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Design questions that are likely to appear on my next Electric Charge and Coulomb’s Law quiz based on the information the slide shows and corresponding quizzes and are similar in format to the quiz questions? Also explain the patterns you observe between the slides and the questions on the quizzes?
Electric Charge and Coulomb’s Law Slides: Law Copyright © 2009 Pearson Education, Inc. Objects can be charged by rubbing Static Electricity; Electric Charge and Its Conservation Copyright © 2009 Pearson Education, Inc. Charge comes in two types, positive and negative; like charges repel and opposite charges attract. Static Electricity; Electric Charge and Its Conservation Copyright © 2009 Pearson Education, Inc. Electric charge is conserved – the arithmetic sum of the total charge cannot change in any interaction. Static Electricity; Electric Charge and Its Conservation Copyright © 2009 Pearson Education, Inc. Atom: Nucleus (small, massive, positive charge) Electron cloud (large, very low density, negative charge) Electric Charge in the Atom Copyright © 2009 Pearson Education, Inc. Polar molecule: neutral overall, but charge not evenly distributed Electric Charge in the Atom Copyright © 2009 Pearson Education, Inc. Conductor: Charge flows freely Metals Insulator: Almost no charge flows Most other materials Some materials are semiconductors. Insulators and Conductors Copyright © 2009 Pearson Education, Inc. Metal objects can be charged by conduction: Induced Charge; the Electroscope Copyright © 2009 Pearson Education, Inc. They can also be charged by induction, either while connected to ground or not: Induced Charge; the Electroscope Copyright © 2009 Pearson Education, Inc. Nonconductors won’t become charged by conduction or induction, but will experience charge separation: Induced Charge; the Electroscope Copyright © 2009 Pearson Education, Inc. The electroscope can be used for detecting charge. Induced Charge; the Electroscope Copyright © 2009 Pearson Education, Inc. The electroscope can be charged either by conduction or by induction. Induced Charge; the Electroscope Copyright © 2009 Pearson Education, Inc. The charged electroscope can then be used to determine the sign of an unknown charge. Induced Charge; the Electroscope Copyright © 2009 Pearson Education, Inc. Experiment shows that the electric force between two charges is proportional to the product of the charges and inversely proportional to the distance between them. Coulomb’s Law Copyright © 2009 Pearson Education, Inc. Coulomb’s law: This equation gives the magnitude of the force between two charges. Coulomb’s Law Copyright © 2009 Pearson Education, Inc. The force is along the line connecting the charges, and is attractive if the charges are opposite, and repulsive if they are the same. Coulomb’s Law Copyright © 2009 Pearson Education, Inc. Unit of charge: coulomb, C. The proportionality constant in Coulomb’s law is then: k = 8.99 x 109 N·m2 /C2 . Charges produced by rubbing are typically around a microcoulomb: 1 μC = 10-6 C. Coulomb’s Law Copyright © 2009 Pearson Education, Inc. Charge on the electron: e = 1.602 x 10-19 C. Electric charge is quantized in units of the electron charge. Coulomb’s Law Copyright © 2009 Pearson Education, Inc. The proportionality constant k can also be written in terms of ε0 , the permittivity of free space: Coulomb’s Law Copyright © 2009 Pearson Education, Inc. Coulomb’s Law Conceptual Example: Which charge exerts the greater force? Two positive point charges, Q1 = 50 μC and Q2 = 1 μC, are separated by a distance . Which is larger in magnitude, the force that Q1 exerts on Q2 or the force that Q2 exerts on Q1? Copyright © 2009 Pearson Education, Inc. Coulomb’s Law Example: Three charges in a line. Three charged particles are arranged in a line, as shown. Calculate the net electrostatic force on particle 3 (the -4.0 μC on the right) due to the other two charges. Copyright © 2009 Pearson Education, Inc. Coulomb’s Law Example: Electric force using vector components. Calculate the net electrostatic force on charge Q3 shown in the figure due to the charges Q1 and Q2 . Copyright © 2009 Pearson Education, Inc. Coulomb’s Law Conceptual Example: Make the force on Q3 zero. In the figure, where could you place a fourth charge, Q4 = -50 μC, so that the net force on Q3 would be zero?
