Physics 107 TUTORIAL ASSIGNMENT #6
Cutnell & Johnson, 7th edition
Chapter 12: Problem 17, 21, 37
Chapter 13: Problems 12, 31, 32, 34
Chapter 16: Problems 13, 27, 97
Chapter 12
*17 The brass bar and the aluminum bar in the drawing are each attached to an immovable wall.
At 28 °C the air gap between the rods is 1.3 x 10-3 m. At what temperature will the gap be
closed?
*21 Consult Conceptual Example 5 for background pertinent to this problem. A lead sphere has
a diameter that is 0.050% larger than the inner diameter of a steel ring when each has a
temperature of 70.0 °C. Thus, the ring will not slip over the sphere. At what common
temperature will the ring just slip over the sphere?
**37 ssm Two identical thermometers made of Pyrex glass contain, respectively, identical
volumes of mercury and methyl alcohol. If the expansion of the glass is taken into account, how
many times greater is the distance between the degree marks on the methyl alcohol thermometer
than that on the mercury thermometer?
Chapter 13
*12 Multiple-Concept Example 3 discusses an approach to problems such as this. The ends of a
thin bar are maintained at different temperatures. The temperature of the cooler end is 11 °C,
while the temperature at a point 0.13 m from the cooler end is 23 °C and the temperature of the
warmer end is 48 °C. Assuming that heat flows only along the length of the bar (the sides are
insulated), find the length of the bar.
31 ssm A car parked in the sun absorbs energy at a rate of 560 watts per square meter of surface
area. The car reaches a temperature at which it radiates energy at this same rate. Treating the car
as a perfect radiator (e = 1), find the temperature.
*32 A solid sphere has a temperature of 773 K. The sphere is melted down and recast into a cube
that has the same emissivity and emits the same radiant power as the sphere. What is the cube’s
temperature?
*34 Three building materials, plasterboard [
], brick
[
], and wood [
], are sandwiched together as the
drawing illustrates. The temperatures at the inside and outside surfaces are 27 °C and 0 °C,
respectively. Each material has the same thickness and cross-sectional area. Find the temperature
(a) at the plasterboard–brick interface and (b) at the brick–wood interface.
Chapter 16
13 ssm The middle C string on a piano is under a tension of 944 N. The period and wavelength
of a wave on this string are 3.82 ms and 1.26 m, respectively. Find the linear density of the string.
*27 ssm A transverse wave is traveling on a string. The displacement y of a particle from its
equilibrium position is given by y = (0.021m)sin(25t – 2.0x). Note that the phase angle
(25t - 2.0x) is in radians, t is in seconds, and x is in meters. The linear density of the string is
1.6 x 10-2 kg/m. What is the tension in the string?
*97 ssm The drawing shows a frictionless incline and pulley. The two blocks are connected by a
wire (mass per unit length = 0.0250 kg/m) and remain stationary. A transverse wave on the wire
has a speed of 75.0 m/s. Neglecting the weight of the wire relative to the tension in the wire, find
the masses m1 and m2 of the blocks.
SOLUTIONS
CHAPTER 12
17. REASONING AND SOLUTION L = L0T gives for the expansion of the aluminum
LA = ALAT
(1)
and for the expansion of the brass
LB = BLBT
(2)
Taking the coefficients of thermal expansion for aluminum and brass from Table 10.1,
adding Equations (1) and (2), and solving for T give
T 
LA  LB
1.3  103 m

 21 C
 A LA   B LB  23  106  C 1  1.0 m   19  106  C 1   2.0 m 




The desired temperature is then
T = 28 °C + 21 C° = 49 C
21. REASONING AND SOLUTION The initial diameter of the sphere, ds, is
ds = (5.0  10–4)dr + dr
(1)
where dr is the initial diameter of the ring. Applying L = L0T to the diameter of the
sphere gives
ds = sdsT
(2)
and to the ring gives
dr = rdrT
If the sphere is just to fit inside the ring, we must have
ds + ds = dr + dr
Using Equations (2) and (3) in this expression and solving for T give
T 
d r  ds
 s ds   r d r
(3)
Substituting Equation (1) into this result and taking values for the coefficients of thermal
expansion of steel and lead from Table 10.1 yield
T 

