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Unformatted text preview: Chapter 17 CHAPTER 17 Temperature, Thermal Expansion, and the Ideal Gas Law 1. The number of atoms in a mass m is given by N = m / Mm atomic . Because the masses of the two rings are the same, for the ratio we have N Au / N Ag = M Ag / M Au = 108/197 = 0.548 . 2. The number of atoms in a mass m is given by N = m / Mm atomic = (3.4 10 3 kg)/(63.5 u)(1.66 10 27 kg/u) = 3.2 10 22 atoms . 3. ( a ) T (C) = (5/9)[ T (F) 32] = (5/9)(68F 32) = 20C . ( b ) T (F) = (9/5) T (C) + 32 = (9/5)(1800C) + 32 = 3272F 3300F . 4. ( a ) T (F) = (9/5) T (C) + 32 = (9/5)( 15C) + 32 = 5F . ( b ) T (C) = (5/9)[ T (F) 32] = (5/9)( 15F 32) = 26C . 5. T (F) = (9/5) T (C) + 32 = (9/5)(40.0C) + 32.0 = 104.0F . 6. Because the temperature and length are linearly related, we have ? T /? L = (100.0C 0.0C)/(22.85 cm 11.82 cm) = 9.067 C/cm. ( a ) ( T 1 0.0C)/(16.70 cm 11.82 cm) = 9.067 C/cm, which gives T 1 = 44.2C . ( b ) ( T 2 0.0C)/(20.50 cm 11.82 cm) = 9.067 C/cm, which gives T 2 = 78.7C . 7. We set T (F) = T (C) = T in the conversion between the temperature scales: T (F) = (9/5) T (C) + 32 T = (9/5) T + 32, which gives T = 40F = 40C . 8. At any temperature below 20C the expansion cracks will increase. Thus the expansion from 20C to 50C must eliminate the cracks. Any higher temperature will cause stress in the concrete. If the cracks have a width ? L , we have ? L = L ? T = [12 10 6 (C) 1 ](12 m)(50C 20C) = 4.3 10 3 m = 0.43 cm . 9. For the expansion ? L , we have ? L Invar = Invar L ? T = [0.2 10 6 (C) 1 ](2.0 m)(5.0 C) = 2.0 10 6 m . For the other materials we have ? L steel = steel L ? T = [12 10 6 (C) 1 ](2.0 m)(5.0 C) = 1.2 10 4 m . ? L marble = marble L ? T = [2.5 10 6 (C) 1 ](2.0 m)(5.0 C) = 2.5 10 5 m . 10. We find the height change from ? L = L ? T = [12 10 6 (C) 1 ](300 m)(25C 2C) = 8.3 10 2 m = 8.3 cm ....
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This homework help was uploaded on 03/26/2008 for the course PHYS 208 taught by Professor Ross during the Spring '08 term at Texas A&M.
 Spring '08
 ROSS
 Mass

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