516_PartUniversity Physics Solution

# 516_PartUniversity Physics Solution - 17-2 17.7. Chapter 17...

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17-2 Chapter 17 17.7. IDENTIFY: Convert T in &C to &F. SET UP: 9 FC 5 (3 2 ) TT =+ ° EXECUTE: (a) 9 F 5 (40.2 ) 32 104.4 F. T = °° ° Yes, you should be concerned. (b) 99 55 ( 32) (12C ) 32 54F . =+= + = °°° ° EVALUATE: In doing the temperature conversion we account for both the size of the degrees and the different zero points on the two temperature scales. 17.8. IDENTIFY: Set CF = and FK . = SET UP: 9 5 32 C ° and 5 KC F 9 273.15 ( 32 ) 273.15 T = − + ° EXECUTE: (a) TTT == gives 9 5 32 ° and 40 ; T =− ° 40 C 40 F. −= (b) gives 5 9 ( 32 ) 273.15 =− + ° and ( ) ( ) 95 49 (32 ) 273.15 575 ; T + = 575 F 575 K. = ° EVALUATE: Since 273.15 there is no temperature at which Celsius and Kelvin thermometers agree. 17.9. IDENTIFY: Convert to the Celsius scale and then to the Kelvin scale. SET UP: Combining Eq.(17.2) and Eq.(17.3), ( ) 5 KF 9 32 273.15, ° + EXECUTE: Substitution of the given Fahrenheit temperatures gives (a) 216.5 K (b) 325.9 K (c) 205.4 K EVALUATE: All temperatures on the Kelvin scale are positive. 17.10. IDENTIFY: Convert K T to C T and then convert C T to F . T SET UP: 273.15 and 9 5 32 °. EXECUTE: (a) C 400 273.15 127 C, T = ° F (9/5)(126.85) 32 260 F T = += ° (b) C 95 273.15 178 C, T = −° F (9/5)( 178.15) 32 289 F T + = ° (c) 77 C 1.55 10 273.15 1.55 10 C, T ° F (9/5)(1.55 10 ) 32 2.79 10 F T = ×+ ° EVALUATE: All temperatures on the Kelvin scale are positive. C T is negative if the temperature is below the freezing point of water. 17.11. IDENTIFY: Convert F T to C T and then convert C T to K . T SET UP: 5 9 2 ) . ° 273.15. EXECUTE: (a) 5 C 9 ( 346 32 ) 210 C T = ° (b) K 210 273.15 63 K T + = ° EVALUATE: The temperature is negative on the Celsius and Fahrenheit scales but all temperatures are positive on the Kelvin scale. 17.12. IDENTIFY: Apply Eq.(17.5) and solve for p . SET UP: triple 325 mm of mercury p = EXECUTE: ( ) 373.15 K (325.0 mm of mercury) 444 mm of mercury 273.16 K p EVALUATE: mm of mercury is a unit of pressure. Since Eq.(17.5) involves a ratio of pressures, it is not necessary to convert the pressure to units of Pa. 17.13. IDENTIFY: When the volume is constant, 22 11 , Tp = for T in kelvins. SET UP: triple 273.16 K. T = Figure 17.7 in the textbook gives that the temperature at which 2 CO solidifies is 2 CO 195 K. T = EXECUTE: 2 21 1 195 K (1.35 atm) 0.964 atm 273.16 K T pp T ⎛⎞ = ⎜⎟ ⎝⎠ EVALUATE: The pressure decreases when T decreases. 17.14. IDENTIFY: 1 K 1 C = ° and 9 5 F , = so 9 5 R = °. SET UP: On the Kelvin scale, the triple point is 273.16 K. EXECUTE: triple (9/5)273.16 K 491.69 R. T ° EVALUATE: One could also look at Figure 17.7 in the textbook and note that the Fahrenheit scale extends from 460 F to 32 F −° +° and conclude that the triple point is about 492 R. °

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Temperature and Heat 17-3 17.15. IDENTIFY and SET UP: Fit the data to a straight line for ( ) p T and use this equation to find T when 0. p = EXECUTE: (a) If the pressure varies linearly with temperature, then 21 2 1 () . p pT T γ =+ − 44 6.50 10 Pa 4.80 10 Pa 170.0 Pa/C 100 C 0.01 C pp TT −× × == = ° −° ° Apply 11 p T with 1 0.01 C T and 0 p = to solve for T . 0( ) p 4 1 1 0.01 C 282 C. 170 Pa/C p × =− = °− = − ° ° (b) Let 1 100 C T and 2 0.01 C; T use Eq.(17.4) to calculate 2 . p Eq.(17.4) says // , TT pp = where T is in kelvins.
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## This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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516_PartUniversity Physics Solution - 17-2 17.7. Chapter 17...

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