Preview Objectives Defining Temperature Thermal Equilibrium Thermal Expansion Measuring Temperature Chapter 9 Section 1 Temperature and Thermal Equilibrium

Chapter 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.

Kinetic Theory of Matter All matter is composed of small particles –Atoms, molecules, ions These particles are in constant, random motion These particles are constantly colliding with each other and with the walls of the container.

Thermal Energy / Internal Energy Particles in matter are in constant motion –Solids- the particles vibrate –Liquids – held loosely together –Gases – move about independently Kinetic Energy due to motion of the particles –Translational, Rotational, Vibrational Potential Energy due to the attractive forces between the particles

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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. Section 1 Temperature and Thermal Equilibrium

Temperature Temperature tells how hot or cold an object is Temperature is a measure of the average kinetic energy of the particles of a substance Ice molecules are moving slower than molecules of liquid water which are moving slower than molecules of steam. Absolute zero is the temperature at which no thermal energy can be removed from an object o C also known as 0 K

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. Section 1 Temperature and Thermal Equilibrium

Thermal Equilibrium Hot object in contact with a cooler object. After a period of time the two reach the same temperature. When two objects in contact no longer change temperature, the two objects are said to be in thermal equilibrium. Temperature is meaningful only when the object is in thermal equilibrium. Basis for using thermometers to measure temperature.

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Thermal Expansion Thermal Expansion is an increase in the size of an object when the temperature increases. Kinetic Theory Explanation –Particles move faster the hotter they are –Fast moving particles move farther apart Thermal Contraction is a decrease in the size of an object when the temperature decreases –Particles move slower when cooled –Slower moving particles move closer together

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. Section 1 Temperature and Thermal Equilibrium

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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. Section 1 Temperature and Thermal Equilibrium

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. Section 1 Temperature and Thermal Equilibrium

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: Section 1 Temperature and Thermal Equilibrium The number 32.0 indicates the difference between the ice point value in each scale: 0.0ºC and 32.0ºF.

Figure it Out Between freezing point and boiling point Fahrenheit = = 180 Celsius = 100 – 0 = 100 a degree Celsius is 9/5 a degree Fahrenheit Freezing point on Fahrenheit is 32 above zero T f = 9/5 T c + 32 Kelvin units = Celsius degrees Kelvin starts with 0 = C T k = T c

Chapter 9 Measuring Temperature, continued Section 1 Temperature and Thermal Equilibrium 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 Measuring Temperature, continued Section 1 Temperature and Thermal Equilibrium 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 Measuring Temperature, continued Section 1 Temperature and Thermal Equilibrium 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 K, and the steam point (100.00°C) equals K. The Celsius temperature can therefore be converted to the Kelvin temperature by adding :

Chapter 9 Temperature Scales and Their Uses Section 1 Temperature and Thermal Equilibrium

Assignments Chapter 9 Practice A Page 303: 1,2,3 T f = 9/5T c + 32 T c = 5/9 (T f - 32)

Preview Objectives Heat and Energy Thermal Conduction Conservation of Energy Sample Problem Chapter 9 Section 2 Defining Heat

Chapter 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.

Section 2 Defining Heat Chapter 9 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 Section 2 Defining 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.

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Section 2 Defining Heat Chapter 9 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.

Section 2 Defining Heat Chapter 9 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 Section 2 Defining Heat

Chapter 9 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.

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Section 2 Defining Heat Chapter 9 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 ENERGY  PE +  KE +  U = 0 the change in potential energy + the change in kinetic energy + the change in internal energy = 0

Energy Conservation Closed system- a set of conditions and objects for which energy does not enter or leave the system. In a closed system the total of all types of energy remains the same. ∆PE + ∆KE + ∆U = 0 U is internal energy

Caloric Theory of Heat Heat is a substance that flows from a hot object to a cooler object. Calorie is the amount of heat required to raise the temperature of 1 gram of pure water (at STP) by 1 0 C

Conservation of Energy ∆PE + ∆KE + ∆U = 0 U is internal energy

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Section 2 Defining Heat Chapter 9 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.

Section 2 Defining Heat Chapter 9 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.)

Section 2 Defining Heat Chapter 9 Sample Problem, continued 1. Define Given: m = 11.5 kg h = 1.3 m g = 9.81 m/s 2 Unknown:  U = ?

Section 2 Defining Heat Chapter 9 Sample Problem, continued 2. Plan Choose an equation or situation: Use the conservation of energy, and solve for  U.  PE +  KE +  U = 0 (PE f – PE i ) + (KE f – KE i ) +  U = 0  U = –PE f + PE i – KE f + KE i Tip: Don’t forget that a change in any quantity, indicated by the symbol ∆, equals the final value minus the initial value.

