 + (19 × 10 –6 /C ° ) (140 C °) = 2.0027 s
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22. REASONING The length of either heated strip is L0 + ∆L, where L0 is the initial length and
∆L is the amount by which it expands. The expansion ∆L can be expressed in terms of the
coefficient of linear expansion α, the initial length L0, and the change in temperature ∆T,
according to Equation 12.2. To find the change in temperature, we will set the length of the
heated steel strip equal to the length of the heated aluminum strip and solve the resulting
equation for ∆T.
SOLUTION According to Equation 12.2, the expansion is ∆L = α L0∆T.
equation we have Using this 638 TEMPERATURE AND HEAT L0, Steel + αSteel L0, Steel ∆T = L0, Aluminum + α Aluminum L0, Aluminum ∆T
144
244
3
1444 24444
4
3
∆LSteel
∆LAluminum
L0, Steel (1 + αSteel ∆T ) = L0, Aluminum (1 + α Aluminum ∆T )
We know that the steel strip is 0.10 % longer than the aluminum strip, so that
L0, Steel = (1.0010 ) L0, Aluminum . Substituting this result into the equation above, solving for
∆T, and taking values for the coefficients of linear expansion for aluminum and steel from
Table 12.1give (1.0010 ) L0, Aluminum (1 + αSteel ∆T ) = L0,...
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 Spring '13
 CHASTAIN
 Physics, Pythagorean Theorem, Force, The Lottery, Right triangle, triangle

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