Temperature and Kinetic Theory Quiz:
Two systems are in thermal equilibrium. The two systems are at the same a. Pressure. b. Volume. C. Height. d. pressure and temperature. e.)temperature
The three phases of matter can exist together in equilibrium at the a. critical point. (b) triplepoint. C. melting point. d. boiling point. e. evaporation point.
A student finds a blank liquid bulb thermometer and calibrates the scale by placing it in ice water and placing a reference mark at the level of the fluid. The student places the thermometer in boiling water and finds that the fluid level is 12.0 cm above the reference mark. What is the temperature when the fluid level is 2.4 cm above the reference mark?
Internal human body temperature is often stated to be normal at 98.6°F. What is this temperature on the Kelvin scale?
Temperature, and Ideal Gas Law Slides: Copyright © 2009 Pearson Education, Inc. Temperature is a measure of how hot or cold something is. Most materials expand when heated. Temperature and Thermometers Copyright © 2009 Pearson Education, Inc. Thermometers are instruments designed to measure temperature. In order to do this, they take advantage of some property of matter that changes with temperature. Early thermometers: Temperature and Thermometers Copyright © 2009 Pearson Education, Inc. Common thermometers used today include the liquid-in-glass type and the bimetallic strip. Temperature and Thermometers Copyright © 2009 Pearson Education, Inc. Temperature is generally measured using either the Fahrenheit or the Celsius scale. The freezing point of water is 0°C, or 32°F; the boiling point of water is 100°C, or 212°F. Temperature and Thermometers Copyright © 2009 Pearson Education, Inc. Temperature and Thermometers Example: Taking your temperature. Normal body temperature is 98.6°F. What is this on the Celsius scale? Copyright © 2009 Pearson Education, Inc. The relationship between the volume, pressure, temperature, and mass of a gas is called an equation of state. We will deal here with gases that are not too dense. Boyle’s law: the volume of a given amount of gas is inversely proportional to the pressure as long as the temperature is constant. The Gas Laws and Absolute Temperature Copyright © 2009 Pearson Education, Inc. The volume is linearly proportional to the temperature, as long as the temperature is somewhat above the condensation point and the pressure is constant. Extrapolating, the volume becomes zero at −273.15°C; this temperature is called absolute zero. The Gas Laws and Absolute Temperature Copyright © 2009 Pearson Education, Inc. The concept of absolute zero allows us to define a third temperature scale—the absolute, or Kelvin, scale. This scale starts with 0 K at absolute zero, but otherwise is the same as the Celsius scale. Therefore, the freezing point of water is 273.15 K, and the boiling point is 373.15 K. Finally, when the volume is constant, the pressure is directly proportional to the temperature. The Gas Laws and Absolute Temperature Copyright © 2009 Pearson Education, Inc. The Gas Laws and Absolute Temperature Conceptual Example: Why you should not throw a closed glass jar into a campfire. What can happen if you did throw an empty glass jar, with the lid on tight, into a fire, and why? Copyright © 2009 Pearson Education, Inc. We can combine the three relations just derived into a single relation: What about the amount of gas present? If the temperature and pressure are constant, the volume is proportional to the amount of gas: The Ideal Gas Law Copyright © 2009 Pearson Education, Inc. A mole (mol) is defined as the number of grams of a substance that is numerically equal to the molecular mass of the substance: 1 mol H2 has a mass of 2 g. 1 mol Ne has a mass of 20 g. 1 mol CO2 has a mass of 44 g. The number of moles in a certain mass of material: The Ideal Gas Law Copyright © 2009 Pearson Education, Inc. We can