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FLUIDS
What is a fluid? A substance that can flow. A fluid is a substance that cannot maintain its
own shape but takes the shape of its container. Gases and liquids are called fluids because
neither have an orderly arrangement.
In rigid bodies we expressed Newton's laws in terms of mass and force. We describe
fluids in terms of density and pressure
Pressure
force per surface area; symbol is P; SI unit is Pa (Pascal), or N/m2.
P=F/A
where F is force or weight in Newtons
A is cross-sectional area in m2
The pressure at any point in a fluid acts equally in all directions. Also, the force
due to the fluid pressure always acts perpendicularly to any surface the fluid is in
contact with.
Atmospheric pressure= 1.013 x 105 Pa (measured at sea level). This can be
approximated as 101 kPa.
Gauge Pressure Pressure gauges measure the pressure over and above
atmospheric pressure. This is called gauge pressure.
Absolute Pressure To get the absolute pressure at a point, one must add the
atmospheric pressure to the gauge pressure. For example, if gauge pressure is 100
kPa, the absolute pressure at that point is the sum of 100 kPa and 101 kPa, or 201
kPa.
Manometer A U-shaped tube partially filled with liquid used to measure
pressure. The pressure is equal to the difference in height of the two levels of the
liquid according to P = Patm + gh.
Density
the ratio of mass to volume; density is a characteristic property of a any pure
substance. Its SI unit is kg/m3
=m/V
where  is density in kg/m3
m is mass in kilograms
V is volume in m3
(H2O) = 1000 kg/m3
Sometimes densities are given in g/cm3. To convert to the SI unit of kg/m3 simply
multiply by 1000.
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Density Lab - Float or Sink: You find out!"
http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&Resou
rceID=17
Specific Gravity The ratio of the density of that substance to the density of water
at 4.0. It has no units.
substance density, kg/m3
substance
density, kg/m3
aluminum
2700
iron and steel
7800
copper
8900
lead
11300
mercury
13600
ethyl alcohol
790
Hydrostatics
the study of fluids at rest
1. Pascal’s principle - any change in pressure at any point in a fluid is
transmitted unchanged throughout the fluid. Or, pressure applied to a
confined fluid increases the pressure throughout the fluid by the same
amount.
F1/A1 = F2/A2
This is the basis for squeezing a tube of toothpaste, hydraulic brakes, and for the
Heimlich maneuver.
2.
Hydrostatic pressure - pressure due to a fluid’s depth
P=gh
where  is density of fluid in kg/m3
h is the height (depth) of fluid
Pressure increases with depth. The pressure at any depth depends only upon that
depth and not upon any horizontal dimension. For example, Hoover Dam holds
back Lake Mead which is 700 ft deep. The bottom of Hoover Dam must
withstand the same pressure if it were only holding back a few thousand gallons
of water 700 ft deep. Also, the pressure at equal depths within a fluid is the same.
3.
Archimede’s principle - an object immersed in a fluid is buoyed up by a force
equal to the weight of the displaced fluid. A fluid provides some support for any object
placed in it. The upward force on an object placed in a fluid is called the buoyant force.
FB =  g V
where  is the density of the fluid in kg/m3
V is the volume of the displace fluid
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FB is the buoyant force (The buoyant force occurs because the pressure in a fluid
increases with depth. The upward force on the bottom surface of a submerged
object is greater than the downward pressure on its top surface.)
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If an solid floats partially submerged in a liquid, the volume of liquid
displaced is less than the volume of the object. According to Archimedes
Principle, the weights of the object and its displaced fluid are the same.
The fractional part of an object that is submerged is equal to the ratio of
the density of the solid to the density of the liquid in which is floats (for
example, about 90% of an iceberg is submerged because the density of the
ice is about 90% that of sea water).
An object floats when the buoyant force is equal to its weight.
Objects submerged in a fluid appear to weigh less than they do when
outside the fluid.
Applet demonstrating buoyant force
http://www.sciencejoywagon.com/physicszone/lesson/otherpub/wfendt/buoyforce.htm
Fluid Dynamics
study of fluids in motion
Hydrodynamics
study of water in motion
Aerodynamics
study of air in motion
Lift on an airplane wing
http://www.explorelearning.com/index.cfm?method=cResource.dspDetail&Resou
rceID=24
Equation of Continuity
the volume of fluid passing two points per second is equal
A1 V1 = A2 V2
In a narrow tube, the velocity of the liquid is high; in a wide tube, the velocity of
the liquid is low.
Bernoulli’s principle
as the velocity of a fluid increases, the pressure exerted by that fluid decreases
on top of an airfoil there is low pressure due to high velocity airflow
on the bottom of an airfoil there is high pressure due to low velocity airflow
Fluids in Motion
There are two type of fluid flow, streamline (laminar) and turbulant flow. If the
slow is smooth (layers of fluid slide by each other smoothly), the flow is said to
be steamlined, or laminar. Above a certain speed (which depends upon many
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factors), a flow becomes turbulent. Turbulent flow is characterized by the
formation of eddies.
Viscosity Internal friction in a fluid.
Mass Flow Rate the ratio of the mass of a fluid that passes a certain point in a
certain interval of time (or, m/t)
Volume Rate of Flow the ratio of the volume of a fluid that passes a certain point
in a certain interval of time (or, V/t). In SI units, this is m3/sec (or the same
thing as the product of area, A, and velocity, v.)
Bernoulli's Equation Where the velocity of a fluid is high, the pressure is low;
where the velocity is low, the pressure is high. Bernoulli's equation is an
expression of the law of conservation of energy.
P1 + 1/2 v12 + gh1 = P2 + 1/2 v22 + gh2
Torricelli's theorem A liquid leaves a spigot at the bottom os a reservoir with the
same speed that a freely falling object falling through the same height.
v1 = (2g(h2-h1))1/2
AP Multiple Choice Questions Fluid questions were added to the AP B test
beginning in 2002.
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Be able to determine the buoyant force acting on an object when given the
weight in air and the weight in the fluid.
Be able to recognize that the pressure acting on the bottom of a container
is due to the weight of the fluid above it.
