Physics Pp Presentation Ch 9

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HeatChapter 9

Table of Contents

Section 1 Temperature and Thermal Equilibrium

Section 2 Defining Heat

Section 3 Changes in Temperature and Phase



Section 1 Temperature and Thermal EquilibriumChapter 9

Objectives

• Relate temperature to the kinetic energy of atoms and molecules.

• Describe the changes in the temperatures of two objects reaching thermal equilibrium.

• Identify the various temperature scales, and convert from one scale to another.



Chapter 9

Defining Temperature

• Temperature is a measure of the average kinetic energy of the particles in a substance.

• Adding or removing energy usually changes temperature.

• Internal energy is the energy of a substance due to both the random motions of its particles and to the potential energy that results from the distances and alignments between the particles.




Chapter 9

Forms of Internal Energy




Chapter 9

Thermal Equilibrium

• Thermal equilibrium is the state in which two bodies in physical contact with each other have identical temperatures.– By placing a thermometer in contact with an object and waiting

until the column of liquid in the thermometer stops rising or falling, you can find the temperature of the object.

– The reason is that the thermometer is in thermal equilibrium with the object.

• The temperature of any two objects in thermal equilibrium always lies between their initial temperatures.




Chapter 9

Thermal Equilibrium




Chapter 9

Thermal Expansion

• In general, if the temperature of a substance increases, so does its volume. This phenomenon is known as thermal expansion.

• Different substances undergo different amounts of expansion for a given temperature change.

• The thermal expansion characteristics of a material are indicated by a quantity called the coefficient of volume expansion.

• Gases have the largest values for this coefficient. Solids typically have the smallest values.




Chapter 9

Thermal Expansion




Chapter 9

Measuring Temperature

• The most common thermometers use a glass tube containing a thin column of mercury, colored alcohol, or colored mineral spirits.

• When the thermometer is heated, the volume of the liquid expands.

• The change in length of the liquid column is proportional to the temperature.




Chapter 9

Measuring Temperature, continued

• When a thermometer is in thermal equilibrium with a mixture of water and ice at one atmosphere of pressure, the temperature is called the ice point or melting point of water. This is defined as zero degrees Celsius, or 0°C.

• When the thermometer is in thermal equilibrium with a mixture of steam and water at one atmosphere of pressure, the temperature is called the steam point or boiling point of water. This is defined as 100°C.




Chapter 9

Measuring Temperature, continued• The temperature scales most widely used today are the Fahrenheit, Celsius, and Kelvin scales. • Celsius and Fahrenheit temperature measurements can be converted to each other using this equation:


9 32.05

9 Fahrenheit temperatu

re

Celsius temperature 32.05

F CT T

• The number 32.0 indicates the difference between the ice point value in each scale: 0.0ºC and 32.0ºF.



Chapter 9



• Temperature values in the Celsius and Fahrenheit scales can have positive, negative, or zero values.

• But because the kinetic energy of the atoms in a substance must be positive, the absolute temperature that is proportional to that energy should be positive also.

• A temperature scale with only positive values is suggested by the graph on the next slide. This scale is called the Kelvin scale.



Chapter 9



• The graph suggests that if the temperature could be lowered to –273.15°C, the pressure would be zero.

• This temperature is designated in the Kelvin scale as 0.00 K, where K represents the temperature unit called the kelvin.

• Temperatures in the Kelvin scale are indicated by the symbol T.



Chapter 9



• A temperature difference of one degree is the same on the Celsius and Kelvin scales. The two scales differ only in the choice of zero point. • Thus, the ice point (0.00°C) equals 273.15 K, and the steam point (100.00°C) equals 373.15 K. • The Celsius temperature can therefore be converted to the Kelvin temperature by adding 273.15:

T TC 273.15 Kelvin temperature Celsius temperature 273.15



Chapter 9

Temperature Scales and Their Uses




Section 2 Defining HeatChapter 9

Objectives

• Explain heat as the energy transferred between substances that are at different temperatures.