now write the ideal gas law: where n is the number of moles and R is the universal gas constant. The Ideal Gas Law Copyright © 2009 Pearson Education, Inc. Standard temperature and pressure (STP): T = 273 K (0°C) P = 1.00 atm = 1.013 N/m2 = 101.3 kPa. Problem Solving with the Ideal Gas Law Example: Volume of one mole at STP. Determine the volume of 1.00 mol of any gas, assuming it behaves like an ideal gas, at STP. Copyright © 2009 Pearson Education, Inc. Problem Solving with the Ideal Gas Law Example: Helium balloon. A helium party balloon, assumed to be a perfect sphere, has a radius of 18.0 cm. At room temperature (20°C), its internal pressure is 1.05 atm. Find the number of moles of helium in the balloon and the mass of helium needed to inflate the balloon to these values. Copyright © 2009 Pearson Education, Inc. Problem Solving with the Ideal Gas Law Example: Mass of air in a room. Estimate the mass of air in a room whose dimensions are 5.0 m x 3.0 m x 2.5 m high, at STP. Copyright © 2009 Pearson Education, Inc. Problem Solving with the Ideal Gas Law • Volume of 1 mol of an ideal gas is 22.4 L • If the amount of gas does not change: • Always measure T in kelvins • P must be the absolute pressure Copyright © 2009 Pearson Education, Inc. Problem Solving with the Ideal Gas Law Example: Check tires cold. An automobile tire is filled to a gauge pressure of 200 kPa at 10°C. After a drive of 100 km, the temperature within the tire rises to 40°C. What is the pressure within the tire now? Copyright © 2009 Pearson Education, Inc. Since the gas constant is universal, the number of molecules in one mole is the same for all gases. That number is called Avogadro’s number: Ideal Gas Law in Terms of Molecules: Avogadro’s Number Copyright © 2009 Pearson Education, Inc. Therefore we can write: where k is called Boltzmann’s constant. Ideal Gas Law in Terms of Molecules: Avogadro’s Number or Copyright © 2009 Pearson Education, Inc. Ideal Gas Law in Terms of Molecules: Avogadro’s Number Example: Hydrogen atom mass. Use Avogadro’s number to determine the mass of a hydrogen atom. Example: How many molecules in one breath? Estimate how many molecules you breathe in with a 1.0-L breath of air. Copyright © 2009 Pearson Education, Inc. Ideal Gas Temperature Scale—a Standard This standard uses the constant-volume gas thermometer and the ideal gas law. There are two fixed points: Absolute zero—the pressure is zero here The triple point of water (where all three phases coexist), defined to be 273.16 K—the pressure here is 4.58 torr. Copyright © 2009 Pearson Education, Inc. Ideal Gas Temperature Scale—a Standard Then the temperature is defined as: In order to determine temperature using a real gas, the pressure must be as low as possible.
Ideal Gas Law, Thermal Expansion Quiz: 1. As shown in the figure, a bimetallic strip, consisting of metal G on the top and metal H on the bottom, is rigidlyattached to a wall at the left. The coefficient of linear thermal expansion for metal G isgreater han that of metal H. If the strip is uniformly heated, it will G H V curve upward. curve downward. C remain horizontal, but get longer. e. bend in the middle. d. remain horizontal, but get shorter. Oxygen molecules are l6 times more massive than hydrogen molecules. At a given temperature, the average molecular kinetic energy of oxygen molecules, compared to that of hydrogenmolecules, a/isgreater. 6isless. C.is thesame. d. cannot be determined without knowing the pressure and volume. Kg -hnye By what length will a slab of concrete that is originally 18 m long contract when the temperature drops from 24°C to -16°C? The coefficient of linear thermal expansion for this concrete is 1.0 x 10 K!
A quantity of an idealgasis keptn a rigidcontainerofconstantvolume. Ii thegas is originally at a temperature of 19°C, at what temperature will the pressure of the gas double from its original value?
