# D27183 Chapter 17 Problems Solutions

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```					                                                                        CHAPTER 17
Homework Problems & Solutions
Problems 6, 10, 12, 26, 38, & 64

6. If 3.25 × 10–3 kg of gold is deposited on the negative electrode of an electrolytic cell in a period
of 2.78 h, what is the current through the cell in this period? Assume that the gold ions carry one
elementary unit of positive charge.

17.6    The mass of a single gold atom is

M       197 g m ol
m atom                         3.  10-22 g  3.  10-25 kg
27             27
N A 6.  10 at s m ol
02    23
om

The number of atoms deposited, and hence the number of ions moving to the
negative electrode, is

m          3.  103 kg
25
n                              9.  1021
93
m atom       3.  10-25 kg
27

Thus, the current in the cell is

I        
93     
Q ne 9.  10 1.  10 C
21
60   19

 0. A  159 m A
159
t t     2. h  3600 s 1 h
78

10. A lightbulb has a resistance of 240 Ω when operating at a voltage of 120 V. What is the
current through the lightbulb?

V 120 V
17.10    I              0. A  500 m A
500
R   240 
12. Suppose that you wish to fabricate a uniform wire out of 1.00 g of copper. If the wire is to
have a resistance of R = 0.500 Ω, and if all of the copper is to be used, what will be (a) the length
and (b) the diameter of this wire?

17.12   The volume of the copper is

m      1.  103 kg
00
V                                              1.  107 m
12                   3

de iy 8.  103 kg m
ns t   92                           3

Since, V  A  L , this gives A  L  1.  107 m 3 .
12                                       (1)

L
(a) From R               , we find that
A

    1.  108   m                 
A   L
 R  
70
0. 
500                        
     8
 L  3.  10 m L .
40                         

Inserting this expression for A into Equation 1 gives

 3.  10
40           8

m L2  1.  107 m 3 , which yields
12                                                      82
L 1. m

 d2        1.  107 m 3
12
(b) From equation (1), A                                      , or
4            L

d

4 1.  107 m
12                  3
  41.  10
12            7
m   3

L                      1. m 
82

 2.  104 m  0. m m
80            280
26. An aluminum rod has a resistance of 1.234 Ω at 20.0°C. Calculate the resistance of the rod at
120°C by accounting for the changes in both the resistivity and the dimensions of the rod.

For aluminum, the coefficient of linear expansion is   24  106  C 
1
17.26                                                                                             and the
temperature coefficient of resistivity is  e  3.  10
9           3
 C  . At temperature T,
1

the length and cross-sectional area may be expressed as L  L0 1  T  T0   and
               
A  A 0 1  T  T0   , respectively.
2
               

L                           L0 1   T  T0    L0 
                          1  e  T  T0  
                  .
Thus, R        0 1  e  T  T0  
                                             0A 
A                           A 0 1   T  T0  
2
   0   1   T  T0  
                 
                

At T  120C , this gives

1   e  T  T0             1 3.  10  C  120  20.  C  
1
               9     3
0     
R  R0                       1.  
234
1   T  T0  
                              1 24  106  C  1 120  20.  C 
0
                                       
1. 
71

38. A certain toaster has a heating element made of Nichrome resistance wire. When the toaster is
first connected to a 120-V source of potential difference (and the wire is at a temperature of
20.0°C) the initial current is 1.80 A. However, the current begins to decrease as the resistive
element warms up. When the toaster has reached its final operating temperature, the current has
dropped to 1.53 A. (a) Find the power the toaster converts when it is at its operating temperature.
(b) What is the final temperature of the heating element?

17.38   (a) At the operating temperature,

  V  I 120 V 1. A   184 W
53

R  R0
(b) From R  R0 1  T  T0   , the temperature T is given by T  T0 
                                                                               .
 R0
The resistances are given by Ohm’s law as

 V        120 V              V 0 120 V
R                      , and R 0         
I           53
1. A                 I0     80
1. A

Therefore, the operating temperature is

T  20. C 
0
120 1.   120 1.  
53          80
461C
0.400  10 C   120 1.80
3     1
64. A 50.0-g sample of a conducting material is all that is available. The resistivity of the material
is measured to be 11 × 10–8 Ω · m, and the density is 7.86 g/cm3. The material is to be shaped into
a solid cylindrical wire that has a total resistance of 1.5 Ω. (a) What length is required? (b) What
must be the diameter of the wire?

17.64   The volume of the material is

m ass    50. g  1 m 3 
0
V                                 6.  106 m
36                              3

de iy 7. g cm 3  106 cm 3 
ns t   86               

Since V  A  L , the cross-sectional area of the wire is A  V L .

L        L        L2
(a) From R                              , the length of the wire is given by
A        V L        V

R V        1.   6.  106 m 3 
5       36
L                                                 9. m
3
                   11 10-8   m

 d2        V
(b) The cross-sectional area of the wire is A                             . Thus, the diameter is
4       L

d
4V


4 6.  106 m
36                3
  9.  10
3    4
m  0. m m
93
L                     9. m 
3

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