Download Solucionario Capitulo 16 - Paul E. Tippens PDF

TitleSolucionario Capitulo 16 - Paul E. Tippens
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Page 1

Chapter 16. Temperature and Expansion Physics, 6th Edition

Chapter 16. Temperature and Expansion

16-1. Body temperature is normal at 98.60F. What is the corresponding temperature on the Celsius

scale?

0 0 05 5
9 9( 32 ) (98.6 32 )C Ft t= − = − ; tC = 370C

16-2. The boiling point of sulfur is 444.50C. What is the corresponding temperature on the

Fahrenheit scale?

0 0 09 9
5 532 (444.5 ) 32F Ct t= + = + ; tF = 8320F

16-3. A steel rail cools from 70 to 300C in 1 h. What is the change of temperature in Fahrenheit

degrees for the same time period?

∆t = 700C – 300C = 40 C0;
0

0
0

9 F
40

5 C
t C

 
∆ =  

 
; ∆ t = 72 F0

*16-4. At what temperature will the Celsius and Fahrenheit scales have the same numerical

reading?

0 05 9
9 5( 32 ) 32x x− = + ; x = -400C or –400F

16-5. A piece of charcoal initially at 1800F experiences a decrease in temperature of 120 F0.

Express this change of temperature in Celsius degrees. What is the final temperature on the

Celsius scale?

∆t = 120 F0;
0

0 0
0

5 C
120 F 66.7 C

9 F

 
= 

 
; ∆t = 66.7 C0

The final temperature is 1800F – 120 F0 = 600F which must be converted to 0C:

0 0 05 5
9 9( 32 ) (60 32 )C Ft t= − = − ; tC = 15.6 0C

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Page 5

Chapter 16. Temperature and Expansion Physics, 6th Edition

16-19. What is the increase in volume of 16.0 liters of ethyl alcohol when the temperature is

increased by 30 C0?

∆V = βVo∆t = (11 x 10-4/C0)(16 L)(500C – 200C); ∆V = 0.528 L

16-20. A Pyrex beaker has an inside volume of 600 mL at 200C. At what temperature will the

inside volume be 603 mL? [ ∆V = 603 mL – 300 mL = 3 mL ]

0
0 5 0

0

3 mL
; 556 C

(0.3 x 10 /C )(600 mL)

V
V V t t

V
β

β −


∆ = ∆ ∆ = = =

t = 200 C + 5560C; t = 5760C

16-21. If 200 cm3 of benzene exactly fills an aluminum cup at 400C, and the system cools to 180C,

how much benzene (at 180C) can be added to the cup without overflowing?

0 0(3 ) ;B AL BV V V V t V tβ α∆ = ∆ − ∆ = ∆ − ∆ ∆t = (18 – 40) = -22 C0

∆V = (12.4 x 10-4 /C0)(200 cm3)(-22 C0) – 3(2.4 x 10-5/C0)(200 cm3)(-22 C)

∆V = -5.456 cm3 + 0.3168 cm3 = -5.14 cm3; VB = 5.14 cm3

16-22. A Pyrex glass beaker is filled to the top with 200 cm3 of mercury at 200C. How much

mercury will overflow if the temperature of the system is increased to 680C?

Vo = 200 cm3; βm = 1.8 x 10-4/C0; αp = 0.3 x 10-5/C0

0 0(3 ) ;m p m pV V V V t V tβ α∆ = ∆ − ∆ = ∆ − ∆ ∆t = 680C – 200C = 48 C0;

∆V = (1.8 x 10-4 /C0)(200 cm3)(48 C0) – 3(0.3 x 10-5/C0)(200 cm3)(48 C0)

∆V = 1.728 cm3 – 0.0864 cm3 = 5.14 cm3; VB = 1.64 cm3

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Page 6

Chapter 16. Temperature and Expansion Physics, 6th Edition

Challenge Problems

*16-23. The diameter of the hole in a copper plate at 200C is 3.00 mm. To what temperature must

the copper be cooled if its diameter is to be 2.99 mm. [∆L = (2.99 – 3.00) = -0.01 mm ]

0
0 -5 0

0.010 m
; ; -196 C

(1.7 x 10 /C )(3.00 m)o

L
L L t t t

L
α

α
∆ −

∆ = ∆ ∆ = = ∆ =

t = to + ∆t = 200C - 196 C0; t = -1760C

16-24. A rectangular sheet of aluminum measures 6 by 9 cm at 280C. What is its area at 00C?

Ao = (6 cm)(9 cm) = 54 cm2; ∆t = 00 – 280C = -280C ; γ = 2α

∆A = 2αAo∆t = 2(2.4 x 10-5/C0)(54 cm2)(-28 C0) = -0.0726 cm2

A = 54 cm2 – 0.0726 cm2; A = 53.9 cm2

*16-25. A steel tape measures the length of an aluminum rod as 60 cm when both are at 80C. What

will the tape read for the length of the rod if both are at 380?

The aluminum rod will expand more than does the steel tape. Thus the tape will

give a smaller reading based on the difference in the change of length.

∆LAL = αAlL0∆t = (2.4 x 10-5/C0)(60 cm)(30 C0); ∆LAl = 0.0432 cm

∆Ls = αsL0∆t = (1.2 x 10-5/C0)(60 cm)(30 C0); ∆LAl = 0.0216 cm

The reading will be less by the difference in the expansions.

Reading = 60 cm + (0.0432 cm - 0.0216 cm); Reading = 60.02 cm

16-26. At 200C, a copper cube measures 40 cm on a side. What is the volume of the cube when

the temperature reaches 1500C? [ V0 = (40 cm)3 = 64,000 cm3 ; ∆t = 150 – 20 = 130 C0 ]

V = V0 + 3αV0∆t = 64,000 cm3 + 3(1.7 x 10-5/C0)(64,000 cm3)(130 C0)

V = 64,000 cm3 + 424 cm3; V = 64,420 cm3

228

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