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Unformatted text preview: Chapter 17 Page 1 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 . 11. We can treat the change in diameter as a simple change in length, so we have ? L = L ? T ; 1.869 cm 1.871 cm = [12 10 6 (C) 1 ](1.871 cm)( T 20C), which gives T = 69C . 12. For the expanded dimensions, we have = (1 + ? T ); w = w (1 + ? T ). Thus the change in area is ? A = A A = w w = w (1 + ? T ) 2 w = w [2 ? T + ( ? T ) 2 ] = w ? T (2 + ? T ). Because ? T 2, we have ? A = 2 w ? T . Chapter 17 Page 2 13. The contraction of the glass causes the enclosed volume to decrease as if it were glass. The volume of water that can be added is ? V = ? V glass ? V water = V glass ? T V water ? T = V ( glass water )? T = (350 mL)[27 10 6 (C) 1 210 10 6 (C) 1 ](20C 100C) = 5.1 mL ....
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics, Mass

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