# Download Physics 41 Chapter 19 &amp; 20 HW Solutions

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```Physics 41 Chapter 19 & 20 HW Solutions
1. Is it possible to convert internal energy to mechanical energy? Explain with examples.
2. The pendulum of a certain pendulum clock is made of brass. When the temperature increases, does the period of
the clock increase, decrease or remain the same? Explain.
3. You need to pick up a very hot cooking pot in your kitchen. Which will protect your hands more: dry pads or wet
pads. Explain.
4. Why are icebergs often surrounded by fog? Air becomes saturated, water vapor condense into clouds.
5. Some folks can walk on hot coals with bare feet and not burn their feet. Explain how they manage this. Magic or
physics? Wet feet. Specific heat of water and feet is high.
6. The power radiated by a distant star is 4.2 x10 27 W. The radius of the star, which may be considered a perfect radiator, is 1.06
x 1010 m. Determine the surface temperature of the star.
Solution: This is a problem of thermal radiation. The rate at which energy is being radiated is proportional to
the “fourth power of the absolute temperature. This behavior is known as Stefan’s Law,” From this behavior
we can see that the energy radiated is also proportional to the surface area of the star.
   AeT 4
T (

)
 Ae
1
4
e is the emissivity of the object; since we make the assumption that the star is a
perfect radiator, we can conclude that it also is a perfect absorber or a black body;
therefore e=1.
Another assumption is that the star is a perfect sphere, and therefore the surface area is:
A  4 r 2 | r  radius
T (
1

4.2 x1027 W
4
)

(
4
4 r 2 e
4 (1.06 x1010 m) 2 (1)(5.6696 x108 WK
T  2691.3K   2690 K 
1
)4
m2
)
7. A 905.0 g iron meteor impacts the earth at a speed of 1629.0 m/s. If its energy is entirely converted to heat of the meteorite,
what will the resultant temperature rise be? (The specific heat for iron is 113 cal/kg•°C .)
Answer:
2810°C
8
3
8. Assume that the sun is a sphere of radius 6.96 x 10 m and that its surface temperature is 5.8 x 10 K. a) If the sun is a perfect
emitter, at what rate is energy emitted from the surface of the sun? b) What is the rate per square meter at the sun's surface - that
is, the Intensity? c)What is the Intensity at which energy is received on Earth in one hour? Ignore effects of absorption due to the
11
atmosphere. The average distance from the Earth to the sun is 1.50 x 10 m.
The intensity of solar radiation reaching the top of the Earth’s atmosphere is 1 340 W/m 2
9. A granite wall has a thickness of 0.61 m and a thermal conductivity of 2.1 W/(m • C°). The temperature on one face of the
wall is 3.2 °C and 20.0 °C on the opposite face. How much heat is transferred in one hour through each square meter of the
granite wall?
Solution: Power describes the amount of energy given off over a period of time. Since the energy being given off
is only in the form of heat, we can conclude that:
Q
P
t
dT
The law of thermal conduction can be described as P  kA
where k is the thermal conductivity of the
dx
1
1min
material. The change in time is: 1hour (
)(
)  3600s
60 min 60s
dT Th  Tc
The Temperature gradient is

| l  length
dx
l
Th  Tc
Q
20.0C o  3.2C o
W
 tk (
)  (3600s)(2.1
)(
)  208209.8 J 2 
mC o
m 

A
l
0.61m
J
s( )
o
W C
sW
J
)( )  2  s2  2
Unit analysis: s(
o
mC
m
m
m
m
So we have that 208209.8 Joules transferred per meter square.
10.
What is the total mass of the Earth's atmosphere? (The radius of the Earth is 6.37  106 m, and atmospheric
pressure at the surface is 1.013  105 N/m2.)
The Earth’s surface area is 4 R 2 . The force pushing inward over this area amounts to




F  P0A  P0 4 R 2 .
This force is the weight of the air:
Fg  m g  P0 4 R 2
so the mass of the air is
m 

P0 4 R 2
g
  1.013 10
5
N m
2
 4 6.37  10 m 
6
2
9.80 m s
2
  5.27  1018 kg .
11
A swimming pool has dimensions 30.0 m  10.0 m and a flat bottom. When the pool is filled to a depth of
2.00 m with fresh water, what is the force caused by the water on the bottom? On each end? On each side?
Pb   gz  1.96  104 Pa
The pressure on the bottom due to the water is
So,
Fb  PbA  5.88  106 N
On each end,
F  PA  9.80  103 Pa 20.0 m 2  196 kN
3
F  PA  9.80  10
On the side,


Pa 60.0 m  
2
588 kN
12. A constant-volume gas thermometer is calibrated in dry ice (that is, carbon dioxide in the solid state, which has a
temperature of –80.0°C) and in boiling ethyl alcohol (78.0°C). The two pressures are 0.900 atm and 1.635 atm. (a)
What Celsius value of absolute zero does the calibration yield? What is the pressure at (b) the freezing point of
water and (c) the boiling point of water?
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)
1.635 atm  A  78.0CB
(2)
Solving (1) and (2) simultaneously,
we find A  1.272 atm
and
B  4.652  103 atm C
Therefore,
(a)

3
P  0  1.272 atm  4.652  10
At absolute zero
3

atm C T

atm C T
T  274C .
which gives


P  1.272 atm  4.652  10
(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 .


