Name:
Skill Sheet 25.2
Temperature Scales
Temperature, a measure of the average kinetic energy of the molecules of a substance, has an important role in
our daily lives. Whether we are cooking dinner, dressing for school, or suffering from a cold, we often wish to
know something about the temperature of our environment or some body of matter. To report values for
temperature, the Fahrenheit and Celsius scales are commonly used by scientists and society. In this exercise,
you will examine these scales and practice converting temperature values from one scale to the other.
1. Fahrenheit and Celsius
The Fahrenheit and Celsius temperature scales are the most commonly used scales for reporting temperature
values. Scientists use the Celsius scale almost exclusively, as do many countries of the world. Some countries,
such as the United States, still rely heavily on the Fahrenheit scale for reporting temperature information. When
the weatherman on television tells you the forecast for the week, the scale he is using for temperature is
Fahrenheit. Your oven is calibrated in Fahrenheit values, as is the thermometer your doctor uses to assess your
health. In the United States, we are comfortable with the Fahrenheit scale and design our appliances and tools
using this system of measurement.
However, if you were in Europe, you would see temperatures given in degrees Celsius. Scientists use the Celsius
scale for their experiments and report their results in degrees Celsius. Therefore, it may be necessary at some
point in time to convert information reported in degrees Celsius to degrees Fahrenheit or vice versa. To
accomplish this, we use conversion formulas.
2. Converting Fahrenheit values into Celsius
If you have been given temperature information in degrees Fahrenheit (°F) and need the values to be reported in
degrees Celsius (°C), you would use the following formula:
5
°C = --- ( °F – 32 )
9
Example:
What is the Celsius value for 65° Fahrenheit?
5
°C = --- ( 65°F – 32 )
9
°C = 18.3
1
Skill Sheet 25.2 Temperature Scales
3. Practice converting Fahrenheit to Celsius values
1.
The weatherman tells you that today will reach a high of 45°F. Your friend from Sweden asks what the
temperature will be in degrees Celsius. What value would you report to your friend?
2.
Your father orders a fancy oven from England. When it arrives, you notice that the temperature dial is
calibrated in degrees Celsius. You wish to bake a cake at 350°F. At what temperature will you have to set the
dial on this new oven?
3.
Your new German automobile's engine temperature gauge reads in Celsius, not Fahrenheit. You know that
the engine temperature should not rise above about 225°F. What is the corresponding Celsius temperature on
your new car's gauge?
4. Converting Celsius to Fahrenheit values
To convert Celsius temperature values to degrees Fahrenheit, you must again use a conversion formula:
9
°F =  --- × °C + 32
5
Example:
200°C is the same temperature as what value on the Fahrenheit scale?
9
°F = --- ( 200°C ) + 32
5
°F = 392
2
Skill Sheet 25.2 Temperature Scales
5. Practice converting Celsius to Fahrenheit values
1.
Your grandmother in Ireland sends you her favorite cookie recipe. Her instructions say to bake the cookies at
190.5°C. To what Fahrenheit temperature would you set the oven to bake the cookies?
2.
A scientist wishes to generate a chemical reaction in his laboratory. The temperature values in his laboratory
manual are given in degrees Celsius. However, his lab thermometers are calibrated in degrees Fahrenheit. If
he needs to heat his reactants to 232°C, what temperature will he need to monitor on his lab thermometers?
3.
You phone a friend who lives in Denmark and tell him that the temperature today only rose as high as 15°F.
He replies that you must have enjoyed the warm weather. Explain his answer using your knowledge of the
Fahrenheit and Celsius scales and conversion formulas.