• Relate heat and temperature change on the macroscopic level to particle motion on the microscopic level.

• Apply the principle of energy conservation to calculate changes in potential, kinetic, and internal energy.




Heat and Energy

• Heat is the energy transferred between objects because of a difference in their temperatures.

• From a macroscopic viewpoint, energy transferred as heat tends to move from an object at higher temperature to an object at lower temperature.

• The direction in which energy travels as heat can be explained at the atomic level, as shown on the next slide.



Chapter 9

Transfer of Particles’ Kinetic Energy as Heat


Energy is transferred as heat from the higher-energy particles to the lower-energy particles, as shown on the left. The net energy transferred is zero when thermal equilibrium is reached, as shown on the right.



Chapter 9

Temperature and Heat





Heat and Energy, continued

• The atoms of all objects are in continuous motion, so all objects have some internal energy. – Because temperature is a measure of that energy,

all objects have some temperature.

• Heat, on the other hand, is the energy transferred from one object to another because of the temperature difference between them. – When there is no temperature difference between

a substance and its surroundings, no net energy is transferred as heat.




Heat and Energy, continued

• Just as other forms of energy have a symbol that identifies them (PE for potential energy, KE for kinetic energy, U for internal energy, W for work), heat is indicated by the symbol Q.

• Because heat, like work, is energy in transit, all heat units can be converted to joules, the SI unit for energy.



Chapter 9

Thermal Units and Their Values in Joules





Thermal Conduction

• The type of energy transfer that is due to atoms transferring vibrations to neighboring atoms is called thermal conduction.

• The rate of thermal conduction depends on the substance.

• Two other mechanisms for transferring energy as heat are convection and electromagnetic radiation.

When this burner is turned on, the skillet’s handle heats up because of conduction.



Chapter 9

Convection, Conduction, and Radiation





Conservation of Energy

• If changes in internal energy are taken into account along with changes in mechanical energy, the total energy is a universally conserved property.

• In other words, the sum of the changes in potential, kinetic, and internal energy is equal to zero.

CONSERVATION OF ENERGYPE + KE + U = 0

the change in potential energy + the change in kinetic energy + the change in internal energy = 0



Chapter 9

Conservation of Energy





Sample Problem

Conservation of Energy An arrangement similar to the one

used to demonstrate energy conservation is shown in the figure. A vessel contains water. Paddles that are propelled by falling masses turn in the water. This agitation warms the water and increases its internal energy. The temperature of the water is then measured, giving an indication of the water’s internal energy increase.




Sample Problem, continued

Conservation of Energy, continued If a total mass of 11.5 kg falls 1.3 m

and all of the mechanical energy is converted to internal energy, by how much will the internal energy of the water increase? (Assume no energy is transferred as heat out of the vessel to the surroundings or from the surroundings to the vessel’s interior.)





1. DefineGiven:

m = 11.5 kgh = 1.3 mg = 9.81 m/s2

Unknown:U = ?





2. PlanChoose an equation or situation: Use the conservation of energy, and solve for U.

PE + KE + U = 0(PEf – PEi) + (KEf – KEi) + U = 0

U = –PEf + PEi – KEf + KEi

Tip: Don’t forget that a change in any quantity, indicated by the symbol ∆, equals the final value minus the initial value.




Sample Problem, continuedBecause the masses begin at rest, KEi equals zero. If we assume that KEf is small compared to the loss of PE, we can set KEf equal to zero also.

KEf = 0 KEi = 0Because all of the potential energy is assumed to be converted to internal energy, PEi can be set equal to mgh if PEf is set equal to zero.

PEi = mgh PEf = 0Substitute each quantity into the equation for ∆U:

∆U = –PEf + PEi – KEf + KEi

∆U = 0 + mgh + 0 + 0 = mgh





4. EvaluateThe answer can be estimated using rounded values. If m ≈ 10 kg and g ≈ 10 m/s2, then ∆U ≈ 130 J, which is close to the actual value calculated.