Recognize that high velocity fluid has low pressure. Predict the
consequences of low pressure due to high velocity fluid.
AP Free Response Questions Fluid questions were added to the AP B test
beginning in 2002.
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All the ones that have occurred since 2002 have dealt with either a fluidbased laboratory exercise or with fluids at rest.
Be able to calculate gauge pressure.
Be able to calculate absolute pressure.
Be able to calculate pressure.
Kinematics is frequently integrated into fluid problems.
AP Fluids Objectives
AP Physics B - Fluids Objectives
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Students should understand that fluids exert pressure in all directions.
Students should understand that a fluid at rest exerts pressure perpendicular to any
surface that it contacts.
Students should understand and be able to use the relationship between pressure
and depth in a fluid, P = gh.
Students should understand that the difference in pressure on the upper and lower
surfaces of an object immersed in liquid results in an upward force on the object.
Students should understand and be able to apply Archimedes Principle: the
buoyant force on a submersed object is equal to the weight of the liquid it
displaces.
Students should understand that for laminar flow, the flow rate of a liquid through
its cross section is the same at any point along its path.
Students should understand and be able to apply the equation of continuity,
1A1v1 = 2A2v2.
Students should understand that the pressure of a flowing liquid is low when the
velocity is high, and vice versa.
Students should understand and be able to apply Bernoulli's equation: P + gy +
1/2 v2 = constant.
AP Fluids Problems
AP Fluids Sample Problems-Hydrostatics
1. Estimate the pressure exerted on a floor by a 50 kg model standing momentarily
on a single spiked heel (area = 0.05 cm2) and compare it to the pressure exerted
by a 1500 kg elephant standing on one foot (area = 800 cm2). Ans: 9.8 x 107 Pa;
1.8375 x 105 Pa
2. A tire gauge reads 220 kPa. What is the absolute pressure within the tire? Ans:
321 kPa
3. What is the pressure due to a column of water 100 m high? Ans: 980,000 Pa
4. The area of the output piston in a hydraulic lift is 20 times that of the input
cylinder. What force would it take to lift a 4000 lb car? Ans: 200 lb
5. A geologist finds that a moon rock whose mass is 8.20 kg has an apparent mass of
6.18 kg when submerged in water. What is the density of the rock? Ans: 4059.41
kg/m3
6. Water and then oil are poured into a U-shaped tube, open at both ends, and do not
mix. They come to equilibrium as shown. What is the density of the oil? The
height of the oil column is 27.2 cm; the distance from the top of the water column
to the top of its tube is 9.41 cm; the top of its tube is even with the top of the oil
column. (Hint: pressures at a and b are equal.) Ans: 654.04 kg/m3
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AP Fluids Sample Problems-Hydrodynamics
1. A stream of water emerges from a pipe. The cross-sectional area of the water
stream at the tap is 1.2 cm2 and 43 mm lower than the tap is 0.35 cm2. At what
volume rate of flow does the water flow from the tap? Ans: 3.21 x 10-5 m3/s
2. Water is pumped from a pipeline 2 m above the ground to a water tower 15 m
above the ground. If the pipeline velocity is 8 m/s, its pressure is 3.103 x 105 Pa,
and water enters the tank at a pressure of one atmosphere, with what velocity does
the water enter the tank? Ans: 15.09 m/s
3. A gas is flowing through a pipe whose cross-sectional area is 0.07 m2. The gas
has a density of 1.30 kg/m3. A Venturi meter is used to measure the speed of the
gas. It has a cross-sectional area of 0.05 m2. The pressure difference between the
pipe and the Venturi meter is found to be 120 Pa. Find the speed of the gas in the
pipe and the volume rate of flow of the gas. Ans: 13.87 m/s; 0.97 m3/s
4. An aneurysm is an abnormal enlargement of a blood vessel such as the aorta.
Suppose that, because of an aneurysm, the cross-sectional area of the aorta
increases to a value 1.7 times greater than the original. The speed of the blood
(density of blood is =1060 kg/m3) through a normal portion of the aorta is 0.40
m/s. Assuming the aorta is horizontal (the person is lying down), determine the
amount by which the pressure in the enlarged region exceeds that in the normal
region. Ans: 55.53 Pa
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Thermal Energy & States of Matter
Kinetic theory of gases
particles in a hot body have more kinetic energy than those in a cold body; as
temperature increases, kinetic energy increases. If the temperature of rises, the gas
molecules move at greater speeds. If the volume remains the same, the hotter
molecules would be expected to hit the walls of the container more frequently
than the cooler ones, resulting in a rise in pressure.
An advanced look at the kinetic theory: The assumptions describing an ideal
gas make up the postulates of the kinetic theory:
1. An ideal gas is made up of a large number of gas molecules N each with
mass m moving in random directions with a variety of speeds.
2. The gas molecules are separated from each other by an average distance
that is much greater than the molecule's diameter.
3. The molecules obey laws of mechanics, interacting only when they
collide.
4. Collisions between the walls of the container or with other gas molecules
are assumed to be perfectly elastic.
Entropy disorder; the higher the temperature, the more disorder (or entropy) a
substance has
Temperature
measure of an object’s kinetic energy; temperature measures how hot or how cold
an object is with respect to a standard
Temperature Scales The most common scale is the Celsius (or Centigrade,
though in the United States the Fahrenheit scale is common. Both of these scales
use the freezing point and boiling point of water at atmospheric pressure as fixed
points. On the Celcius scale, the freezing point of water corresponds to 0C and
the boiling point of water corresponds to 100C. On the Farenheit scale, the
freezing point of water is defined to be 32F and the boiling point 212F. It is
easy to convert between these two scales by remembering that 0C = 32F and
that 5C = 9F. The Kelvin scale is based upon absolute zero
(-273.15 C), or 0 K.
Triple Point The triple point of water serves as a point of reference. It is only at
this point (273.16 K) that the three phases of water (gas, liquid, and solid) exist
together at a unique value of temperature and pressure.
Temperature is a property of a system that determines whether the system
will be in thermal equilibrium with other systems.
