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Unformatted text preview: †T for the change in temperature and substituting the
result into T = T0 + âˆ†T gives T = T0 + âˆ†T = T0 + âˆ†L
0.000 16 m
= 17 Â°C +
= 25 Â°C
Î± L0 23 Ã— 10âˆ’6 ( CÂ° )âˆ’1 ( 0.86 m ) _____________________________________________________________________________________________ Chapter 12 Problems 11. 633 SSM WWW REASONING AND SOLUTION The steel in the bridge expands
according to Equation 12.2, âˆ† L = Î±L0 âˆ†T . Solving for L0 and using the value for the
coefficient of thermal expansion of steel given in Table 12.1, we find that the approximate
length of the Golden Gate bridge is âˆ†L
0.53 m
=
= 1500 m
Î± âˆ† T 12 Ã— 10 â€“6 (C Â° ) â€“1 (32 Â° C âˆ’ 2 Â° C)
______________________________________________________________________________
L0 = [ 12. REASONING AND SOLUTION Using Equation 12.2 and the value for the coefficient of
thermal expansion of steel given in Table 12.1, we find that the linear expansion of the
aircraft carrier is
âˆ†L = Î±L0 âˆ†T = (12 Ã— 10 â€“6 â€“1 CÂ° )(370 m)(21 Â°C âˆ’ 2.0 Â°C) = 0.084 m (12.2)
______________________________________________________________________________
13. REASONING AND SOLUTION
a. The radius of the hole will be larger when the plate is heated, because the hole expands as if it were made of copper.
b. According to Equation 12.2, the expansion of the radius is âˆ†r = Î± r0âˆ†T. Using the value
for the coefficient of thermal expansion of copper given in Table 12.1, we find that the
fractional change in the radius is
âˆ†r/r0 = Î± âˆ†T = [17 Ã— 10â€“6 (CÂ°)â€“1](110 Â°C âˆ’ 11 Â°C) = 0.0017 ______________________________________________________________________________
14. REASONING The disk will fall into the pipe when it reaches a temperature T0 at which its
diameter is equal to that of the inner diameter of the pipe. At the higher temperature of
T = 85 Â°C, the diameter of the disk is âˆ†L = 3.9 Ã— 10âˆ’5 m greater than the diameter of the
pipe. We will use âˆ†L = Î± L0 âˆ†T (Equation 12.2) to determine the required change in
temperature âˆ†T for this to occur. The lower temperature T0 will then be the higher
temperature minus the difference âˆ†T between the higher and lower temperatures: T0 = T âˆ’ âˆ†T (1) SOLUTION Solving Equation 12.2 for the change âˆ†T in temperature, we obtain âˆ†T = âˆ†L
Î± L0 (2) 634 TEMPERATURE AND HEAT Substituting Equation (2) into Equation (1) yields the final temperature of the aluminum
disk:
âˆ†L
3.9 Ã—10âˆ’5 m
T0 = T âˆ’
= 85 o C âˆ’
= 59 o C
o âˆ’1 âˆ’6
âˆ’5
Î± L0 23 Ã—10 (C ) 0.065 m + 3.9 Ã—10 m ( ) We have taken the value for Î± (the coefficient of thermal expansion) for aluminum from
Table 12.1 in the text. 15. SSM REASONING AND SOLUTION The change in the coinâ€™s diameter is âˆ†d = Î± d0âˆ†T, according to Equation 12.2. Solving for Î± gives âˆ†d
2.3 Ã— 10 âˆ’5 m
=
= 1.7 Ã— 10 â€“5 (C Â°) â€“1
(12.2)
d 0 âˆ† T (1.8 Ã— 10 âˆ’2 m)(75 C Â° )
______________________________________________________________________________ Î±= âˆ†L
âˆ†t
(Equation 2.1), where âˆ†L is the amount by which it contracts, and âˆ†t is the elapsed time.
The amount of contraction the pole undergoes is found from âˆ†L = Î± L0 âˆ†T (Equation 12.2), 16. REASONING The average speed v of the flagpoleâ€™s contraction is given by v = where Î± is the coefficient of thermal expansion for aluminum (see Table 12.1 in the text), L0
is the length of the pole before it begins contracting, and âˆ†T is the difference between the
higher and lower temperatures of the pole.
SOLUTION Substituting âˆ†L = Î± L0 âˆ†T (Equation 12.2) into v =
v= âˆ†L
(Equation 2.1) yields
âˆ†t Î± L0 âˆ†T (1) âˆ†t The elapsed time âˆ†t is given in minutes, which must be converted to SI units (seconds)
before employing Equation (1): 60 s âˆ†t = 27.0 min = 1620 s 1 min ( ) Substituting this result and the given values into Equation (1), we obtain ( ) 23 Ã— 10 âˆ’6 (Co ) âˆ’1 (19 m ) 12.0 o C âˆ’ âˆ’20.0 o C = 8.6 Ã— 10 âˆ’6 m/s
v=
1620 s Chapter 12 Problems 635 17. REASONING According to âˆ†L = Î± L0 âˆ†T (Equation 12.2), the factors that determine the amount âˆ†L by which the length of a rod changes are the coefficient of thermal expansion Î±,
its initial length L0, and the change âˆ†T in temperature. The materials from which the rods
are made have different coefficients of thermal expansion (see Table 12.1 in the text). Also,
the change in length is the same for each rod when the change in temperature is the same.
Therefore, the initial lengths must be different to compensate for the fact that the expansion
coefficients are different.
SOLUTION The change in length of the lead rod is (from Equation 12.2) âˆ†LL = Î± L L0, L âˆ†T (1) Similarly, the change in length of the quartz rod is âˆ†LQ = Î± Q L0, Q âˆ†T (2) In Equations (1) and (2) the temperature change âˆ†T is the same for both rods. Since each
changes length by the same amount, âˆ†LL = âˆ†LQ. Equating Equations (1) and (2) and solving
for L0, Q yields
Î±
L0, Q = L
Î±
Q 29 Ã— 10âˆ’6 (CÂ°) âˆ’1 0.10 m ) = 5.8 m L0, L = âˆ’1 (
âˆ’6 0.50 Ã— 10 (CÂ°) The values for the coefficie...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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