6. Extension: the Kelvin temperature scale
For some scientific applications, a third temperature scale is used: the Kelvin scale. The Kelvin scale is calibrated
so that raising the temperature one degree Kelvin raises it by the same amount as one degree Celsius. The
difference between the scales is that 0°C is the freezing point of water, while 0 K is much, much colder. On the
Kelvin scale, 0 K (degree symbols are not used for Kelvin values) represents absolute zero. Absolute zero is the
temperature when the average kinetic energy of a perfect gas is zero—the molecules display no energy of
motion. Absolute zero is equal to -273°C, or -459°F. When scientists are conducting research, they often obtain
or report their temperature values in Celsius, and other scientists must convert these values into Kelvin for their
own use, or vice versa. To convert Celsius values to their Kelvin equivalents, you would use the formula:
K = °C + 273
Example:
Water boils at a temperature of 100°C. What would be the corresponding temperature for the Kelvin scale?
K = °C + 273
K = 100°C + 273
K = 373
3
Skill Sheet 25.2 Temperature Scales
To convert Kelvin values to Celsius, you would perform the opposite operation; subtract 273 from the Kelvin
value to find the Celsius equivalent.
Example:
A substance has a melting point of 625 K. At what Celsius temperature would this substance melt?
°C = K – 273
°C = 625 K – 273
°C = 352
Although we rarely need to convert between Kelvin and Fahrenheit, use the following formulas to do so:
5
K = --- ( °F + 460 )
9
9
°F =  --- × K – 460
5

Solving problems
1.
A gas has a boiling point of -175°C. At what Kelvin temperature would this gas boil?
2.
A chemist notices some silvery liquid on the floor in her lab. She wonders if someone accidentally broke a
mercury thermometer, but did not thoroughly clean up the mess. She decides to find out of the silver stuff is
really mercury. From her tests with the substance, she finds out that the melting point for the liquid is 275 K.
A reference book says that the melting point for mercury is -38.87°C. Is this substance mercury? Explain
your answer and show all relevant calculations.
3.
You are at a Science Camp in Florida. It’s August 1st. Today’s activity is an outdoor science quiz. The first
question on the quiz involves a thermometer that reports the current temperature as 90°. You need to state the
temperature scale in which this thermometer is calibrated: Kelvin, Fahrenheit, or Celsius. Which scale is
correct? Defend your answer with your knowledge of the temperature scales.
4
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Skill Sheet 26.3
Heat Transfer
In this skill sheet, you will calculate the heat transfer in watts for conduction, convection, and radiation in
simple systems. You will make your calculations using the heat conduction equation, the convection equation,
and the Stefan-Boltzmann equation.
1. The heat conduction equation
Solve the following problems using the heat conduction equation. The first problem is done for you.
1.
A copper bar connects two beakers. The water in one beaker is 25°C. The water in the other beaker is 75°C.
The cross-sectional area of the 0.75-meter bar is 0.0004 m2. How many watts of heat are conducted through
the bar from the hot to the cold beaker? The thermal conductivity of copper is 401 W/m°C.
°
2
°
°
( 401 W/m C ) × ( 0.0004 m ) × ( 75 C – 25 C )- = 10.7 watts
PH = -------------------------------------------------------------------------------------------------------------0.75 m
2.
A copper bar that is 0.75 m long and has a cross section of 0.0004 m2 connects two beakers. If 20 watts of
heat are conducted through the bar, what is the temperature difference between the two beakers?
3.
A piece of aluminum and a piece of steel have the same length and the same cross-sectional area.
Aluminum’s thermal conductivity is 226 W/m°C and the thermal conductivity of steel is 43 W/m°C. The
piece of aluminum connects two regions with a temperature difference of 10°C. For the piece of steel to
conduct the same amount of heat as the aluminum, what would the temperature difference have to be
between two regions connected by the steel? (HINT: Plug the values into a heat equation for each material
and set the equations equal to each other.)
4.
The air temperature is 5°C and the temperature of coffee inside a styrofoam cup is 35°C. The thermal
conductivity of styrofoam is 0.025 W/m°C. Let’s say the length of the styrofoam is the thickness of the cup
or 0.003 meter (3 millimeters). The cross-sectional area that we will use is 0.0004 m2 (2 centimeters × 2
centimeters). How many watts of heat flow through the styrofoam coffee cup in this area?
