Solucionario de fisica serway i capitulo 19

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Chapter 19 Solutions
*19.1 (a) To convert from Fahrenheit to Celsius, we use TC = 5 5 (T – 32.0) = (98.6 – 32.0) = 37.0°C 9 F 9

and the Kelvin temperature is found as T = TC + 273 = 310 K (b) In a fashion identical to that used in (a), we find TC = –20.6°C 19.2 and T = 253 K

P1V = nRT1 and P2V = nRT2 imply that P2 T2 = P1 T1 P1T2 T1 T1P3 P1 = (0.980 atm)(273 + 45.0)K = 1.06 atm (273 +20.0)K (293 K)(0.500 atm) = 149 K = –124°C (0.980 atm)

(a)

P2 =

(b) 19.3

T3 =

=

Since we have a linear graph, the pressure is related to the temperature as P = A + BT, where A and B are constants. To find A and B, we use the data 0.900 atm = A + (–80.0°C)B 1.635 atm = A + (78.0°C)B (1) (2)

Solving (1) and (2) simultaneously, we find A = 1.272 atm and B = 4.652 × 10–3 atm/°CTherefore, P = 1.272 atm + (4.652 × 10–3 atm/°C)T (a) At absolute zero P = 0 = 1.272 atm + (4.652 × 10–3 atm/°C)T which gives T = –274°C

© 2000 by Harcourt College Publishers. All rights reserved.

2

Chapter 19 Solutions

(b)

At the freezing point of water P = 1.272 atm + 0 = 1.27 atm

(c)

and at the boiling point P = 1.272 atm + (4.652 × 10–3 atm/°C)(100°C) = 1.74 atm

19.4

Letus use TC =

5 (T – 32.0) with TF = – 40.0°C. We find 9 F

TC =

5 (– 40.0 – 32.0) = – 40.0°C 9 9 9 TC + 32.0°F = (–195.81) + 32.0 = –320°F 5 5

19.5

(a) (b)

TF =

T = TC + 273.15 = –195.81 + 273.15 = 77.3 K 0.00°C = a(–15.0°S) + b 100°C = a(60.0°S) + b

19.6

Require

Subtracting, 100°C = a(75.0°S) a = 1.33 C°/S° Then 0.00°C = 1.33(–15.0°S)C° + b b = 20.0°C So the conversionis TC = (1.33 C°/S°)TS + 20.0°C ∆T = 450 C° = 450 C°  ∆T = 450 C° = 450 K T = 1064 + 273 = 1337 K melting point T = 2660 + 273 = 2933 K boiling point (b) ∆T = 1596 C° = 1596 K The differences are the same. 212°F – 32.0°F  = 810 F° 100°C – 0.00°C

19.7

(a)

(b) 19.8 (a)

© 2000 by Harcourt College Publishers. All rights reserved.

Chapter 19 Solutions
19.9 The wire is 35.0 m longwhen TC = –20.0°C – ∆L = Li α (T – Ti) – α ≈ α(20.0°C) = 1.70 × 10–5 (C°)–1 for Cu. ∆L = (35.0 m)(1.70 × 10–5 (C°)–1(35.0°C – (–20.0°C)) = +3.27 cm

3

Goal Solution G: Based on everyday observations of telephone wires, we might expect the wire to expand by less than a meter since the change in length of these wires is generally not noticeable. O: The change in length can be found from thelinear expansion of copper wire (we will assume that the insulation around the copper wire can stretch more easily than the wire itself). From Table 19.2, the coefficient of linear expansion for copper is 17 × 10–6 (°C)–1. A: The change in length between cold and hot conditions is ∆L = αL0∆T = [17 × 10–6 (°C)–1](35.0 m)(35.0°C – (–20.0°C)) ∆L = 3.27 × 10–2 m or ∆L = 3.27 cm

L: This expansion iswell under our expected limit of a meter. From ∆L, we can find that the wire sags 0.757 m at its midpoint on the hot summer day, which also seems reasonable based on everyday observations.

19.10 19.11

∆L = Liα ∆T = (25.0 m)(12.0 × 10–6/C°)(40.0 C°) = 1.20 cm (a) ∆L = αLi ∆T = 24.0 × 10-6(C°)-1(3.0000 m)(80.0°C) = 0.00576 m Lf = 3.0058 m (b) ∆L = 24.0 × 10-6(C°)-1(3.0000 m)(–20.0°C) = – 0.0014Lf = 2.9986 m

19.12

(a)

LAl(1 + αAl ∆T) = LBrass (1 + αBrass ∆T) ∆T = ∆T = LAl – LBrass LBrass α Brass – LAl α Al (10.01 – 10.00) (10.00)(19.0 × 10 –6) – (10.01)(24.0 × 10 –6)

∆T = –199 C° so T = –179°C This is attainable.

© 2000 by Harcourt College Publishers. All rights reserved.

4

Chapter 19 Solutions

(b)

∆T =

(10.02 – 10.00) (10.00)(19.0 × 10 –6) – (10.02)(24.0 ×10 –6)

∆T = –396 C° so T = –376°C which is below 0 K so it cannot be reached 19.13 For the dimensions to increase, ∆L = αLi ∆T 1.00 × 10-2 cm = (1.30 × 10–4/°C)(2.20 cm)(T – 20.0°C) T = 55.0°C 19.14

α = 1.10 × 10–5 deg–1 for steel
∆L = (518 m)(1.10 × 10–5 deg–1)[35.0°C – (–20.0°C)] = 0.313 m

*19.15

(a)

∆A = 2α A i(∆T) ∆A = 2(17.0 × 10–6/°C)(0.0800 m)2(50.0°C) ∆A = 1.09 × 10-5 m2...
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