5.0  104

 29  106  C 1  5.0 104  1  12  106  C 1


  29 C
The final temperature is
Tf = 70.0 °C  29 C° = 41 C
37.
REASONING The cavity that contains the liquid in either Pyrex
thermometer expands according to Equation 12.3, Vg  gV0 T . On the other hand, the
volume of mercury expands by an amount Vm  mV0 T , while the volume of alcohol
expands by an amount Va  aV0 T . Therefore, the net change in volume for the mercury
thermometer is
Vm  Vg  (m  g )V0 T
SSM
WWW
while the net change in volume for the alcohol thermometer is
Va  Vg  ( a  g )V0 T
In each case, this volume change is related to a movement of the liquid into a cylindrical
region of the thermometer with volume r2 h , where r is the radius of the region and h is the
height of the region. For the mercury thermometer, therefore,
hm 
( m  g )V0 T
r2
Similarly, for the alcohol thermometer
ha 
(a  g )V0T
r2
These two expressions can be combined to give the ratio of the heights, ha /hm .
SOLUTION Taking the values for the coefficients of volumetric expansion for methyl
alcohol, Pyrex glass, and mercury from Table 12.1, we divide the two expressions for the
heights of the liquids in the thermometers and find that
ha
hm

a  g
 m  g

1200  10 –6 (C) 1  9.9  10 –6 (C) 1
 6.9
182  10 –6 (C) 1  9.9  10 –6 (C) 1
Therefore, the degree marks are 6.9 times further apart on the alcohol thermometer than
on the mercury thermometer.
CHAPTER 13
12. REASONING
Heat Q flows along the length L of the bar via conduction, so that
 k AT  t , where k is the thermal conductivity of the material
Equation 13.1 applies: Q 
L
from which the bar is made, A is the cross-sectional area of the bar, T is the difference in
temperature between the ends of the bar, and t is the time during which the heat flows. We
will apply this expression twice in determining the length of the bar.
SOLUTION Solving Equation 13.1 for the length L of the bar gives
L
 k AT  t  k A  TW  TC  t
Q
Q
(1)
where TW and TC, respectively are the temperatures at the warmer and cooler ends of the bar.
In this result, we do not know the terms k, A, t, or Q. However, we can evaluate the heat Q
by recognizing that it flows through the entire length of the bar. This means that we can also
apply Equation 13.1 to the 0.13 m of the bar at its cooler end and thereby obtain an
expression for Q:
k A  T  TC  t
Q
D
where the length of the bar through which the heat flows is D = 0.13 m and the temperature
at the 0.13-m point is T = 23 C, so that T  T  TC . Substituting this result into
Equation (1) and noting that the terms k, A, and t can be eliminated algebraically, we find
L
k A  TW  TC  t
Q

k A  TW  TC  t
k A  T  TC  t

k A  TW  TC  t D
k A  T  TC  t
D

 TW  TC  D   48 C  11 C  0.13 m   0.40 m
23 C  11 C
 T  TC 
31. SSM REASONING AND SOLUTION The power radiated per square meter by the car
when it has reached a temperature T is given by the Stefan-Boltzmann law, Equation 13.2,
Pradiated / A  e T 4 , where Pradiated  Q / t . Solving for T we have
1/4


560 W/m2



  320 K
–8
2
4


(1.00)
5.67

10
J/(s

m

K
)





______________________________________________________________________________
1/ 4
/ A) 
 (P
T   radiated

e


32. REASONING AND SOLUTION According to Equation 13.2, for the sphere we have
Q/t = eAsTs4, and for the cube Q/t = eAcTc4. Equating and solving we get
Tc4 = (As/Ac)Ts4
Now
As/Ac = (4 R2)/(6L2)
1/ 3
 3 
The volume of the sphere and the cube are the same, (4/3)  R = L , so R  

