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Unformatted text preview: 4. (a) Let the reading on the Celsius scale be x and the reading on the Fahrenheit scale be y . Then 9 5 32 y x = + . If we require y = 2 x , then we have 9 2 32 (5)(32) 160 C 5 x x x = + ⇒ = = ° which yields y = 2 x = 320°F. (b) In this case, we require 1 2 y x = and find 1 9 (10)(32) 32 24.6 C 2 5 13 x x x = + ⇒ =  ≈  ° which yields y = x /2 = –12.3°F. 7. We assume scale X is a linear scale in the sense that if its reading is x then it is related to a reading y on the Kelvin scale by a linear relationship y = mx + b . We determine the constants m and b by solving the simultaneous equations: 373.15 ( 53.5) 273.15 ( 170) m b m b = + = + which yield the solutions m = 100/(170 – 53.5) = 0.858 and b = 419. With these values, we find x for y = 340: 340 419 92.1 . 0.858 y b x X m = = =  ° 8. (a) The coefficient of linear expansion α for the alloy is 5 10.015cm 10.000cm / 1.88 10 /C . (10.01cm)(100 C 20.000 C) L L T = ∆ ∆ = = × ° ° ° α Thus, from 100°C to 0°C we have 5 2 (10.015cm)(1.88 10 / C )(0 C 100 C) = 1.88 10 cm. L L T ∆ = ∆ = × ° ° ° × α The length at 0°C is therefore L ′ = L + ∆ L = (10.015 cm – 0.0188 cm) = 9.996 cm. (b) Let the temperature be T x . Then from 20°C to T x we have 5 10.009cm 10.000cm = (1.88 10 /C )(10.000cm) , L L T T ∆ = ∆ = × ° ∆ α giving ∆ T = 48 °C. Thus, T x = (20°C + 48 °C )= 68°C. 14. The change in length for the section of the steel ruler between its 20.05 cm mark and 20.11 cm mark is 6 (20.11cm)(11 10 / C )(270 C 20 C) = 0.055cm....
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 Spring '07
 Giammanico
 Heat, Eint

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