13. A copper telephone wire has essentially no sag between poles 35.0 m apart on a winter day when the
temperature is –20.0°C. How much longer is the wire on a summer day when TC = 35.0°C?
The wire is 35.0 m long when TC  20.0C .
L  LiT  Ti
   20.0C  1.70  10 5 C for Cu.
1

L  35.0 m 1.70 10
5
C135.0C  20.0C 
3.27 cm
14. A square hole 8.00 cm along each side is cut in a sheet of copper. (a) Calculate the change in the area of
this hole if the temperature of the sheet is increased by 50.0 K. (b) Does this change represent an increase
or a decrease in the area enclosed by the hole?
(a)
A  2A iT :

6
A  2 17.0  10 C
5
A  1.09 10
(b)
0.080 0 m2 50.0C
1
m  0.109 cm
2
2
The length of each side of the hole has increased. Thus, this represents an increase
in the area of
the hole.
15. A tank having a volume of 0.100 m3 contains helium gas at 150 atm. How many balloons can the tank blow up if
each filled balloon is a sphere 0.300 m in diameter at an absolute pressure of 1.20 atm?
PV  NP V  
4 3
 r N P :
3
N 
3PV

31500.100
 884 balloons
3
4 0.150 1.20
If we have no special means for squeezing the last 100 L of helium out of the tank, the tank will be full of helium at
3
0.100 m 3
1.20 atm when the last balloon is inflated. The number of balloons is then reduced to to 884 
3  877 .
4 0.15 m
4 r P
3


16. Just 9.00 g of water is placed in a 2.00-L pressure cooker and heated to 500°C. What is the pressure inside the
container?
P
nRT  9.00 g 8.314 J
773 K

 


3
3  1.61 MPa  15.9 atm
V
18.0 g mol  mol K 2.00 10 m 
17. At 25.0 m below the surface of the sea (  = 1 025 kg/m3), where the temperature is 5.00°C, a diver exhales an air
bubble having a volume of 1.00 cm3. If the surface temperature of the sea is 20.0°C, what is the volume of the bubble
just before it breaks the surface?
At depth,

P  P0  gh
and
At the surface,
P0 V f  nRT f :
Therefore
T f P   gh 
V f  V i  0

 Ti  P0 
PVi  nRTi
P0V f
P0  ghVi



Tf
Ti

5
3
2
293 K 1.013 10 Pa  1 025 kg m 9.80 m s 25.0 m 

V f  1.00 cm3 

5

278 K 
1.013
10
Pa


V f  3.67 cm3

18. An aluminum pipe, 0.655 m long at 20.0C and open at both ends, is used as a flute. The pipe is
cooled to a low temperature, but then filled with air at 20.0C as soon as you start to play it. After that, by
how much does its fundamental frequency change as the metal rises in temperature from 5.00C to
20.0C?
The frequency played by the cold-walled flute is fi 
ff 
When the instrument warms up
v
f
v

i

v
.
2Li
fi
v
v
.


2Lf 2Li 1  T 1  T
The final frequency is lower. The change in frequency is

1 
f  fi  f f  fi 1

 1  T 
f 
f 
v  T  v

 
T 
2Li 1  T  2Li
343


m s 24.0 106 C 15.0C
20.655 m
 0.094 3 Hz
This change in frequency is imperceptibly small.
19 A clock with a brass pendulum has a period of 1.000 s at 20.0°C. If the temperature increases to 30.0°C, (a) by how
much does the period change, and (b) how much time does the clock gain or lose in one week?
(a)
Ti  2 
Li
g
so
Li 
Ti2 g
4

2
1.000 s2 9.80
4
m s
2
2
  0.248 2 m
L  Li T  19.0 106 C 1 0.284 2 m 10.0C   4.72  105 m
0.248 3 m
Li  L
 2
2  1.000 095 0 s
g
9.80 m s
Tf  2 
T  9.50  10 5 s
(b)

In one week, the time lost is time lost  1 week 9.50 10
86 400 s 
5 s lost 

time lost  7.00 d week 
9.50 10
s 
 1.00 d 
,
5
s lost per second

time lost  57.5 s lost
20. Two concrete spans of a 250-m-long bridge are placed
end to end so that no room is allowed for expansion (Fig.
P19.61a). If a temperature increase of 20.0°C occurs, what is
the height y to which the spans rise when they buckle?
After expansion, the length of one of the spans is

L f  Li 1  T  125 m 1 12  10 C
6
1
20.0C 125.03 m .
L f , y, and the original 125 m length of this span form a right triangle with y as the altitude. Using the Pythagorean
theorem gives:
125.03 m2  y2  125 m2
yielding y  2.74 m .
```