 4 
3
4 R 2 4  3 
The ratio of the areas is




Ac
6  4 
6 L2
then
As
3
L.
2/3
 0.806 . The temperature of the cube is,
1/4
A 
1/ 4
Tc   s  Ts   0.806   773 K   732 K
A 
 c
34. REASONING AND SOLUTION The rate of heat transfer is the same for all three materials
so
Q/t = kpATp/L = kbATb/L = kwATw/L
Let Ti be the inside temperature, T1 be the temperature at the plasterboard-brick interface, T2
be the temperature at the brick-wood interface, and To be the outside temperature. Then
kpTi  kpT1 = kbT1  kbT2
(1)
kbT1  kbT2 = kwT2  kwTo
(2)
and
Solving (1) for T2 gives
T2 = (kp + kb)T1/kb  (kp/kb)Ti
a. Substituting this into (2) and solving for T1 yields
T1 
 kp /kb  1  kw /kb  Ti   kw /kb T0 
1  kw /kb  1  kp /kb   1
21 C
b. Using this value in (1) yields
T2  18 C
CHAPTER 16
13.
SSM WWW REASONING According to Equation 16.2, the linear density of the string
is given by (m / L)  F / v 2 , where the speed v of waves on the middle C string is given by
Equation 16.1, v  f   / T .
SOLUTION Combining Equations 16.2 and 16.1 and using the given data, we obtain
m/ L 
27.
F FT 2 (944 N)(3.82 10 –3 s) 2
 2 
 8.68 10 –3 kg/m
2
2
v

(1.26 m)
SSM REASONING According to Equation 16.2, the tension F in the string is given by
F  v 2 (m / L) . Since v   f from Equation 16.1, the expression for F can be written
m 
F  (  f ) 2  
 L 
(1)
where the quantity m / L is the linear density of the string. In order to use Equation (1), we
must first obtain values for f and  ; these quantities can be found by analyzing the
expression for the displacement of a string particle.
SOLUTION The displacement is given by y  (0.021 m)sin(25t  2.0x) . Inspection of this

2  x 
, gives
equation and comparison with Equation 16.3, y  A sin 2  f t –

 
2 f  25 rad/s
and
2

 2.0 m –1
or
or
f 
25
Hz
2
=
Substituting these values f and  into Equation (1) gives
2
m
2.0
2

m   2
25 Hz 
 (1.6  10 –2 kg/m) = 2.5 N
F  (  f ) 2    
m 
 L  2.0
 2  
97.
SSM WWW REASONING Using the procedures developed in Chapter 4 for using
Newton's second law to analyze the motion of bodies and neglecting the weight of the wire
relative to the tension in the wire lead to the following equations of motion for the two
blocks:
 Fx  F  m1g (sin 30.0) = 0
(1)
 Fy  F  m2 g = 0
(2)
where F is the tension in the wire. In Equation (1) we have taken the direction of the +x axis
for block 1 to be parallel to and up the incline. In Equation (2) we have taken the direction
of the +y axis to be upward for block 2. This set of equations consists of two equations in
three unknowns, m1, m2, and F. Thus, a third equation is needed in order to solve for any of
the unknowns. A useful third equation can be obtained by solving Equation 16.2 for F:
F   m / L  v2
(3)
Combining Equation (3) with Equations (1) and (2) leads to
(m / L)v 2  m1g sin 30.0 = 0
(4)
(m / L)v 2  m2 g = 0
(5)
Equations (4) and (5) can be solved directly for the masses m1 and m2.
SOLUTION Substituting values into Equation (4), we obtain
m1 
( m / L )v 2
(0.0250 kg/m)(75.0 m/s) 2

= 28.7 kg
g sin 30.0
(9.80 m/s 2 ) sin 30.0
Similarly, substituting values into Equation (5), we obtain
(m / L)v 2 (0.0250 kg/m)(75.0 m/s) 2

= 14.3 kg
g
(9.80 m/s 2 )
______________________________________________________________________________
m2 