Physics Solution Manual for 1100 and 2101

10 longer than the aluminum strip so that l0 steel

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Unformatted text preview: T for the change in temperature and substituting the result into T = T0 + ∆T gives T = T0 + ∆T = T0 + ∆L 0.000 16 m = 17 °C + = 25 °C α L0 23 × 10−6 ( C° )−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, ∆ L = αL0 ∆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 ∆L 0.53 m = = 1500 m α ∆ T 12 × 10 –6 (C ° ) –1 (32 ° C − 2 ° 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 ∆L = αL0 ∆T = (12 × 10 –6 –1 C° )(370 m)(21 °C − 2.0 °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 ∆r = α r0∆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 ∆r/r0 = α ∆T = [17 × 10–6 (C°)–1](110 °C − 11 °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 °C, the diameter of the disk is ∆L = 3.9 × 10−5 m greater than the diameter of the pipe. We will use ∆L = α L0 ∆T (Equation 12.2) to determine the required change in temperature ∆T for this to occur. The lower temperature T0 will then be the higher temperature minus the difference ∆T between the higher and lower temperatures: T0 = T − ∆T (1) SOLUTION Solving Equation 12.2 for the change ∆T in temperature, we obtain ∆T = ∆L α L0 (2) 634 TEMPERATURE AND HEAT Substituting Equation (2) into Equation (1) yields the final temperature of the aluminum disk: ∆L 3.9 ×10−5 m T0 = T − = 85 o C − = 59 o C o −1 −6 −5 α L0 23 ×10 (C ) 0.065 m + 3.9 ×10 m ( ) We have taken the value for α (the coefficient of thermal expansion) for aluminum from Table 12.1 in the text. 15. SSM REASONING AND SOLUTION The change in the coin’s diameter is ∆d = α d0∆T, according to Equation 12.2. Solving for α gives ∆d 2.3 × 10 −5 m = = 1.7 × 10 –5 (C °) –1 (12.2) d 0 ∆ T (1.8 × 10 −2 m)(75 C ° ) ______________________________________________________________________________ α= ∆L ∆t (Equation 2.1), where ∆L is the amount by which it contracts, and ∆t is the elapsed time. The amount of contraction the pole undergoes is found from ∆L = α L0 ∆T (Equation 12.2), 16. REASONING The average speed v of the flagpole’s contraction is given by v = where α 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 ∆T is the difference between the higher and lower temperatures of the pole. SOLUTION Substituting ∆L = α L0 ∆T (Equation 12.2) into v = v= ∆L (Equation 2.1) yields ∆t α L0 ∆T (1) ∆t The elapsed time ∆t is given in minutes, which must be converted to SI units (seconds) before employing Equation (1): 60 s ∆t = 27.0 min = 1620 s 1 min ( ) Substituting this result and the given values into Equation (1), we obtain ( ) 23 × 10 −6 (Co ) −1 (19 m ) 12.0 o C − −20.0 o C = 8.6 × 10 −6 m/s v= 1620 s Chapter 12 Problems 635 17. REASONING According to ∆L = α L0 ∆T (Equation 12.2), the factors that determine the amount ∆L by which the length of a rod changes are the coefficient of thermal expansion α, its initial length L0, and the change ∆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) ∆LL = α L L0, L ∆T (1) Similarly, the change in length of the quartz rod is ∆LQ = α Q L0, Q ∆T (2) In Equations (1) and (2) the temperature change ∆T is the same for both rods. Since each changes length by the same amount, ∆LL = ∆LQ. Equating Equations (1) and (2) and solving for L0, Q yields α L0, Q = L α Q 29 × 10−6 (C°) −1 0.10 m ) = 5.8 m L0, L = −1 ( −6 0.50 × 10 (C°) 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.

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