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Molecular Interpretation of Temperature The concept that matter is made up
of atoms in continual random motion is called the kinetic theory. We assume that
we are dealing with an ideal gas. In an ideal gas, there are a large number of
molecules moving in random directions at different speeds, the gas molecules are
far apart, the molecules interact with one another only when they collide, and
collisions between gas molecules and the wall of the container are assumed to be
perfectly elastic. The average translational kinetic energy of molecules in a gas is
directly proportional to the absolute temperature. If the average translational
kinetic energy is doubled, the absolute temperature is doubled.
KEav = 1/2 mvav2 = 3/2 kT
where T is the temperature in Kelvin and k is Boltzmann's constant
k = 1.38 x 10-23 J/K
The relationship between Boltzmann's constant (k), Avogadro's number (N), and
the gas constant (R) is given by:
k = R/N
An advanced look at the relationship between pressure and the kinetic
theory: The pressure exerted by an ideal gas on its container is due to the force
exerted on the walls of the container by the collisions of the molecules with the
walls of area A. The collisions cause a change in momentum of the gas molecules.
These assumptions can be used to derive an expression between pressure and the
average kinetic energy of the gas molecules. The pressure is directly proportional
to the square of the average velocity. Since the average kinetic energy is directly
proportional to the temperature, pressure is also directly proportional to the
temperature (for a fixed volume).
PV = 2/3 N (1/2 mvav2)
The higher the temperature, according to kinetic theory, the faster the molecules
are moving, on average.
rms speed The square root of the average speed speed in the kinetic energy
expression is called the rms speed.
vrms = (3RT/M)1/2
where R is the ideal gas constant, T is temperature in Kelvin, and M is the
molecular mass
Heat(symbol is Q; SI unit is Joule)
amount of thermal energy transferred from one object to another due to
temperature differences (we will learn in thermodynamics why heat flows from a
hot to a cold body).
Q = m c T
where m is mass in kg
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c is specific heat of the material
T = Tf - Ti in C
Heat with moles of gas Typically, moles of gas are given instead of the mass of
the gas. In that case, heat can be calculated using
Q = n c T
where n is the number of moles
c is the molar specific heat of the gas
Specific Heats of Gases Since the volume of a gas changes significantly a
change in temperature or a change in pressure, molar specific heats of gases are
also expressed in terms of constant pressure or constant volume conditions. In the
case of a constant volume process, the constant would be expressed as cv; in the
case of a constant pressure process, the constant would be expressed as cp. Use
these constants in Q = n c T.
Mechanical Equivalent of Heat
James Joule described the reversible conversion of heat energy and work. The
calorie is defined as the amount of energy needed to raise the temperature of one
gram of water at 14.5 one degree Celcius. The SI unit for work and energy is the
Joule.
1 calorie = 4.186 J
1000 calories is equal to 1 food Calorie
Specific heat capacity(c)
a characteristic of a material; the amount of energy (measured in Joules) that must
be added to raise the temperature of one kilogram of the material one degree
Celcius or one Kelvin
specific of heat of water: c = 4180 J/kg K (at a temperature of 15C and a pressure
of 1 atmosphere)
Please note, the units J/kg K are the same as J/kg C
Specific Heat Capacity of Gases at Constant Pressure (Cp) is defined as the
amount of heat required to raise the temperature of one mole of a gas through 1 K
at a constant pressure, or Q = nCpT
For monoatomic gases such as Ar, He, Ne, H, O, etc, CP = 20.78 J/molK
Specific Heat Capacity of Gases at Constant Volume (CV) is defined as the
amount of heat required to raise the temperature of one mole of a gas through 1 K
at a constant volume, or Q = nCVT
For monoatomic gases such as Ar, He, Ne, H, O, etc, CV = 12.47 J/molK
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substance
specific
specific
specific
substance
substance
heat
heat
heat
900 J
aluminum
kg-1K-1
copper
390 J
kg-1K-1
iron
450 J
kg-1K-1
140 J
mercury kg1 -1
K
silver
230 J
kg-1K-1
ice
2060 J
kg-1K-1
steam
2020 J
kg-1K-1
sodium
1230
J kg1 -1
K
zinc
388 J
kg-1K-1
lead
128 J
kg-1K-1
glass
837 J
kg-1K-1
water
4180
J kg1 -1
K
Notice that the specific heat of water is very high - higher than ice and steam.
Water has a very high specific heat, meaning that it heats slowly and cools slowly.
The specific heat of a material yields information about how the material heats
and cools. If you add ten joules of heat to two materials, the one with the lowest
specific heat will show the greatest temperature change. If you cool two materials
ten degrees, the material with the greatest specific heat loses the most energy.
Measurement of heat capacity (c) In an experiment, a substance is heated over a
period of time. If V is the voltage, i is the current, t is the change in time in
seconds, T is the difference in temperature, and n is the number of moles, the
specific heat capacity can be found by
Molar heat capacity The molar heat capacity is based upon the number of moles
of the substance. Heat can be expressed in terms of molar heat capacity by:
Q = n C T
Dulong-Petit Law The average molar heat capacities for all metals is
approximately the same and is equal to about 25 J/mole K, or approximately 3 R.
Thus the specific heat of a metal can be calculated using c = C/M, where M is the
molecular mass of the substance.
Energy transfer mechanisms:
1. conduction (solids)-KE transfer due to collisions of particles; heat transfer
occurs only when there is a difference in temperature
10
o
Thermal Conductivity It is found experimentally that the heat
flow per unit of time (Q/t)is proportional to the corss-sectional
area of the object (A), the distance (d) between the two ends of the
object, the temperatures of each end of the object (T1 and T2), and
a proportionality constant, k, called the thermal conductivity of the
substance.
Q/t = kA(T1 - T2)/d
o Substances that have large values for k are good thermal
conductors. Those with low values for k are good insulators.
2. convection (fluids)-KE transfer due to movements of fluids over large
distances caused by different densities at different temperatures
3. radiation-energy transfer through a vacuum. Conduction and convection
require the prescence of matter. Radiation consists of electromagentic
waves.
o Stefan-Boltzmann equation The rate at which an object radiates
energy is proportional to the fourth power of the Kelvin
temperature, T.