Section 2 Defining Heat Chapter 9 Sample Problem, continued Because the masses begin at rest, KE i equals zero. If we assume that KE f is small compared to the loss of PE, we can set KE f equal to zero also. KE f = 0 KE i = 0 Because all of the potential energy is assumed to be converted to internal energy, PE i can be set equal to mgh if PE f is set equal to zero. PE i = mgh PE f = 0 Substitute each quantity into the equation for ∆U: ∆U = –PE f + PE i – KE f + KE i ∆U = 0 + mgh = mgh

Section 2 Defining Heat Chapter 9 Sample Problem, continued 4. Evaluate The answer can be estimated using rounded values. If m ≈ 10 kg and g ≈ 10 m/s 2, then ∆U ≈ 130 J, which is close to the actual value calculated. 3. Calculate Substitute the values into the equation and solve:  U = mgh  U = (11.5 kg)(9.81 m/s 2 )(1.3 m)  U = 1.5  10 2 J

Assignments Chapter 9 Page 303 Practice A 1,2,3 (8) Page 311 Practice B 1,2,3, SR 3 (4)

Preview Objectives Specific Heat Capacity Calorimetry Sample Problem Latent Heat Chapter 9 Section 3 Changes in Temperature and Phase

Chapter 9 Objectives Perform calculations with specific heat capacity. Interpret the various sections of a heating curve.

Section 3 Changes in Temperature and Phase Chapter 9 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.

Section 3 Changes in Temperature and Phase Chapter 9 Specific Heat Capacity, continued Specific heat capacity is expressed mathematically as follows: 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 Section 3 Changes in Temperature and Phase

Chapter 9 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.

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Calculating Heat lost or gained Q = m (T f – T i ) c p Q = m ΔT c p If T f > T i heat is gained Q= + If T f < T i heat is lost Q= - If energy is conserved, then the heat gained by one object must equal the heat lost by another object.

Application Problems Based on Page 311 In sample problem B the internal energy of the water was increased by 150 J when the masses fell. If the apparatus contained.250 kg of water, what would be the temperature change of the water? In practice B 2, the total internal energy increase was 1760 J. By how many degrees did the temperature of the hammer and spike increase? In Section Review 3, the increase in internal energy of the water was 2.48 X 10 5 J. How much did the water increase in temperature?

Assignments Chapter 9 Page 303 Practice A 1,2,3 (8) Page 311 Practice B 1,2,3, SR 3 (4) Practical Problems(3)

Schedule Tuesday- Practice Problems and Calorimeter Wednesday- Specific Heat lab Thursday- Latent Heat Friday- Heat of Fusion Lab Monday- Lab Calculations Tuesday- Latent Heat Problems Wednesday – Chapter Review (1/2) Thursday – Chapter Review (1/2) Friday - Test

Thermal Equilibrium When two objects are at the same temperature then heat is no longer gained nor lost.

Section 3 Changes in Temperature and Phase Chapter 9 Calorimetry, continued Because the specific heat capacity of water is well known (c p,w = 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 Q w = –Q x c p,w m w ∆T w = –c p,x m x ∆T x

Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem Calorimetry A 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.

Assignments Chapter 9 Page 303 Practice A 1,2,3 (8) Page 311 Practice B 1,2,3, SR 3 (4) Practical Problems(3) Page 316 Practice C 1,2,3 (3)

Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem, continued 1. Define Given: m m = kg c p,m = 899 J/kg°C m w = 0.15 kg c p,w = 4186 J/kg°C T w = 21.0°C T f = 25.0°C Unknown: T m = ? Diagram:

Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem, continued 2. Plan Choose an equation or situation: The energy absorbed by the water equals the energy removed from the bolt. Rearrange the equation to isolate the unknown:

Section 3 Changes in Temperature and Phase Chapter 9 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: 4. Evaluate T m is greater than T f, as expected. Tip: Because T w is less than T f, you know that T m must be greater than T f.

Section 3 Changes in Temperature and Phase P317 Chapter 9 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

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Section 3 Changes in Temperature and Phase Chapter 9 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 L f. 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 L v.

Latent Heat of Fusion Latent Heat of Vaporization Latent Heat of Fusion (Melting) L f = Quantity of heat need to melt mass in kilograms Quantity of heat need to vaporize mass in kilograms L v = Latent Heat of Vaporization (Condensing)

Heating Curve Water 10.0g Quantity of Heat kJ solid liquid Gas (vapor) Boiling Point Melting Point Phase Change C 100

Heating Curve Not Water Quantity of Heat kJ solid liquid Gas (vapor ) Boiling Point Melting Point Phase Change

Assignments Chapter 9 Page 303 Practice A 1,2,3 (8) Page 311 PPB 1,2,3, SR 3 (4) Practical Problems (3) Page 316 PP C 1,2,3 (3) Page 319 SR 1,2,6abcde(7) Page 323- Chapter Review 25a,b,31,32,37,39, 40(7)