Physics Solution Manual for 1100 and 2101

3 as v v0t where is the coefficient of volume

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Unformatted text preview: Aluminum (1 + α Aluminum ∆T ) 0.0010 + (1.0010 ) αSteel ∆T = α Aluminum ∆T ∆T = 0.0010 0.0010 = = 91 C° −1 α Aluminum − (1.0010 ) αSteel 23 × 10−6 ( C° ) − (1.0010 ) 12 ×10−6 ( C° )−1 23. REASONING AND SOLUTION The initial diameter of the sphere, ds, is ds = (5.0 × 10–4)dr + dr (1) where dr is the initial diameter of the ring. Applying ∆L = αL0∆T to the diameter of the sphere gives ∆ds = αsds∆T (2) and to the ring gives ∆dr = αrdr∆T If the sphere is just to fit inside the ring, we must have ds + ∆ds = dr + ∆dr Using Equations (2) and (3) in this expression and solving for ∆T give ∆T = d r − ds α s ds − α r d r (3) Chapter 12 Problems 639 Substituting Equation (1) into this result and taking values for the coefficients of thermal expansion of steel and lead from Table 10.1 yield ∆T = ( −5.0 × 10−4 ) 29 × 10−6 ( C° ) −1 5.0 × 10 −4 + 1 − 12 × 10 −6 ( C° )−1 = − 29 C° The final temperature is Tf = 70.0 °C − 29 C° = 41 °C ______________________________________________________________________________ 24. REASONING When the ball and the plate are both heated to a higher common temperature, the ball passes through the hole. Since the ball’s diameter is greater than the hole’s diameter to start with, this must mean that the hole expands more than the ball for the same temperature change. The hole expands as if it were filled with the material that surrounds it. We conclude, therefore, that the coefficient of linear expansion for the plate is greater than that for the ball. In each arrangement, the ball’s diameter exceeds the hole’s diameter by the same amount, the diameters of the holes are the same, the diameters of the balls are virtually the same, and the initial temperatures are the same. The only significant difference between the various arrangements, then, is in the coefficients of linear expansion. The ball and the hole are both expanding. However, the hole is expanding more than the ball, to the extent that the coefficient of linear expansion of the material of the plate exceeds that of the ball. Thus, if we examine the difference between the two coefficients, we can anticipate the order in which the balls fall through the holes as the temperature increases. Referring to Table 12.1, we find the following differences for each of the arrangements: Arrangement A αLead − αGold = 29 × 10−6 (Cº)−1 − 14 × 10−6 (Cº)−1 = 15 × 10−6 (Cº)−1 Arrangement B αAluminum − αSteel = 23 × 10−6 (Cº)−1 − 12 × 10−6 (Cº)−1 = 11 × 10−6 (Cº)−1 Arrangement C αSilver − αQuartz = 19 × 10−6 (Cº)−1 − 0.50 × 10−6 (Cº)−1 = 18.5 × 10−6 (Cº)−1 In Arrangement C the coefficient of linear expansion of the plate-material exceeds that of the ball-material by the greatest amount. Therefore, the quartz ball will fall through the hole first. Next in sequence is Arrangement A, so the gold ball will fall second. Last is Arrangement B, so the steel ball will be the last to fall. SOLUTION According to Equation 12.2, the diameter increases by an amount ∆D = αD0∆T when the temperature increases by an amount ∆T, where D0 is the initial diameter and α is the coefficient of linear expansion. Thus, we can write the final diameter as D = D0 + αD0∆T. Since the diameters of the ball and the hole are the same when the ball falls through the hole, we have 640 TEMPERATURE AND HEAT D0, Ball + α Ball D0, Ball ∆T = D0, Hole + α Hole D0, Hole ∆T 1444 24444 14444 4 3 24444 3 Final diameter of ball Final diameter of hole Solving for the change in temperature, we obtain ∆T = D0, Ball − D0, Hole α Hole D0, Hole − α Ball D0, Ball Arrangement A D0, Gold − D0, Lead ∆T = α Lead D0, Lead − α Gold D0, Gold = 1.0 ×105 m = 6.7 C° 29 ×10−6 ( C° )−1 ( 0.10 m ) − 14 × 10−6 ( C° )−1 0.10 m + 1.0 ×10−5 m ( ) Arrangement Β D0, Steel − D0, Aluminum ∆T = α Aluminum D0, Aluminum − αSteel D0, Steel = 1.0 × 10−5 m = 9.1 C° 23 ×10−6 ( C° )−1 ( 0.10 m ) − 12 × 10−6 ( C° ) −1 0.10 m + 1.0 ×10−5 m ( ) Arrangement C D0, Quartz − D0, Silver ∆T = αSilver D0, Silver − α Quartz D0, Quartz = 1.0 ×10−5 m = 5.4 C° 19 × 10−6 ( C° ) −1 ( 0.10 m ) − 0.50 ×10−6 ( C° )−1 0.10 m + 1.0 ×10−5 m ( ) Since each arrangement has an initial temperature of 25.0 ºC, the temperatures at which the balls fall through the holes are as follows: Arrangement A T = 25.0 ºC + 6.7 Cº = 31.7 ºC Arrangement B T = 25.0 ºC + 9.1 Cº = 34.1 ºC Arrangement C T = 25.0 ºC + 5.4 Cº = 30.4 ºC The order in which the balls fall through the hole as the temperature increases is C, A, B, as we anticipated. Chapter 12 Problems 25. 641 SSM REASONING AND SOLUTION Let L0 = 0.50 m and L be the true length of the line at 40.0 °C. The ruler has expanded an amount ∆Lr = L − L0 = αrL0∆Tr (1) The copper plate must shrink by an amount ∆Lp = L0 − L = αpL∆Tp...
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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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