Q/t = eAT4
where A is the area,  is the Stefan-Boltzmann constant which has the value of
=5.67 x 10-8 W/m2 K4, and e is the emissivity (a number between 0 and 1)
o
Very black surfaces has emissivities close to 1 and very shiny surfaces
have emissivities close to 0. A good absorber of radiation is also a good emitter of
radiation.
When different parts of an isolated system are at different temperatures, heat will flow
from the part at a higher temperature to that at the lower temperature until they are at
thermal equilibrium
Law of heat exchange
the sum of heat losses and gains in a closed system is zero. When two bodies of
unequal temperature are mixed, the cold body absorbs heat (raising its
temperature) and the hot body loses heat (lowering its temperature) until an
equilibrium temperature is reached. Thermal equilibrium exists when two objects
that are in themal contact with one another no longer affect each other's
temperature.
Qloss + Qgain = 0
Objects are in thermal equilibrium when they are at the same temperature.
Calorimeter
device used to measure changes in thermal energy
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Changes of State
The three most common states of matter are solid, liquid, and gas. When heat is
added to a substance, one of two things can occur. The temperature can increase
or the material can change to a different state. There is a fourth state of matter plasma. A plasma is a state of matter in which atoms are stripped of their
electrons. In a plasma, atoms are separated into their electrons and bare nuclei.
When a material changes phases from solid to liquid or from liquid to gas, a
certain amount of energy is absorbed (in the reverse process, the heat is given
off). Let's look at ice (a solid) at a temperature of -5. When heat is added to ice,
its temperature increases until it reaches 0. At this point, ice begins to melt--it
changes its state from a solid to a liquid. The temperature remains constant at 0
until all the ice has melted. Now we have water at 0. As heat is added to the
water, its temperature increases until it reaches 100. At this point, the water
begins to boil, changing its state from liquid to gas. The temperature remains
constant at 100 until all the water boils, turning into steam. Now we have steam
at 100. If you continue to add heat, the temperature of the steam begins to
increase.
Latent heat of fusion, (Hf or Lf)
amount of energy needed to change 1 kg of a substance from a solid to a liquid.
for water, Hf = 333,000 J/kg (333 x 103 J/kg) or 3.33 x 105 J/kg
Latent heat of vaporization,(Hv or Lv)
amount of energy needed to change 1 kg of a substance from a liquid to a gas.
for water, Hv = 2,260,000 J/kg K (or 2.26 x 106 J/kg) or 22.6 x 105 J/kg
If energy is added to a system heating it and causing an increase in temperature,
energy is positive; if energy is removed from a system cooling it and causing a
decrease in temperature, energy is negative. If energy is added to a system
causing a change in the state of matter from a solid to a liquid or from a liquid to a
solid, that energy is positive. If energy is removed from a system causing a
change in the state of matter from a gas to a liquid or from a liquid to a solid, that
energy is negative.
At a phase change, the amount of heat given off or absorbed is found using:
Q = m H or Q = mL (where L is the latent heat)
where m is mass in kg and H is heat of transformation. No temperature change
occurs at a phase change.
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Example: How much heat is added to 10 kg of ice at -20C to convert it to steam
at 120C?
Using the above phase diagram, one sees that the ice absorbs heat and changes
temperature until it reaches 0C. The amount of heat absorbed is given by Q =
mciceT. At 0C, ice changes phase, from the solid phase (ice) to the liquid phase
(water). Heat must be added to change the phase. Using the phase diagram, one
sees that no temperature change occurs until all the ice is melted into water. The
amount of heat added is given by Q = mHf. Once all the ice is melted into water,
the temperature agains to rise as heat is added until it reaches 100C. The amount
of heat absorbed is given by Q = mcwaterT. At 100C, water changes phase, from
the liquid phase (water) to the gaseous phase (steam). The amount of heat added
to accomplish this phase change is given by Q = mHv. Using the phase diagram,
one sees that no temperature change occurs until all the water is converted into
steam. Once all the water has been converted into steam, the temperature again
begins to rise. The amount of heat added is given by Q = mcsteamT.
The amount of heat added to change ice at -20C to steam at 120C is given by:
Q = (10 kg)(2060 J/kgC)(0C + 20C) + (10 kg)(333 x 103 J/kg) + (10 kg)(4180
J/kgC)(100C - 0C) + (10 kg)(2.26 x 106 J/kg) + (10 kg)(2020 J/kgC)(120C 100C)
Experimentally determining the amount of heat added
In real life, you cannot directly measure the amount of heat added in Joules.
Typically, you graph temperature vs time. If you know the rate at which heat is
being added and how long it is added, you can determine the amount of heat
added. If you use an electric device to heat the substance, you can determine how
much electrical energy was transferred to the substance knowing that:
EE (electrical energy) = Pt = Vit
where P is power in watts (remember, 1 W=1 J/sec), t is time, V is voltage, and i
is current
You can also convert gravitational potential energy (GPE) into thermal energy.
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Remember: work and energy are equivalent!
Sublimation The process whereby a solid changes directly to a gas without
passing through the liquid phase.
Evaporation Evaportation can be explained in terms of the kinetic theory. The
fastest moving molecules in a liquid escape from the surface, decreasing the
average speed of those remaining. When the average speed is less, the absolute
temperature is less. Thus evaporation, the escaping of the fastest moving
molecules from the surface of a liquid, is a cooling process.
Boiling When the temperature of a liquid equals the point where the saturated
vapor pressure equals the external pressure, boiling occurs.
Thermal Expansion
Most substances expand when heated and contract when cooled. The exception is
water. The maximum density of water occurs at 4. This explains why a lake
freezes at the surface, and not from the bottom up. If water at 0C is heated, its
volume decreases until it reaches 4C. Above 4C, water behaves normally and
expands in volume as it is heated. Water expands as it is cooled from 4C to 0C
and expands even more as it freezes. That is why ice cubes float in water and
pipes break when the water inside of them freezes.
