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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)(68°F – 32) = 20°C . ( b ) T (°F) = (9/5) T (°C) + 32 = (9/5)(1800°C) + 32 = 3272°F ˜ 3300°F . 4. ( a ) T (°F) = (9/5) T (°C) + 32 = (9/5)(– 15°C) + 32 = 5°F . ( b ) T (°C) = (5/9)[ T (°F) – 32] = (5/9)(– 15°F – 32) = – 26°C . 5. T (°F) = (9/5) T (°C) + 32 = (9/5)(40.0°C) + 32.0 = 104.0°F . 6. Because the temperature and length are linearly related, we have ? T /? L = (100.0°C – 0.0°C)/(22.85 cm – 11.82 cm) = 9.067 C°/cm. ( a ) ( T 1 – 0.0°C)/(16.70 cm – 11.82 cm) = 9.067 C°/cm, which gives T 1 = 44.2°C . ( b ) ( T 2 – 0.0°C)/(20.50 cm – 11.82 cm) = 9.067 C°/cm, which gives T 2 = 78.7°C . 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 = – 40°F = – 40°C . 8. At any temperature below 20°C the expansion cracks will increase. Thus the expansion from 20°C to 50°C must eliminate the cracks. Any higher temperature will cause stress in the concrete. If the cracks have a width ? L , we have ? L = α L ? T = [12 × 10 –6 (C°) –1 ](12 m)(50°C – 20°C) = 4.3 × 10 –3 m = 0.43 cm . 9. For the expansion ? L , we have ? L Invar = α 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 = α steel L ? T = [12 × 10 –6 (C°) –1 ](2.0 m)(5.0 C°) = 1.2 × 10 –4 m . ? L marble = α 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 = α L ? T = [12 × 10 –6 (C°) –1 ](300 m)(25°C – 2°C) = 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 = α L ? T ; 1.869 cm – 1.871 cm = [12 × 10 –6 (C°) –1 ](1.871 cm)( T – 20°C), which gives T = – 69°C . 12. For the expanded dimensions, we have ¬ ′ = ¬ (1 + α ? T ); w ′ = w (1 + α ? T ). Thus the change in area is ? A = A ′ – A = ¬ ′ w ′ – ¬ w = ¬ w (1 + α ? T ) 2 – ¬ w = ¬ w [2 α ? T + ( α ? T ) 2 ] = ¬ w α ? T (2 + α ? T ). Because α ? T « 2, we have ? A = 2 α ¬ 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 β glass ? T – V β water ? T = V ( β glass – β water )? T = (350 mL)[27 × 10 –6 (C°) –1 – 210 × 10 –6 (C°) –1 ](20°C – 100°C) = 5.1 mL ....
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 Spring '09
 Fielding
 Physics, Mass, Coefficient of thermal expansion

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