Physics 41 Chapter 14 & 19 HW

Physics 41 Chapter 14 & 19 HW
1.
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.)
P14.5
The Earth’s surface area is 4π R2 . The force pushing inward over this area amounts to
(
)
(
)
=
F P=
P0 4π R2 .
0A
This force is the weight of the air:
=
Fg mg
= P0 4π R2
so the mass of the air is
(
) (1.013 × 10
P0 4π R2
=
=
m
g
5
)
(
)
2
N m 2  4π 6.37 × 106 m 


=
2
9.80 m s
5.27 × 1018 kg .
2.
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?
P14.12 The pressure on the bottom due to the water is
So,
On each end,
On the side,
=
Pb ρ=
gz 1.96 × 104 Pa
=
Fb P=
bA
5.88 × 106 N
(
)
F=
PA =
9.80 × 10 Pa ( 60.0 m ) =
588 kN
F=
PA =
9.80 × 103 Pa 20.0 m 2 =
196 kN
3
2
3.
Mercury is poured into a U-tube as in Figure P14.18a. The left arm of the tube has cross-sectional area A1 of
10.0 cm2, and the right arm has a cross-sectional area
A2 of 5.00 cm2. One hundred grams of water are then poured into the right arm as in Figure P14.18b. (a) Determine
the length of the water column in the right arm of the U-tube. (b) Given that the density of mercury is 13.6 g/cm3,
what distance h does the mercury rise in the left arm?
P14.18
(a)
Using the definition of density, we have
=
hw
(b)
100 g
mw ater
=
=
2
A2ρw ater 5.00 cm 1.00 g cm 3
(
)
20.0 cm
Sketch (b) at the right represents the situation after
the water is added. A volume ( A2 h2 ) of mercury
has been displaced by water in the right tube. The
additional volume of mercury now in the left tube
is A1h . Since the total volume of mercury has not
changed,
FIG. P14.18
A1
(1)
h
A2
At the level of the mercury–water interface in the right tube, we may write the absolute pressure
as:
P
= P0 + ρw ater ghw
The pressure at this same level in the left tube is given by
P =P0 + ρH g g ( h + h2 ) =P0 + ρw ater ghw
or
A2 h2 = A1h
h2 =
which, using equation (1) above, reduces to

A 
ρH g h 1 + 1  =
ρw ater hw
 A2 
ρw ater hw
or h =
.
A
ρH g 1 + A1
(
2
)
1.00 g cm ) ( 20.0 cm )
(=
(13.6 g cm ) (1 + )
3
Thus, the level of mercury has
risen a distance of h
=
3
10.0
5.00
0.490 cm
above the original level.
Chapter 19
4. 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?
P19.1 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
Therefore,
(a)
At absolute zero
−3
B = 4.652 × 10
atm °C
(
P = 1.272 atm + 4.652 × 10
(
)
−3
P = 0 = 1.272 atm + 4.652 × 10−3 atm °C T
)
atm °C T
which gives
T = −274°C .
(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 .
(
)
5. 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?
P19.9
The wire is 35.0 m long when TC = −20.0°C .
∆L = L iα (T − Ti )
α = α (20.0°C) = 1.70 × 10 −5 (C°) for Cu.
−1
(
)(35.0°C − (−20.0°C)) =
∆L = (35.0 m) 1.70× 10−5 (C°)
−1
+3.27 cm
6.
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?
P19.16
(a)
(
)
∆A = 2 17.0 × 10 −6 °C −1 (0.080 0 m) (50.0°C)
∆A = 2αA i ∆T :
−5
∆A = 1.09× 10
(b)
2
m2 = 0.109 cm
2
The length of each side of the hole has increased. Thus, this represents an increase
in the area of
the hole.
7. A hollow aluminum cylinder 20.0 cm deep has an internal capacity of 2.000 L at 20.0°C. It is completely filled with
turpentine, and then slowly warmed to 80.0°C. (a) How much turpentine overflows? (b) If the cylinder is then cooled
back to 20.0°C, how far below the cylinder's rim does the turpentine’s surface recede?
P19.21 (a)
∆V = Vt β t ∆T − V Al β Al ∆T = (β t − 3α Al )V i ∆T
(
)
(
)
= 9.00× 10−4 − 0.720× 10−4 °C −1 2 000 cm3 (60.0°C)
∆V = 99.4 cm
(b)
3
overflows.
The whole new volume of turpentine is
(
)
2 000 cm 3 + 9.00 × 10 −4 °C −1 2 000 cm 3 (60.0°C) = 2 108 cm 3
so the fraction lost is
99.4 cm3
= 4.71× 10−2
2 108 cm 3
and this fraction of the cylinder’s depth will be empty upon cooling:
4.71× 10−2(20.0 cm ) = 0.943 cm .
8.
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?
4 3
π r N P′ :
3
N=
3PV
=
3(150)(0.100)
= 884 balloons
3
4π (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
0.100 m3 3
1.20 atm when the last balloon is inflated. The number of balloons is then reduced to to 884 −
3 = 877 .
4π (0.15 m)
P19.28 PV = NP ′V ′ =
4π r P′
3
(
)
9. 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?
P19.31 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 


10. At 25.0 m below the surface of the sea ( ρ = 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?
P19.38
At depth,
P = P0 + ρgh
At the surface,
P0V f = nRT f :
Therefore
 Tf
V f = Vi 
 Ti
and
PVi = nRTi
P0V f
(P0 + ρ gh)Vi
=
Tf
Ti
  P0 + ρ gh 


P0 

(
)(
)
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 


 278 K 
1.013× 105 Pa

V f = 3.67 cm3
11.
Review problem. 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?
P19.49 The frequency played by the cold-walled flute is fi =
v
λi
=
v
.
2Li
When the instrument warms up
ff =
v
λf
=
fi
v
v
.
=
=
2L f 2Li (1 + α∆T) 1+ α ∆T
The final frequency is lower. The change in frequency is

1 
∆f = fi − f f = fi  1−

α∆T 
1
+

v
v  α∆T 
(α∆T )

 ≈
∆f =
2Li  1 + α∆T  2Li
∆f ≈
(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.
13. Review problem. 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?
P19.56
(a)
Li
Ti = 2 π
g
so
Li =
Ti2 g
4π
2
=
(1.000 s)2 (9.80
4π
m s2
2
) = 0.248 2 m
∆L = αLi ∆T = 19.0× 10−6 °C −1(0.284 2 m )(10.0°C ) = 4.72 × 10−5 m
Tf = 2 π
0.248 3 m
Li + ∆L
= 2π
2 = 1.000 095 0 s
g
9.80 m s
∆T = 9.50 × 10
(b)
−5
s
(
In one week, the time lost is time lost = 1 week 9.50× 10 −5 s lost per second
)
 86 400 s 
−5 s lost 
time lost = (7.00 d week )

  9.50× 10
s 
 1.00 d  
time lost = 57.5 s lost
14.
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 (Fig. P19.61b)?
P19.61 After expansion, the length of one of the spans is
[
]
−6
−1
L f = Li (1 + α∆T) = 125 m 1+ 12 × 10 °C (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 .