The change in length in almost all solids when heated is directly proportional to
the change in themperatuer and to its original length. A solid expands when
heated and contracts when cooled: The length of a material decreases as the
temperature decreases; its length increases as the temperature increases. So a rod
that is 2 m long expands twice as much as a rod which is 1 m long for the same
ten degree increase in temperature.
L =  L T
where L is the length of the material
 is the coefficient of linear expansion
T is the temperature change in  C
A gas expands when heated and contracts when cooled: The volume of a gas
decreases as the temperature decreases; its volume increases as the temperature
increases.
V =  V T
where V is the volume of the material
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 is the coefficient of volume expansion
T is the temperature change in C
material
coefficient of linear expansion coefficient of volume expansion
aluminum
25 x 10-6 /C
75 x 10-6 /C
brass
19 x 10-6 /C
56 x 10-6 /C
iron or steel
12 x 10-6 /C
35 x 10-6 /C
lead
29 x 10-6 /C
87 x 10-6 /C
concrete
12 x 10-6 /C
36 x 10-6 /C
gasoline
950 x 10-6 /C
mercury
180 x 10-6 /C
ethyl alcohol
1100 x 10-6 /C
water
210 x 10-6 /C
air
3400 x 10-6 /C
Thermal Stress In many buildings and roads, the ends of a beam or other
material are held rigidly fixed. If the temperature should change, large
compressive or tensile forces develop, called thermal stresses. Elastic modulus
can be used to calculate these thermal stresses.
L = (1/E) (F/A)L0
AP Multiple Choice Questions on Heat
1. You should be able to use Q=mcT to predict temperature changes. For
example, an object's potential energy change is directly converted into
heat. Or, frictional losses result in an increase in temperature of the
surface.
2. Be able to interpret phase diagrams. Use them to predict melting (freezing)
points, boiling points and the relationship between specific heats for the
substance for different phases.
3. You do not need to know the formulas for length or volume expansion.
You need to know that the heat transferred to a substance is directly
proportional to its cross-sectional area and indirectly proportional to its
length.
4. Be able to solve easy calorimetry problems to predict a final temperature
of a mixture or the specific heat of a metal.
5. Be able to calculate the rate at which heat is transferred. Remember, this
would be in the units of J/sec.
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6. Be able to describe what happens to the temperature and to heat when a
substance freezes (or melts) or boils (or condenses).
7. Be able to predict which would be the hotter substance (of unequal
specific heats) when equal amounts of heat are added.
8. Be able to predict the appearance of an object when it is heated.
AP Free Response Questions on Heat
1. Be able to interpret phase diagrams. Use them to predict melting (freezing)
points, boiling points, latent heats, and the specific heats for the substance
for different phases.
2. All you need to know about entropy is that it is disorder. Entropy increases
when the temperature increases.
3. Know that the temperature of a substance is directly proportional to its
kinetic energy.
4. Be able to convert losses of energy to temperature increases using
Q=mcT.
5. They ask lab-type questions on heat. These are some of the ones that I
have seen:
o Use calorimetry data (which you graph) to determine the specific
heat of a substance.
o Use calorimetry data (which you graph) to determine the latent
heat of a substance.
o Be able to relate the mechanical equivalent of heat to thermal
energy in a lab situation. In other words, if you increase the
amount of potential energy (kinetic energy) of a substance, you do
work. This is equal to Q=mcT.
o Or, you take temperature data over a period of time for a substance
that is heating on a heater. Remember, if you know the wattage of
the heater and how long the heater was on, you can use Electrical
Energy=Vit to calculate energy. This energy would be transferred
to the substance as heat.
o Think about places in an experiment where heat could be lost,
instead of all be absorbed by the substance.
o Be able to graphically describe how temperature varies with time
for different parts of an experiment in which a substance undergoes
multiple phase changes.
6. Be able to solve calorimetry problems.
Thermal Properties of Matter
Physical quantities such as pressure, temperature, volume, and the amount of a
substance describe the conditions in which a particular material exists. They
describe the state of the mateterial and are referred to as state variables. These
state variables are interrelated; one cannot be changed without changing the other.
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The relationship between these variables can be described using an equation of
state.
V = V0[1 + (T - T0) - k(P - P0)]
Where V0, P0, and T0 represent the initial state of the material and V, P, and T
represent the final states of the material.  represents the temperature coefficient
of volume expansion and k is the isothermal compressibility of the material.
In physics, we use an ideal gas to repesent the material and thus simplifying the
equation of state.
Ideal Gas Law The volume of a gas is proportional to the number of moles of the
gas, n. The volume varies inversely with the pressure. The pressure is
proportional to the absolute temperature of the gas. Combining these relationships
yields the following equation of state for an ideal gas,
PV = nRT
Where T is measured in Kelvin and R is the ideal gas constant
Ideal Gas Constant In SI units, R = 8.314 J/ mol K
Ideal Gas Real gases do not follow the ideal gas law exactly. An ideal gas is one
for which the ideal gas law holds precisely for all pressures and temperatures. Gas
behavior approximates the ideal gas model at very low pressures when the gas
molecules are far apart and at temperatures close to that at which the gas liquefies.
pV-diagram A graph of pressure vs volume for a particular temperature for an
ideal gas. Each curve, representing a specific constant temperature, is called an
isotherm. The area under the isotherm represents the work done by the system
during a volume change.
When a system undergoes a change of state from an initial state to a final state,
the system passes through a series of intermediate staes. This series of states is
called a path. Points 1 and 2 represent an initial state (1) with pressure P1 and
volume V1 and a final state (2) with pressure P2 and volume V2. If the pressure is
kept constant at P1, the system expands to volume V2 (point 3 on the diagram).
The pressure is then reduced to P2 (probably by decreasing the temperature)and
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the volume is kept constant at V2 to reach point 2 on the diagram. The work done
by the systemd during this process is the area under the line from state 1 to state 3.
There is no work done during the constant volume process from state 3 to state 2.
Or, the system might traverse the path state 1 to state 4 to state 2, in which case
the work done is the area under the line from state 4 to state 2. Or, the system
might traverse the path represented by the curved line from state 1 to state 2, in
which case, the work is represented by the area underneath the curve from state 1
to state 2. The work is different for each path.