1
Skill Sheet 26.3 Heat Transfer
5.
Write a short paragraph, describing an example (not mentioned in the text book) of how the heat conduction
equation is used in the real world. You may need to do some research in your library or on the Internet.
2. The convection equation
Solve the following problems using the convection equation. The first problem is done for you.
1.
A wind at 15°C blows on the surface of a window that is 21°C. The heat transfer coefficient is 70 W/m2-°C.
If the window is 0.5 meter2, what is the heat transfer in watts?
2°
2
°
°
P H = ( 70 W/m C ) × ( 0.5 m ) × ( 21 C – 15 C ) = 210 watts
2.
On a hot day, a window is 25°C and the air temperature outside 32°C. The heat transfer coefficient is 10
W/m2-°C. If the area of the window is 0.5 meter2, what is the heat transfer in watts?
3.
When comparing the free and forced convection of gases, oil, and water, water has the highest values. The
lowest values are for gases. The values for oil are closer to the values for gas than for water. Come up with a
hypothesis that explains why this may be so. (HINT: See Table 26.2 in chapter 26.)
4.
The range of heat transfer coefficients for water when convection is ‘free’ is 100 - 1,000 W/m2-°C. The
range when convection is forced is 100 - 20,000 W/m2-°C. Explain the large difference between these
values. What is the difference between free and forced convection?
5.
Write a short paragraph, describing an example (not mentioned in the text book) of how the convection
equation is used in the real world. You may need to do some research in your library or on the Internet.
2
Skill Sheet 26.3 Heat Transfer
3. The Stefan-Boltzmann equation
Solve the following problems using the Stefan-Boltzmann equation. The first problem is done for you.
1.
The power radiated by a filament is 0.3 watts. The diameter of the filament is 0.5 millimeters and it has a
length of 50 millimeters. If the surface area is 4 × 10-8 m2, what is the temperature?
P = 0.3 watts
–8
5.7 × 10 W–8 2
4
0.3 watts = ----------------------------×
(
4
×
10
m
)
×
(
Temperature
)
2
m K
4
14
1.3 × 10 = Temperature
3377 K = Temperature
2.
The surface area of a filament is 4 × 10-8 m2. What is the power in watts if the temperature is 1,727°C?
3.
The power radiated by a surface is 0.125 watts. The temperature of this surface is 1500°C. What is the area
of the surface?
4.
The area of a surface is one square meter and its temperature is 273 K. How much heat is this surface
radiating in units of watts of power? What is the temperature of this surface in degrees Celsius?
5.
Regarding radiation, is heat transfer at 0°C less than, greater than, or the same to heat transfer at absolute
zero? Explain your answer.
6.
Write a short paragraph, describing an example (not mentioned in the text book) of how the StefanBoltzmann conduction equation is used in the real world. You may need to do some research in your library
or on the Internet.
3
Name:
Skill Sheet 27.1
Stress and Strain
The stress in a material is the ratio of the force acting through the material divided by the cross-section area
through which the force is carried. The cross-section area is the area perpendicular to the direction of the
force. Dividing force by cross-section area (mostly) separates out the effects of size and shape from the
strength properties of the material itself. Stress (signified by the Greek letter sigma, σ) is force (F) divided by
cross-section area (A).
1. Introduction to stress
Stress (σ) is defined as force (F) divided by cross-section area (A).
Force is what we apply, while stress is what the material experiences.
For example, when a 10-kilogram ball hangs from a steel wire that has a cross-sectional area of 12 mm2, the
stress experienced by the wire is given below. One pascal (Pa) is equal to one newton of force per square meter of
area. A megapascal (MPa) is equal to one million pascals.
