Chapter 17 - 17.1 From Eq 29 29 29 29 29 F 1 134 32 7 56 5...

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Unformatted text preview: 17.1: From Eq. ( 29 ( 29 ( 29 ( 29 ( 29 F. 1 . 134 32 7 . 56 5 9 b) F. . 81 32 8 . 62 5 9 a) , 1 . 17 ° = + °- = +- ( 29 ( 29 F. . 88 32 1 . 31 5 9 c) ° = + 17.2: From Eq. ( 29 ( 29 ( 29 ( 29 ( 29 C. 7 . 41 32 107 9 5 b) C. . 5 32 . 41 9 5 a) , 2 . 17 ° =- ° =- ( 29 ( 29 C. 8 . 27 32 18 9 / 5 c) °- =-- 17.3: ° = ° = ° F . 72 . 40 so , F C 1 5 9 F 140.2 F . 70 1 2 ° = ° + = T T 17.4: a) C. 6 . 55 ) 44 . 56 ( 9) 5 ( b) C. 2 . 27 )) . 4 ( (45.0 ) 9 5 ( °- =-- ° =-- 17.5: ( 29 ( 29 F, 4 . 104 32 2 . 40 5 9 (17.1), Eq. From a) ° = + which is cause for worry. ( 29 ( 29 F 54 or F, 6 . 53 32 12 5 9 b) ° ° = + to two figures. 17.6: ( 29 ( 29 ° = F 2 . 21 8 . 11 5 9 17.7: ° = ° = F C 1 K 1 5 9 , so a temperature increase of 10 K corresponds to an increase of 18 ° F . Beaker B has the higher temperature. 17.8: For ( 29 °- = ∆ = ∆ °- = ∆ = ∆ C . 10 (a), for Then . C . 10 ), b ( 5 9 C 5 9 F K C T T T T . F . 18 °- = 17.9: Combining Eq. (17.2) and Eq. (17.3), ( 29 , 15 . 273 32 9 5 + °- = F K T T and substitution of the given Fahrenheit temperatures gives a) 216.5 K, b) 325.9 K, c) 205.4 K. 17.10: (In these calculations, extra figures were kept in the intermediate calculations to arrive at the numerical results.) + = ° =- = ) 85 . 126 )( 5 / 9 ( C, 127 15 . 273 400 a) F C T T F. 10 79 . 2 32 ) 10 55 . 1 )( 5 / 9 ( C, 10 55 . 1 15 . 273 10 55 . 1 c) F. 289 32 ) 15 . 178 )( 5 / 9 ( C, 178 15 . 273 95 b) F. 260 32 7 7 F 7 7 C F C ° × = + × = ° × =- × = °- = +- = °- =- = ° = T T T T 17.11: From Eq. K. 23 . 27 15 . 273 ) C 92 . 245 ( ), 3 . 17 ( K = + °- = T 17.12: From Eq. C. 1769 273.15 K 2042.14 K) 3.16 (7.476)(27 ), 4 . 17 ( ° =- = 17.13: From Eq. ( 29 mm. 444 ) mm . 325 ( ), 4 . 17 ( K 273.16 K 15 . 373 = 17.14: On the Kelvin scale, the triple point is 273.16 K, so R. 491.69 K 5 (9/5)273.1 R ° = = ° One could also look at Figure 17.7 and note that the Fahrenheit scale extends from F 32 to F 460 ° + °- and conclude that the triple point is about 492 R. ° 17.15: From the point-slope formula for a straight line (or linear regression, which, while perhaps not appropriate, may be convenient for some calculators), C, 33 . 282 Pa 10 80 . 4 Pa 10 6.50 Pa 10 80 . 4 ) C . 100 ( ) C 01 . ( 4 4 4 °- = ×- × × °- ° which is C 282 °- to three figures. b) Equation (17.4) was not obeyed precisely. If it were, the pressure at the triple point would be ( 29 Pa. 10 76 . 4 ) 16 . 273 ( 4 15 . 373 Pa 10 50 . 6 4 × = = × P 17.16: ( 29 ( 29 ( 29 ( 29 ( ( 29 ( 29 C, 168 m 1 . 62 C 10 4 . 2 m 10 25 1 5 2 ° = ° × × = ∆ = ∆--- αL L T so the temperature is C 183 ° . 17.17: . T L m 39 . C) 5.0) ( C m)(18.0 1410 )( ) C ( 10 2 . 1 ( 1 5 + = °-- ° ° × = ∆ α-- 17.18: ) 1 ( T α d d d ∆ + = ∆ + C))) 78.0 ( C . 23 )( ) C ( 10 (2.4 cm)(1 4500 . ( 1 5 °-- ° ° × + =-- mm. 4.511 cm 4511 . = = 17.19: a) cm, 10 1.4 C) (28.0 cm) 90 . 1 )( ) C ( 10 6 . 2 ( 3 1 5--- × = ° ° × = ∆ T αD so the diameter is 1.9014 cm. diameter is 1....
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This homework help was uploaded on 02/26/2008 for the course PHYS 11 and 21 taught by Professor Hickman during the Spring '08 term at Lehigh University .

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Chapter 17 - 17.1 From Eq 29 29 29 29 29 F 1 134 32 7 56 5...

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