The work done by the system depends not only upon the initial and final states,
but also upon the path taken.
Thermodynamics
Thermodynamics
study of properties of thermal energy
Each of the laws of thermodynamics are associated with a variable. The zeroeth
law is associated with temperature, T; the first law is associated with internal
energy, U; and the second law is associated with entropy, S.
System any object or set of objects we are considering. A closed system is one in
which mass is constant. An open system does not have constant mass. No energy
flows into or out of a closed system which is said to be isolated.
Environment everything else
Thermal Equilibrium
If two objects at different temperatures are placed in thermal contact (so that the
heat energy can transfer from one to the other), the two objects will reach the
same temperature, or become in thermal equilibrium.
Zeroth Law of Thermodynamics If two systems are in thermal equilibirum with
a third system, they are in thermal equilibrium with each other.
Internal or Thermal Energy(symbol is U; unit is J)
sum of all the energy an object possesses; it cannot be measured; only changes in
internal energy can be determined
The kinetic theory can be used to clearly distinguish between temperature
and thermal energy. Temperature is a measure of the average kinetic energy
of individual molecules. Thermal energy refers to the total energy of all the
molecules in an object.
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Internal Energy of an Ideal Gas The internal energy of an ideal gas only
depends upon temperature and the number of moles of the gas (n).
U = 3/2 nRT
where R is the ideal gas constant, R = 8.315 J/mol K
Characteristics of an Ideal Gas:
1. An ideal gas consists of a large number of gas molecules occupying a
negligible volume.
2. Ideal gas molecules have random motion.
3. Ideal gas molecules undergo elastic collisions with the walls of the
container and with other gas molecules.
4. The temperature of an ideal gas is proportional to the kinetic energy of the
gas molecules.
1st law of thermodynamics The total increase in the internal energy of a system
is equal to the sum of the work done on the system or by the system and the heat
added to or removed from the system. It is a restatement of the law of
conservation of energy. Changes in the internal energy of a system are caused by
heat and work.
U = Q + W
where Q is the heat added to the system and W is the net work done on the
system. In other words, heat added is positive; heat lost is negative. Work done on
the system (an example would be compression of a gas) is positive; work done by
the system (an example would be expansion of a gas) is negative.
AP changes for 2002 The 1st law is being expressed in this form to be consistent
with changes in the sign convention for work for the AP Physics Exam for 2002.
Effective in 2002, the symbol W will represent the work done on a system rather
than by a system. According to the College Board, this makes the sign convention
consistent with that used for work in mechanics, as well as with the
thermodynamic convention used in most chemistry and some physics textbooks.
The best way to remember the sign convention for work: if a gas is compressed
(volume decreases), work is positive; if a gas expands (volume increases), work is
negative. It is just like mechanics, if you (the environment) do work on the
system, you would compress it. The work you do is considered to be positive.
1. Isothermal Process temperature (T) is constant. If there is no temperature
change, there is no internal energy change.
U = 0
Q = -W
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The curve shown represents an isotherm.
Since the temperature is constant, no change in internal energy occurs. Internal
energy changes only occur when there are temperature changes. At constant
temperature, the pressure and volume of the system decrease as along the path
state 1 to state 2. The amount of work is given by
Example of an isothermal process: An ideal gas (the system) is contained in a
cylinder with a moveable piston. Since the system is an ideal gas, the ideal gas
law is valid. For constant temperture, PV=nRT becomes PV=constant. At point 1,
the gas is at pressure P1, volume V1, and temperature T. A very slow expansion
occurs, so that the gas stays at the same constant temperature. If heat Q is added,
the gas must expand. As the gas expands, it pushes on the moveable piston, thus
doing work on the environment (or negative work). At point w, the gas now has
volume V2 which is greater than V1, pressure P2 which is less than P1, and
temperature T. The amount of work done by the system on the environment
during its expansion has the same magnitude as the amount of heat added to the
system. The amount of work done is equal to the area under the curve.
How to know if heat was added or removed in an isothermal process: if heat is
added, the volume increases and the pressure decreases. Remember, pressure is
determined by the number of collisions the gas molecules make with the walls of
the container. If the volume increases at constant temperature, the gas molecules
make fewer collisions with the walls of the container, and pressure decreases.
2.
Isobaric Process pressure (P) is constant. If pressure is kept constant, the work
done during the process is given by
W = - P V
U = Q + W
P is held constant, so the amount of work done is represented by the area
underneath the path from 1 to 2. Typically, lab experiments are isobaric
processes.
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Example of isobaric process: An ideal gas is contained in a cylinder with a
moveable piston. The pressure experienced by the gas is always the same, and is
equal to the external atmospheric pressure plus the weight of the piston. The
cylinder is heated, allowing the gas to expand. Heat was added to the system at
constant pressure, thus increasing the volume. The change in internal energy U is
equal to the sum of the work done by the system on the environment during the
volume expansion (negative work) and the amount of heat added to the system.
The amount of work done is equal to the area under the curve.
How to determine if heat was added or removed: in an isobaric process, heat is
added if the gas expands and removed if the gas is compressed.
How to tell if the temperature is increasing or decreasing: in an isobaric process,
adding heat results in an increase in internal energy. If the internal energy
increases, the temperature increases. Typically, volume expansions are small and
all the heat added serves to increase the internal energy. In our graph, point 2 was
at a higher temperature than point 1.
3.
Isochoric Process Volume (V) is constant. Since there is no change in volume,
no work is done.
W=0
U = Q
Since V is constant, no work is done. If heat is added to the system, the internal
energy U increases; if heat is removed from the system, the internal energy U
decreases. In the pV diagram shown, heat is removed along the path 1 to 2, thus
decreasing the pressure at constant volume.
Example of an isochoric process: An ideal gas is contained in a rigid cylinder (one
whose volume cannot change). If the cylinder is heated, no work can be done
even though enormous forces are generated within the cylinder. No work is done
because there is no displacement (the system does not move). The heat added only
increases the internal energy of the system.