2
N4
10 kg ) ( 9.8 m/sec )- = -----------------------98 N - = 8.17 × 10 4 -----σ = (-----------------------------------------------= 8.17 × 10 Pa
2
12 mm
2
0.0012 m
2
m
The wire will break when the stress that it experiences exceeds the tensile strength of the material.
6 N
The tensile strength of steel is: 400 × 10 ------2m
Therefore, we can predict that a mass of 49,000 kilograms would break a steel wire with this cross-sectional area.
2
N6
49, 000 kg ) ( 9.8 m/sec -) = -----------------------4802 N - = 400 × 10 6 -----σ = (-----------------------------------------------------------= 400 × 10 Pa = 4.00 MPa
2
12 mm
2
0.0012 m
2
m
There are two ways to talk about the stress experienced by objects. First, stress can be tensile. For example, the
suspension cables in a bridge experience tensile, or pulling, stress. Second, stress can be compressive. For
example, the pillars supporting that bridge experience compressive stress.
1
Skill Sheet 27.1 Stress and Strain
2. Introduction to strain
When force is applied to a material or, equivalently, when a material is under stress, it deforms. When we pull a
rubber band it stretches. When we pull on a piece of metal wire, it also changes length, but not as noticeably as
does the rubber band. Also, when we compress a material, it deforms. The amount of deformation of a material
under stress is called strain. Strain (signified by the Greek letter epsilon, ε) is related to stress by the formula:
N
where E is called the modulus of elasticity and it is given in units of ------2- , or pascals (Pa).
m
Strain (ε) is a dimensionless quantity. It represents the amount of deformation of the material.
in length- = ∆l
ε = change
-----------------------------------------original length
l
As we know, materials expand and contract as temperature changes. This is called thermal expansion or
contraction, and it is also represented by the thermal strain (εθ), which is given by:
where α is the coefficient of thermal expansion and (T2– T1) represents the temperature change. Note that for this
1
equation, the units of α are given in -------- . If the value is negative, the material is contracting.
°C
These three formulas are very powerful and are used extensively in the design and evaluation of structures. We
will also use them in the examples to analyze and design some simple structural elements. The only information
that we need is properties of the materials used. These properties are represented by the parameters E, α, and the
tensile strength of the material.
2
Skill Sheet 27.1 Stress and Strain
3. Problems
1.
A cylindrical concrete pier with a radius of 0.25 meter is loaded with a force of 700 kilonewtons. Find the
stress (force per unit area) and the strain. Concrete has E = 17 × 109 N/m2. The formula for finding the area
2
of a circle is: πr
2.
Concrete is used in building many structures. Concrete has a coefficient of thermal expansion
α = 1 × 10-5/°C. Two pieces of concrete are used to build a bridge. The bridge has a total length of 50 meters,
and the two concrete slabs are anchored at the ends and meet in the middle of the bridge. The design is such
that the gap between the slabs closes in the summer when the temperature reaches 40°C. What is the
maximum gap in the winter when the temperature drops to -10°C? (HINT: Use the thermal strain formula,
and multiply your answer by the total length of the bridge to find the gap.)
3.
A weight of 1,500 kilograms is to be suspended by a wire. We have been asked to design the system using
steel wire. Steel has a tensile strength of 400 megapascals and a modulus of elasticity E = 200 × 109 Pa. We
will do the design in steps.
a. First, what is the minimum wire diameter that satisfies the tensile strength?
b. What is the minimum diameter if a safety factor of 10 is required? A safety factor of 10 means that you
design the wire to support 10 times the weight — so we need to design the wire to support 15,000 kg.
4.
If the weight is supported by 21.6-millimeter-diameter wire with a total length of 5 meters, what is the strain
on the wire? What is the change in length of the wire in meters?