3. CalculateSubstitute the values into the equation and solve:

U = mghU = (11.5 kg)(9.81 m/s2)(1.3 m)

U = 1.5 102 J



Section 3 Changes in Temperature and PhaseChapter 9

Objectives

• Perform calculations with specific heat capacity.

• Interpret the various sections of a heating curve.




Specific Heat Capacity

• The specific heat capacity of a substance is defined as the energy required to change the temperature of 1 kg of that substance by 1°C.

• Every substance has a unique specific heat capacity.

• This value tells you how much the temperature of a given mass of that substance will increase or decrease, based on how much energy is added or removed as heat.




Specific Heat Capacity, continued

• Specific heat capacity is expressed mathematically as follows:

cp QmT

specific heat capacity = energy transferred as heat

mass change in temperature

• The subscript p indicates that the specific heat capacity is measured at constant pressure.

• In this equation, T can be in degrees Celsius or in degrees Kelvin.



Chapter 9

Specific Heat Capacities





Calorimetry

• Calorimetry is used to determine specific heat capacity.

• Calorimetry is an experimental procedure used to measure the energy transferred from one substance to another as heat.

A simple calorimeter allows the specific heat capacity of a substance to be determined.



Chapter 9

Calorimetry





Calorimetry, continued

Because the specific heat capacity of water is well known (cp,w= 4.186 kJ/kg•°C), the energy transferred as heat between an object of unknown specific heat capacity and a known quantity of water can be measured.

energy absorbed by water = energy released by substance

Qw = –Qx

cp,wmw∆Tw = –cp,xmx∆Tx




Sample Problem

Calorimetry A 0.050 kg metal bolt is heated to an unknown initial

temperature. It is then dropped into a calorimeter containing 0.15 kg of water with an initial temperature of 21.0°C. The bolt and the water then reach a final temperature of 25.0°C. If the metal has a specific heat capacity of 899 J/kg•°C, find the initial temperature of the metal.





1. DefineGiven: mm = 0.050 kg cp,m = 899 J/kg•°C

mw = 0.15 kg cp,w = 4186 J/kg•°C

Tw = 21.0°C Tf = 25.0°CUnknown: Tm = ?

Diagram:





2. PlanChoose an equation or situation: The energy absorbed by the

water equals the energy removed from the bolt.

Qw –Qmcp,wmwTw –cp,mmmTmcp,wmw (T f Tw ) –cp,mmm (T f Tm )

Rearrange the equation to isolate the unknown:

Tm cp,wmw (T f Tw )

cp,mmmT f





3. CalculateSubstitute the values into the equation and solve:

4. EvaluateTm is greater than Tf, as expected.

,

,

(4186 J/kg• C)(0.15 kg)(25.0 C 21.0 C)25.0 C

(899 J/kg C)(0.050 kg)81 C

( )p w w f wm f

p m m

m

m

c m T TT T

c m

T

T

Tip: Because Tw is less than Tf, you know that Tm must be greater than Tf.




Latent Heat

• When substances melt, freeze, boil, condense, or sublime, the energy added or removed changes the internal energy of the substance without changing the substance’s temperature.

• These changes in matter are called phase changes.• The energy per unit mass that is added or removed

during a phase change is called latent heat, abbreviated as L.

Q = mL energy transferred as heat during phase change = mass latent heat



Chapter 9

Latent Heat





Latent Heat, continued

• During melting, the energy that is added to a substance equals the difference between the total potential energies for particles in the solid and the liquid phases. This type of latent heat is called the heat of fusion, abbreviated as Lf.

• During vaporization, the energy that is added to a substance equals the difference in the potential energy of attraction between the liquid particles and between the gas particles. In this case, the latent heat is called the heat of vaporization, abbreviated as Lv.



Multiple Choice

1. What must be true about two given objects for energy to be transferred as heat between them?

A. The objects must be large.B. The objects must be hot.C. The objects must contain a large amount of

energy.D. The objects must have different temperatures.