How to tell if heat is added or removed: in an isochoric process, heat is added
when the pressure increases.
How to tell if the temperature increases or decreases: since U=3/2 nRT, if the
internal energy is increasing, then the temperature is increasing. In our diagram,
point 1 is at a higher temperature than point 2.
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4.
Aidabatic Process No heat (Q) is allowed to flow into or out of the system. This
can occur if the system is well-insulated or the process happens quickly. (in other words,
Q=0)
U = W
The internal energy and the temperature decreases if the gas expands.
In this well-insulated process shown, heat cannot transfer to the environment. The
amount of work done is represented by the area under the path from state 1 to
state 2. In this example, the volume increases along the path from state 1 to state
2, so work is done on the environment by the system (negative work). There is a
decrease in internal energy U.
Example of an adiabatic process: An ideal gas is contained in a cylinder with a
moveable piston. Insulating material surrounds the cylinder, preventing heat flow.
The ideal gas is compressed adiabatically by pushing against the moveable piston.
Work is done on the gas (positive work). Remember, Q=0. The amount of work
done in the adiabatic compression results in an increase in the internal energy of
the system.
How to tell if the temperature increases or decreases: since U=3/2 nRt, if the
internal energy increases, the temperature increases. In our example, the final
temperature would be greater. than the initial temperature. In our pV diagram, the
temperature at point 1 is greater than the temperature at point 2.
2nd law of thermodynamics This law is a statement about which processes can occur in
nature and which cannot.
1. In natural processes, heat cannot flow from a cold to a hot substance
2. Natural processes increase the entropy of the universe; with time, disorder cannot
become order
3. A heat engine cannot convert all its heat to mechanical energy. No machine is
ever 100% efficient.
The second law of thermodynamics explains things that don't happen:
1. Air molecules fill the room evenly, instead of all moving to one corner.
2. A spoon reaches an equilibrium temperature, instead of one end being cold and
one end being hot.
3. Coffee, swirling in your cup, will eventually stop swirling. The coffee doesn't
spontaneously cool down and start to swirl around.
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It is not possible to reach absolute zero (0 K). Since heat can only flow from a hot to a
cold substance, in order to decrease the temperature of a substance, heat must be removed
and transferred to a "heat sink" (something that is colder). Since there is no temperature
less than absolute zero, there is no heat sink to use to remove heat to reach that
temperature.
entropy (S)
disorder, chaos.
S = Q / T
where T is the Kelvin temperature
Determining how entropy changes: When dealing with entropy, it is the change
in entropy which is important.


In a reversible process (one in which there is no friction), if heat is added
to a system, the entropy of the system increases, and vice versa. If entropy
increases for the system, it must decrease for the environment by the same
amount, and vice versa. For reversible processes, the total entropy (the
entropy of the system plus the environment) is constant.
In an irreversible process (those in the real world), the total entropy either
is unchanged or increases.
Heat engines:
1. automobile engines-thermal energy from a high heat source is converted
into mechanical energy (work) and exhaust is expelled
2. refrigerator-thermal energy is removed from a cold body (work is
required) and transferred to a hot body (the room. Another example is a
heat pump.
Drawing of a real engine showing transfer of heat from a high to a low
termperature reservoir, performing work. The figure below shows the overall
operation of a heat engine. During every cycle, heat QH is extracted from a
reservoir at temperature TH; useful work is done and the rest is discharged as heat
QL to a reservoir at a cooler temperature TL. Since an engine is a cycle, there is no
change in internal energy adn the net work done per cycle equals the net heat
transferred per cycle.
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The purpose of an engine is to transform as much QH into work as possible.
So...coffee can't spontaneously start swirling around because heat would be
withdrawn from the coffee and totally transformed into work. A heat engine
converts thermal energy into mechanical energy.
Drawing of a refrigerator showing transfer of heat from a low to a high
temperature reservoir, requiring work. The purpose of a refrigerator is to transfer
heat from the low-temperature to the high-temperature reservoir, doing as little
work on the system as possible.
There is no perfect refrigerator because it is not possible for heat to flow from one
body to another body at a higher temperature with no other change taking place.
The purpose of a heat pump or a refrigerator is to convert mechanical energy into
thermal energy.
Efficiency of a heat engine The efficiency e of any heat engine is defined as the
ratio of the work the engine does (W) to the heat input at the high temperature
(QH).
e = W / QH
or, e = (QH - (QL) / QH
Carnot (ideal) efficiency This is the theoretical limit to efficiency. It is defined in
terms of the operating temperatures.
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eideal = (TH - (TL) / TH
AP Multiple Choice Questions on Thermodynamics
1. Be able to perform simple Carnot efficiency calculations.
2. Be able to describe an ideal gas.
3. Be able to interpret PV diagrams.
o Which part of the cycle has U greater than zero?
o Which part of the cycle has Q greater than zero?
o Which part of the cycle W greater than zero?
o In which part of the cycle is no work done?
o At which part of the cycle is the gas at is highest temperature?
o What is the net work done?
o Which part of the cycle does the most work?
o Compare the temperatures at different points in the cycle.
o Which part of the cycle represents an adiabatic, isothermal,
isochoric (isovolumetric), or isobaric process?
4. Be able to perform simple calculations using the ideal gas law or the
combined gas law. For example, pressure is doubled and volume is
quadrupled. What would be the relationship between the initial and final
temperatures?
5. Know the characteristics of the processes.
6. Predict what would happen to temperature if kinetic energy is changed.
7. Be able to perform simple calculations using the first law of
thermodynamics.
AP Free Response Questions on Thermodynamics
1. Be able to interpret PV diagrams.
o Use ideal gas law (or combined gas law) to calculate the
temperature at different points.
o Sketch the PV diagram from information given.
o Predict whether positive, negative, or no work is done for the
cycle.
o Calculate the work done for the cycle (or part of a cycle).
o Calculate the heat absorbed or given off for a cycle (or part of a
cycle).
o Predict which segments absorb the most heat.
o Predict if the cycle represents a heat engine or refrigerator.