3
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Skill Sheet 27.2
Archimedes
Archimedes was a Greek mathematician who specialized in geometry. He figured out the value of pi and the
volume of a sphere, and has been called “the father of integral calculus.” During his lifetime, he was famous
for using compound pulleys and levers to invent war machines that successfully held off an attack on his city
for three years. Today he is best known for the Archimedean principle, which was the first explanation of how
buoyancy works.
Archimedes was born in Syracuse, on Sicily (then an independent Greek city-state), in
287 B.C. His letters suggest he studied in Alexandria, Egypt, as a young man.
Historians believe it was in Egypt that he invented a device for raising water known as
Archimedes’ screw. The device is still used in many parts of the world.
A famous Greek legend says that King Hieron II of Syracuse asked Archimedes to
figure out if his new crown was pure gold or if the craftsman had mixed some less
expensive silver into it. Archimedes had to determine the answer without destroying
the crown. He thought about it for days and then, as he lowered himself into a bath, the
answer struck him. The legend says Archimedes ran naked through the streets,
shouting “Eureka!”—meaning “I have found it.”
Archimedes realized that if he had equal masses of gold and silver, the denser gold would have a smaller volume.
Therefore, the gold would displace less water than the silver when submerged.
Archimedes found the mass of the crown, and then made a bar of pure gold with the same mass. He submerged
the gold bar and measured the volume of water it displaced. Next, he submerged the crown. He found that the
crown displaced more water than the gold bar had and, therefore, could not be pure gold. The gold had been
mixed with a less dense material. Archimedes had solved the mystery.
Archimedes wrote a treatise titled “On Floating Bodies,” further exploring density and buoyancy. He explained
that an object immersed in a fluid is buoyed upward by a force equal to the weight of the fluid displaced by the
object. Therefore, if an object weighs more than the fluid it displaces, it will sink. If it weighs less than the fluid
it displaces, it will float. This statement is known as the Archimedean principle. Although we commonly assume
the fluid is water, the statement holds true for any fluid, whether liquid or gas. A helium balloon floats because
the air it displaces weighs more than the balloon filled with lightweight gas.
Archimedes wrote several other treatises, including “On the Sphere and the Cylinder,” “On the Measurement of
the Circle,” “On Spirals,” and “The Sand Reckoner.’ In this last treatise, he devised a system of exponents that
allowed him to represent very large numbers on paper—large enough, he said, to count the grains of sand that
would be needed to fill the universe.
Archimedes was killed by a Roman soldier during an invasion of Syracuse in 212 B.C.
Questions
1. Draw a diagram showing how Archimedes solved the problem of determining the gold crown’s purity.
2. Research one of Archimedes’ inventions and create a poster that shows how the device worked.
1
Name:
Skill Sheet 27.3
Gas Laws
This skill sheet reviews the gas laws and includes practice problems that utilize these laws.
1. Boyle’s law: pressure and volume
The relationship between the volume of a gas and the pressure of a gas, at a constant temperature, is known as
Boyle’s law. The equation for Boyle’s law is:
Here’s how you solve a problem using this relationship:
A kit used to fix flat tires consists of an aerosol can containing compressed air and a patch to seal the hole in the
tire. Suppose 10.0 liters of air at atmospheric pressure (101.3 kilopascals, or kPa) is compressed into a 1.0-liter
aerosol can. What is the pressure of the compressed air in the can?
1. Identify what you know and what you are trying to find out from the information given.
P1 = 101.3 kPa
V1 = 10.0 L
P2 = unknown
V2 = 1.0 L
2. Rearrange the variables in the equation to solve for the unknown variable.
Divide each side by V2 to isolate P2 on one side of the equation. The final equation is:
P1 V1
P 2 = ----------V2
3. Plug in the values and solve the problem.
kPa × 10.0 L- = 1,013 kPa
P 2 = 101.3
--------------------------------------------1.0 L
The pressure inside the aerosol can is 1,013 kPa.