Standardized Test PrepChapter 9



Multiple Choice

1. What must be true about two given objects for energy to be transferred as heat between them?

A. The objects must be large.B. The objects must be hot.C. The objects must contain a large amount of

energy.D. The objects must have different temperatures.




Multiple Choice, continued

2. A metal spoon is placed in one of two identical cups of hot coffee. Why does the cup with the spoon have a lower temperature after a few minutes?

F. Energy is removed from the coffee mostly by conduction through the spoon.G. Energy is removed from the coffee mostly by convection through the spoon.H. Energy is removed from the coffee mostly by radiation through the spoon.J. The metal in the spoon has an extremely large specific heat capacity.





2. A metal spoon is placed in one of two identical cups of hot coffee. Why does the cup with the spoon have a lower temperature after a few minutes?

F. Energy is removed from the coffee mostly by conduction through the spoon.G. Energy is removed from the coffee mostly by convection through the spoon.H. Energy is removed from the coffee mostly by radiation through the spoon.J. The metal in the spoon has an extremely large specific heat capacity.





Use the passage below to answer questions 3–4.

The boiling point of liquid hydrogen is –252.87°C.

3. What is the value of this temperature on the Fahrenheit scale?

A. 20.28°FB. –220.87°FC. –423.2°FD. 0°F







3. What is the value of this temperature on the Fahrenheit scale?

A. 20.28°FB. –220.87°FC. –423.2°FD. 0°F







4. What is the value of this temperature in kelvins?

F. 273 KG. 20.28 KH. –423.2 KJ. 0 K







4. What is the value of this temperature in kelvins?

F. 273 KG. 20.28 KH. –423.2 KJ. 0 K





5. A cup of hot chocolate with a temperature of 40°C is placed inside a refrigerator at 5°C. An identical cup of hot chocolate at 90°C is placed on a table in a room at 25°C. A third identical cup of hot chocolate at 80°C is placed on an outdoor table, where the surrounding air has a temperature of 0°C. For which of the three cups has the most energy been transferred as heat when equilibrium has been reached?

A. The first cup has the largest energy transfer.B. The second cup has the largest energy transfer.C. The third cup has the largest energy transfer.D. The same amount of energy is transferred as heat for all three cups.





5. A cup of hot chocolate with a temperature of 40°C is placed inside a refrigerator at 5°C. An identical cup of hot chocolate at 90°C is placed on a table in a room at 25°C. A third identical cup of hot chocolate at 80°C is placed on an outdoor table, where the surrounding air has a temperature of 0°C. For which of the three cups has the most energy been transferred as heat when equilibrium has been reached?

A. The first cup has the largest energy transfer.B. The second cup has the largest energy transfer.C. The third cup has the largest energy transfer.D. The same amount of energy is transferred as heat for all three cups.





6. What data are required in order to determine the specific heat capacity of an unknown substance by means of calorimetry?

F. cp,water, Twater, Tsubstance, Tfinal, Vwater, Vsubstance

G. cp,substance, Twater, Tsubstance, Tfinal, mwater, msubstance

H. cp,water, Tsubstance, mwater, msubstance

J. cp,water, Twater, Tsubstance, Tfinal, mwater, msubstance





6. What data are required in order to determine the specific heat capacity of an unknown substance by means of calorimetry?

F. cp,water, Twater, Tsubstance, Tfinal, Vwater, Vsubstance

G. cp,substance, Twater, Tsubstance, Tfinal, mwater, msubstance

H. cp,water, Tsubstance, mwater, msubstance

J. cp,water, Twater, Tsubstance, Tfinal, mwater, msubstance





7. During a cold spell, Florida orange growers often spray a mist of water over their trees during the night. Why is this done?

A. The large latent heat of vaporization for water keeps the trees from freezing.B. The large latent heat of fusion for water prevents it and thus the trees from freezing.C. The small latent heat of fusion for water prevents the water and thus the trees from freezing.D. The small heat capacity of water makes the water a good insulator.





7. During a cold spell, Florida orange growers often spray a mist of water over their trees during the night. Why is this done?