2. Remember the equation for power (P=W/t or P=Fv) and that P=F/A.
3. Sometimes they ask about the rate at which heat is discarded by an engine.
This can be found knowing that W=QH-QL
4. Be able to perform simple Carnot efficiency calculations.
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Thermal Energy Sample Problems
Thermal Energy Sample Problems
1. How much heat is absorbed by 60 g of copper (c=385 J/kg K) when it is heated
from 20C to 80 C?
2. A 38 kg block of lead (c=130 J/kg K) is heated from -26 C to 180 C. How much
heat is absorbed?
3. A 0.5 kg sample of water is in a calorimeter at 15C. 0.04 kg of zinc (c=388 J/kg
K) is added. What is the final temperature? Initial temperature of the zinc is
300C.
4. 100 g of brass (c=376 J/kg K) at 90 C is added to 200 g of water at 20 C. What
is the final temperature?
5. 100 g of aluminum at 100 C is added to 100 g of water at 10 C. What is the
specific heat of aluminum if the final temperature of the mixture is 25 C?
6. 100 g of ice at 0 C is heated until it becomes water at 20 C. How much heat is
added?
7. 100 g of water at 0 C is frozen, becoming ice (c=2060 J/kg K) at -20 C. How
much heat was removed?
Thermal Energy Homework
Thermal Energy
1. How much heat does 100 g of water at 100C lose as it cools and becomes water
at 0C? Ans: -41,800 J
2. 100 J of heat is needed to raise the temperature of an unknown mass of lead
(c=130 J/kgC) 5C. What is the lead’s mass? Ans: 0.154 kg
3. How much heat must be added to 124 g of brass (c=376 J/kgC) at 12.5C to raise
its temperature to 97C? Ans: 3939.73 J
4. 45 g of water at 4C is added to 150 g of water at 75C. What is the mixture’s
equilibrium temperature? Ans: 58.6C
5. An insulated container contains 100 g of ethyl alcohol (c=2440 J/kgC) at 25C.
A 25 g block of iron (c=450 J/kgC) is heated to 400C and then added to the
alcohol. What is the final temperature? Ans: 42C
6. 250 g of copper (c=385 J/kgC) at 100C is placed in a cup containing 32 g of
water at 20C. What is the final temperature? Ans: 53.47C
7. How much heat is needed to change 50 g of water at 80C to steam at 110C?
Ans: 118,190 J
8. How much heat is removed from 60 g of steam at 100C to change it to 60 g of
water at 20C? Ans: -155,664 J
9. How much heat is absorbed by 100 g of ice at -20C to become water at 0C?
Ans: 37,420 J
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States of Matter Sample Problems
States of Matter Sample Problems
1. A metal bar 2.6 m long at 21C is heated to 93C. Its new length is 2.6034 m.
What is its coefficient of linear expansion?
2. A piece of aluminum ( = 25 x 10-6 /C) is 3.66 m long at -28C. What will its
length be at 39C?
3. An aluminum ( = 75 x 10-6 / C) can has a volume of 354 cm3 at 4.4 C. What
will its volume be at 34.5 C? When it is at 4.4 C, it is filled to the brim with
water ( =210 x 10-6 / C). How much will overflow when at 34.5 C?
4. What is the pressure exerted by a 3 kg mass on a box top 5 cm x 2 cm?
5. What is the force causing a pressure of 3000 Pa over a 0.50 m2 cross-sectional
area?
6. A lake is 30 m deep. What is the pressure at that depth?
7. What is the buoyant force experienced by a box of mass 5 kg with dimensions of
15 cm x 12 cm x 13 cm when it is totally immersed in water? What is the apparent
weight of the object?
8. What is the buoyant force exerted on 0.001 m3 of steel ( = 9000 kg/m3)
immersed in water? What is its apparent weight?
States of Matter Homework
States of Matter Homework
1. Calculate the amount a copper rod (=16 x 10-6/C) 25 m long will expand when
it is heated from 4C to 124C. Ans: 0.048 m
2. What is the volume difference for 2 liters of water (=210 x 10-6/C) at 5°C and
for water at 85C? Ans: 0.034 L
3. An aluminum rod (=25 x 10-6/C) that is 250 m long is heated. What is its final
temperature if the length of the rod increases 0.215 m? Its initial temperature is
20C. Ans: 54.4°C
4. What is the change in length of a 15 m steel rail (=12 x 10-6/C) as it is cooled
from 1535C to 20C? Ans: -0.27 m
5. A concrete sidewalk (=36 x 10-6/C) with dimensions of 8m x 1m x 0.1 m at
exactly 0C is heated to 35C. What is its new volume? Ans: 0.801 m3
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6. What is the mass of a lead sphere (=11,300 kg/m3) with a volume of 0.523 m3?
Ans: 5910 kg
7. What is the pressure due to a 30 m tall column of water? Ans: 2.9 x 105 N/m2
8. A 70 kg object is suspended in a lake. Its volume is 0.03 m3. What is the buoyant
force exerted on the object by the water? Ans: 294 N
9. A 5 N force is applied to a piston 0f 0.5 m2 cross-sectional area of a hydraulic
piston. What force must be applied to a 5 m2 cross-sectional area? Ans: 50 N
10. A solid cube of a certain material has a volume of 0.005 m3. It has a mass of 57.5
kg. What is its density? Ans: 11,500 kg/m3
11. What is the height of a column of water that exerts a pressure of 50,000 Pa? Ans:
5.10 m
12. A 20 N force is exerted on the small piston of a hydraulic system. The crosssectional area of the small piston is 0.50 m2. What weight can be lifted by the
large piston if it has a surface area of 1.00 m2? Ans: 40 N
13. A buoyant force of 20 N acts on an object suspended in alcohol (=800 kg/m3).
What is the volume of the object? Ans: 0.0026 m3
14. Water flows through the bottom of a tapered pipe at the rate of 6 m/s. The bottom
of the tapered pipe has a cross-sectional area of 0.100 m2. What will be its rate of
flow through the top of the tapered pipe where the cross-sectional area is 0.035
m2? Ans: 17.14 m/s
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