1
Skill Sheet 27.3 Gas Laws
2. Charles’ law: pressure and temperature
The French scientist Jacques Charles was a pioneer in hot-air ballooning. He investigated how changing the
temperature of a fixed amount of gas at constant pressure affected its volume. The Charles’ law equation is:
Charles’ law shows a direct relationship between the volume of a gas and the temperature of a gas when the
temperature is given in the Kelvin scale. Zero on the Kelvin scale is the coldest possible temperature, also known
as absolute zero. Absolute zero is equal to -273°C which is 273°C below the freezing point of water. Why do you
think this scale is used to solve these problems?
Converting from degrees Celsius to Kelvin is easy — you add 273 to the Celsius temperature. To convert from
Kelvins to degrees Celsius, you subtract 273 from the Kelvin temperature.
To solve problems with Charles’ law, you can follow the same problem-solving steps you learned for Boyle’s
law, except you use the equation for Charles’ law. You also need to convert degrees Celsius to Kelvin. To practice
both equations, do the problems below.
3. Practice problems with Bolyle’s and Charles’ laws
1.
A truck tire holds 25.0 liters of air at 25°C. If the temperature drops to 0°C, and the pressure remains
constant, what will be the new volume of the tire?
(HINT: Remember to convert degrees Celsius to Kelvins.)
2.
You pump 25.0 liters of air at atmospheric pressure (101.3 kPa) into a soccer ball that has a volume of
4.5 liters. What is the pressure inside the soccer ball if the temperature does not change?
3.
Hyperbaric oxygen chambers (HBO) are used to treat divers with decompression sickness. Research has
shown that HBO can also aid in the healing of broken bones and muscle injuries. As pressure increases
inside the HBO, more oxygen is forced into the bloodstream of the patient inside the chamber. To work
properly, the pressure inside the chamber should be three times greater than atmospheric pressure
(101.3 kPa). What volume of oxygen, held at atmospheric pressure, will need to be pumped into a 190-liter
HBO chamber to make the pressure inside three times greater than atmospheric pressure?
4.
A balloon holds 20.0 liters of helium at 10°C. If the temperature doubles, and the pressure does not change,
what will be the new volume of the balloon?
5.
A scuba tank holds 12.5 liters of oxygen at 101.3 kPa. If the oxygen pumped into the scuba tank was held at
a pressure of 202.6 kPa, what was the original volume of the gas?
2
Skill Sheet 27.3 Gas Laws
4. The ideal gas law
The ideal gas law combines the pressure, volume, and temperature relationships for a gas into one equation that
also includes the mass of the gas. The ideal gas law equation is:
In physics and engineering, mass (m) is used for the quantity of the gas in units of kilograms. The parameter R is
called the gas constant, and it has units of (J/kg-K). Table 27.7 on page 559 of the Student Text shows the value
for R for several common gases. Each different gas has its own value for R. The temperature, T, is given in
Kelvin. The pressure of the gas, P, is given in units of N/m2, also called pascals, and volume, V, is given in units
of m3.
To use the ideal gas law, pressure needs to be absolute pressure, not gauge pressure. Here is how to calculate
absolute pressure from gauge pressure:
absolute pressure = gauge pressure + atmospheric pressure (101,000 N/m2)
5. Practice problems using the ideal gas law
1.
A sample of helium gas occupies 5 liters at 22°C and a gauge pressure of 200,000 pascals. What is the mass
of the gas? (The gas constant for helium is 2.08 × 103 J/kg-K.)
2.
Helium gas of mass m occupies a volume of 2 liters at temperature of 30°C and atmospheric pressure. The
gas is then heated to 100°C and the pressure increases to 5 atmospheres. Find the mass of the gas and the
change in volume, if any.
3.
A gas of mass, m, occupies a volume, V1, at a pressure of 5 atmospheres and a temperature of 25°C. The gas
is heated to 200°C while the volume is increased by 20 percent. What is the final pressure? HINT: Solve the
problem using the formula:
P1
V
------ = -----2P2
V1
3