A. The large latent heat of vaporization for water keeps the trees from freezing.B. The large latent heat of fusion for water prevents it and thus the trees from freezing.C. The small latent heat of fusion for water prevents the water and thus the trees from freezing.D. The small heat capacity of water makes the water a good insulator.





Use the heating curve to answer questions 8–10. The graph shows the change in temperature of a 23 g sample as energy is added to the sample as heat.


8. What is the specific heat capacity of the liquid? F. 4.4 105 J/kg•°CG. 4.0 102 J/kg•°CH. 5.0 102 J/kg•°CJ. 1.1 103 J/kg•°C






8. What is the specific heat capacity of the liquid? F. 4.4 105 J/kg•°CG. 4.0 102 J/kg•°CH. 5.0 102 J/kg•°CJ. 1.1 103 J/kg•°C






9. What is the latent heat of fusion? A. 4.4 105 J/kgB. 4.0 102 J/kg•°CC. 10.15 103 JD. 3.6 107 J/kg






9. What is the latent heat of fusion? A. 4.4 105 J/kgB. 4.0 102 J/kg•°CC. 10.15 103 JD. 3.6 107 J/kg






10. What is the specific heat capacity of the solid? F. 1.85 103 J/kg•°CG. 4.0 102 J/kg•°CH. 5.0 102 J/kg•°CJ. 1.1 103 J/kg•°C






10. What is the specific heat capacity of the solid? F. 1.85 103 J/kg•°CG. 4.0 102 J/kg•°CH. 5.0 102 J/kg•°CJ. 1.1 103 J/kg•°C



Short Response

Base your answers to questions 11–12 on the information below.

The largest of the Great Lakes, Lake Superior, contains 1.20 1016 kg of fresh water, which has a specific heat capacity of 4186 J/kg•°C and a latent heat of fusion of 3.33 105 J/kg.

11. How much energy would be needed to increase the temperature of Lake Superior by 1.0°C?




Short Response



11. How much energy would be needed to increase the temperature of Lake Superior by 1.0°C?

Answer: 5.0 1019 J




Short Response, continued



12. If Lake Superior were still liquid at 0°C, how much energy would need to be removed from the lake for it to become completely frozen?







12. If Lake Superior were still liquid at 0°C, how much energy would need to be removed from the lake for it to become completely frozen?Answer: 5.00 1021 J






13. Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and water in separate beakers at the same temperature are supplied with the same amount of energy, which will have the higher final temperature?




13. Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and water in separate beakers at the same temperature are supplied with the same amount of energy, which will have the higher final temperature?


Answer: the ethyl alcohol




14. A 0.200 kg glass holds 0.300 kg of hot water, as shown in the figure. The glass and water are set on a table to cool. After the temperature has decreased by 2.0°C, how much energy has been removed from the water and glass?


(The specific heat capacity of glass is 837 J/kg•°C, and that of water is 4186 J/kg•°C.)




14. A 0.200 kg glass holds 0.300 kg of hot water, as shown in the figure. The glass and water are set on a table to cool. After the temperature has decreased by 2.0°C, how much energy has been removed from the water and glass?


(The specific heat capacity of glass is 837 J/kg•°C, and that of water is 4186 J/kg•°C.)

Answer: 2900 J



Extended Response

15. How is thermal energy transferred by the process of convection?




Extended Response

15. How is thermal energy transferred by the process of convection?

Answer: The increasing temperature of a liquid or gas causes it to become less dense, so it rises above colder liquid or gas, transferring thermal energy with it.




Extended Response, continued

16. Show that the temperature –40.0° is unique in that it has the same numerical value on the Celsius and Fahrenheit scales. Show all of your work.




Extended Response, continued

16. Show that the temperature –40.0° is unique in that it has the same numerical value on the Celsius and Fahrenheit scales. Show all of your work.

Answer:


TF 95

(–40.0C) 32.0 (–72.0 32.0)F –40.0F



Chapter 9

Measuring Temperature




Chapter 9

Determining Absolute Zero for an Ideal Gas





Calorimetry

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