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**How to Use This Presentation**• To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.” • To advance through the presentation, click the right-arrow key or the space bar. • From the resources slide, click on any resource to see a presentation for that resource. • From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation. • You may exit the slide show at any time by pressing the Esc key.**Resources**Chapter Presentation Visual Concepts Sample Problems Transparencies Standardized Test Prep**Chapter 9**Heat 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 Equilibrium**Chapter 9 Objectives • Relatetemperature to the kinetic energy of atoms and molecules. • Describethe changes in the temperatures of two objects reaching thermal equilibrium. • Identifythe various temperature scales, and convert from one scale to another.**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Defining Temperature • Temperatureis a measure of the average kinetic energy of the particles in a substance. • Addingorremoving energyusually changes temperature. • Internal energyis 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**Chapter 9 Forms of Internal Energy**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Thermal Equilibrium • Thermal equilibriumis 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 liesbetweentheir initial temperatures.**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Thermal Equilibrium**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Thermal Expansion • In general, if the temperature of a substance increases, so does its volume. This phenomenon is known asthermal 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 thecoefficient of volume expansion. • Gases have the largest values for this coefficient. Solids typically have the smallest values.**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Thermal Expansion**Section 1 Temperature and Thermal Equilibrium**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 theice pointormelting pointof water. This is defined as zero degrees Celsius, or0°C. • When the thermometer is in thermal equilibrium with a mixture of steam and water at one atmosphere of pressure, the temperature is called thesteam pointor boiling pointof water. This is defined as100°C.**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Measuring Temperature, continued • The temperature scales most widely used today are theFahrenheit, Celsius,andKelvinscales. • Celsius and Fahrenheit temperature measurements can be converted to each other using this equation: • The number 32.0 indicates the difference between the ice point value in each scale: 0.0ºC and 32.0ºF.**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Measuring Temperature, continued • Temperature values in the Celsius and Fahrenheit scales can havepositive, negative,orzero values. • But because the kinetic energy of the atoms in a substance must be positive, theabsolute temperaturethat 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 theKelvin scale.**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Measuring Temperature, continued • The graph suggests that if the temperature could be lowered to–273.15°C,the pressure would bezero. • This temperature is designated in theKelvin scale as 0.00 K,where Krepresents the temperature unit called thekelvin. • Temperatures in the Kelvin scale are indicated by the symbolT.**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Measuring Temperature, continued • 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:**Section 1 Temperature and Thermal Equilibrium**Chapter 9 Temperature Scales and Their Uses**Section 2 Defining Heat**Chapter 9 Objectives • Explainheat as the energy transferred between substances that are at different temperatures. • Relateheat and temperature change on the macroscopic level to particle motion on the microscopic level. • Applythe principle of energy conservation to calculate changes in potential, kinetic, and internal energy.**Section 2 Defining Heat**Chapter 9 Heat and Energy • Heatis the energy transferred between objects because of adifference in their temperatures. • From a macroscopic viewpoint, energy transferred as heat tends to move from an object athighertemperature to an object atlowertemperature. • The directionin which energy travels as heat can be explained at the atomic level, as shown on the next slide.**Section 2 Defining Heat**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.**Section 2 Defining Heat**Chapter 9 Temperature and Heat**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 isno temperature differencebetween 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),heatis indicated by the symbolQ. • Because heat, like work, is energy in transit, all heat units can be converted tojoules,the SI unit for energy.**Section 2 Defining Heat**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 calledthermal conduction. • The rate of thermal conductiondepends on the substance. • Two other mechanisms for transferring energy as heat areconvectionand electromagnetic radiation. When this burner is turned on, the skillet’s handle heats up because of conduction.**Section 2 Defining Heat**Chapter 9 Convection, Conduction, and Radiation**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 DPE + DKE + DU = 0 the change in potential energy + the change in kinetic energy + the change in internal energy = 0**Section 2 Defining Heat**Chapter 9 Conservation of Energy**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/s2 Unknown: DU = ?**Section 2 Defining Heat**Chapter 9 Sample Problem, continued 2. Plan Choose an equation or situation: Use the conservation of energy, and solve for DU. DPE + DKE + DU = 0 (PEf– PEi) + (KEf– KEi) + DU = 0 DU = –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.**Section 2 Defining Heat**Chapter 9 Sample Problem, continued Because the masses begin at rest, KEi equals zero. If we assume that KEfis small compared to the loss of PE, we can set KEfequal to zero also. KEf = 0 KEi = 0 Because 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 = mghPEf = 0 Substitute each quantity into the equation for ∆U: ∆U = –PEf+ PEi – KEf+ KEi ∆U = 0 + mgh + 0 + 0 = mgh**Section 2 Defining Heat**Chapter 9 Sample Problem, continued 3. Calculate Substitute the values into the equation and solve: DU = mgh DU = (11.5 kg)(9.81 m/s2)(1.3 m) DU = 1.5 102 J 4. Evaluate The 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.**Section 3 Changes in Temperature and Phase**Chapter 9 Objectives • Perform calculations with specific heat capacity. • Interpretthe various sections of a heating curve.**Section 3 Changes in Temperature and Phase**Chapter 9 Specific Heat Capacity • The specific heat capacityof 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, DT can be in degrees Celsius or in degrees Kelvin.**Section 3 Changes in Temperature and Phase**Chapter 9 Specific Heat Capacities**Section 3 Changes in Temperature and Phase**Chapter 9 Calorimetry • Calorimetryis used to determine specific heat capacity. • Calorimetry is anexperimental procedureused to measure theenergy transferredfrom one substance to another as heat. A simple calorimeter allows the specific heat capacity of a substance to be determined.**Section 3 Changes in Temperature and Phase**Chapter 9 Calorimetry**Section 3 Changes in Temperature and Phase**Chapter 9 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**Section 3 Changes in Temperature and Phase**Chapter 9 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.**Section 3 Changes in Temperature and Phase**Chapter 9 Sample Problem, continued 1. Define Given: 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°C Unknown: Tm = ? 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: Tip: Because Tw is less than Tf, you know that Tm must be greater than Tf. 4. Evaluate Tm is greater than Tf, as expected.**Section 3 Changes in Temperature and Phase**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 calledphase changes. • The energy per unit mass that is added or removed during a phase change is calledlatent heat,abbreviated asL. Q = mL energy transferred as heat during phase change = mass latent heat**Section 3 Changes in Temperature and Phase**Chapter 9 Latent Heat**Section 3 Changes in Temperature and Phase**Chapter 9 Latent Heat, continued • Duringmelting,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 theheat of fusion, abbreviated asLf. • Duringvaporization,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 theheat of vaporization, abbreviated asLv.**Chapter 9**Standardized Test Prep 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.**Chapter 9**Standardized Test Prep 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.