Physics Solution Manual for 1100 and 2101

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Unformatted text preview: (2) Eliminating L from Equations (1) and (2), solving for ∆Tp, and using values of the coefficients of thermal expansion for copper and steel from Table 12.1, we find that ∆Tp = = −α r ∆Tr α p (1 + α r ∆Tr ) 17 ×10−6 −1 − 12 × 10−6 ( C° ) ( 40.0 °C − 20.0 °C ) = −14 °C −1 −1 C° ) 1 + 12 ×10−6 ( C° ) ( 40.0 °C − 20.0 °C ) ( { } Therefore, Tp = 40.0 °C − 14 C° = 26 °C . ______________________________________________________________________________ 26. REASONING AND SOLUTION a. As the wheel heats up, it will expand. Its radius, and therefore, its moment of inertia, will increase. Since no net external torque acts on the wheel, conservation of angular momentum applies where, according to Equation 9.10, the angular momentum is given by: L = Iω , where I is the moment of inertia and ω is the angular velocity. When the moment of inertia increases at the higher temperature, the angular speed must decrease in order for the angular momentum to remain the same. Thus, the angular speed of the wheel decreases as the wheel heats up . b. According to the principle of conservation of angular momentum, I 0ω0 = I f ωf { { Initial angular momentum, L0 Final angular momentum, Lf Solving for ω f , we have, treating the bicycle wheel as a thin-walled hollow hoop ( I = MR2 , see Table 9.1) 642 TEMPERATURE AND HEAT I ωf = ω0 0 = ω0 If 2 MR0 = ω0 MRf2 R 0 Rf 2 According to Equation 12.2, ∆ R = α R0 ∆T , and the final radius of the wheel at the higher temperature is, Rf = R0 + ∆R = R0 + α R0 ∆T = R0 (1 + α ∆T ) Therefore, taking the coefficient of thermal expansion α for steel from Table 12.1, we find that the angular speed of the wheel at the higher temperature is R0 ωf = ω0 R0 (1 + α ∆T ) 2 1 = ω0 1 + α∆ T 2 2 1 = (18.00 rad/s) = 17.83 rad/s –6 –1 1 + [12 × 10 (C°) ][300.0 °C – (–100.0 °C)] ______________________________________________________________________________ 27. REASONING AND SOLUTION Let L0 be the length of the wire before heating. After heating the wire will have stretched an amount ∆Lw = αwL0∆T, while the gap in the concrete "stretches" an amount ∆Lc = αc L0∆T. The net change in length of the wire is then ∆Lc − ∆Lw = (αc − αw) L0∆T. The additional stress needed to produce this change is Stress = Y ( ∆Lc − ∆Lw ) L0 = Y (α c − α w ) ∆T Taking values for the coefficients of thermal expansion of concrete and aluminum from Table 12.1 and the value for Young’s modulus Y of aluminum from Table 10.1, we find that the additional tension in the wire is ( ) ( T = ( Stress ) π R 2 = Y (α c − α w )∆T π R 2 ( ) ) ( −1 −1 = 6.9 × 1010 N/m 2 12 × 10−6 ( C° ) − 23 × 10−6 ( C° ) (185 °C − 35 °C ) π 3.0 × 10−4 m ) = − 32 N The new tension in the wire is, then, 50.0 N – 32 N = 18 N . ______________________________________________________________________________ 2 Chapter 12 Problems 643 28. REASONING We can identify the liquid by computing the coefficient of volume expansion β and then comparing the result with the values of β given in Table 12.1. The relation ∆V = β V0 ∆T (Equation 12.3) can be used to calculate β . SOLUTION Solving Equation 12.3 for β, we have that β = ∆V / (V0 ∆T ) . The change in the volume of the liquid is ∆V = 1.500 L –1.383 L = 0.117 L . Therefore, the coefficient of volume expansion for the unknown liquid is β= ∆V 0.117 L = = 9.50 ×10 –4 (C°) –1 = 950 ×10 –6 (C°) –1 V0 ∆T (1.500 L)(97.1 °C –15.0 °C) A comparison with the values of β in Table 12.1 indicates that the liquid is gasoline . ____________________________________________________________________________________________ 29. REASONING The change ∆V in the interior volume of the shell is given by Equation 12.3 as ∆V = βV0∆T, where β is the coefficient of volume expansion, V0 is the initial volume, and ∆T is the increase in temperature. The interior volume behaves as if it were filled with the surrounding silver. The interior volume is spherical, and the volume of a sphere is 4 π r 3 , 3 where r is the radius of the sphere. SOLUTION In applying Equation 12.3, we note that the initial spherical volume of the interior space is V0 = 4 π r 3 , so that we have 3 ∆V = βV0 ∆T = β ( 4 π r 3 ) ∆T 3 ( −1 = 57 × 10−6 ( C° ) 4 π 2.0 ×10−2 m 3 ) 3 (147 °C − 18 °C ) = 2.5 ×10−7 m3 We have taken the coefficient of volume expansion for silver from Table 12.1. 30. REASONING The carbon tetrachloride contracts from its volume V0 = 2.54 × 10−4 m3 at the higher temperature T0 = 75.0 °C to a volume V at the lower temperature T = −13.0 °C . The smaller volume V is the larger volume V0 minus the difference ∆V in volume: V = V0 − ∆V (1) 644 TEMPERATURE AND HEAT We will find the difference in volume from ∆V = β V0 ∆T (Equation 12.3), where β is the coefficient of volume expansion for carbon tetrachloride and ∆T is the difference between the higher and l...
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