Thermal Expansion and Kinetic Theory Slides:
Copyright © 2009 Pearson Education, Inc. Linear expansion occurs when an object is heated. Here, α is the coefficient of linear expansion. Thermal Expansion Copyright © 2009 Pearson Education, Inc. Thermal Expansion Copyright © 2009 Pearson Education, Inc. Thermal Expansion Example: Bridge expansion. The steel bed of a suspension bridge is 200 m long at 20°C. If the extremes of temperature to which it might be exposed are -30°C to +40°C, how much will it contract and expand? Copyright © 2009 Pearson Education, Inc. Thermal Expansion Conceptual Example: Do holes expand or contract? If you heat a thin, circular ring in the oven, does the ring’s hole get larger or smaller? Copyright © 2009 Pearson Education, Inc. Thermal Expansion Example: Ring on a rod. An iron ring is to fit snugly on a cylindrical iron rod. At 20°C, the diameter of the rod is 6.445 cm and the inside diameter of the ring is 6.420 cm. To slip over the rod, the ring must be slightly larger than the rod diameter by about 0.008 cm. To what temperature must the ring be brought if its hole is to be large enough so it will slip over the rod? Copyright © 2009 Pearson Education, Inc. Thermal Expansion Conceptual Example: Opening a tight jar lid. When the lid of a glass jar is tight, holding the lid under hot water for a short time will often make it easier to open. Why? Copyright © 2009 Pearson Education, Inc. Volume expansion is similar, except that it is relevant for liquids and gases as well as solids: Here, β is the coefficient of volume expansion. For uniform solids, β ≈ 3α. Thermal Expansion Copyright © 2009 Pearson Education, Inc. Thermal Expansion Example: Gas tank in the Sun. The 70-liter (L) steel gas tank of a car is filled to the top with gasoline at 20°C. The car sits in the Sun and the tank reaches a temperature of 40°C (104°F). How much gasoline do you expect to overflow from the tank? Copyright © 2009 Pearson Education, Inc. Water behaves differently from most other solids—its minimum volume occurs when its temperature is 4°C. As it cools further, it expands, as anyone who leaves a bottle in the freezer to cool and then forgets about it can testify. Thermal Expansion Copyright © 2009 Pearson Education, Inc. Atomic and molecular masses are measured in unified atomic mass units (u). This unit is defined so that the carbon-12 atom has a mass of exactly 12.0000 u. Expressed in kilograms: 1 u = 1.6605 x 10-27 kg. Brownian motion is the jittery motion of tiny flecks in water; these are the result of collisions with individual water molecules. Atomic Theory of Matter Copyright © 2009 Pearson Education, Inc. On a microscopic scale, the arrangements of molecules in solids (a), liquids (b), and gases (c) are quite different. Atomic Theory of Matter Copyright © 2009 Pearson Education, Inc. Atomic Theory of Matter Example: Distance between atoms. The density of copper is 8.9 x 103 kg/m3 , and each copper atom has a mass of 63 u. Estimate the average distance between the centers of neighboring copper atoms. Copyright © 2009 Pearson Education, Inc. Two objects placed in thermal contact will eventually come to the same temperature. When they do, we say they are in thermal equilibrium. The zeroth law of thermodynamics says that if two objects are each in equilibrium with a third object, they are also in thermal equilibrium with each other. Thermal Equilibrium and the Zeroth Law of Thermodynamics Copyright © 2009 Pearson Education, Inc. The force exerted on the wall by the collision of one molecule is Then the force due to all molecules colliding with that wall is The Ideal Gas Law and the Molecular Interpretation of Temperature Copyright © 2009 Pearson Education, Inc. The averages of the squares of the speeds in all three directions are equal: So the pressure is: The Ideal Gas Law and the Molecular Interpretation of Temperature Copyright © 2009 Pearson Education, Inc. Rewriting, so The average translational kinetic energy of the molecules in an ideal gas is directly proportional to the temperature of the gas. The Ideal Gas Law and the Molecular Interpretation of Temperature Copyright © 2009 Pearson Education, Inc. Example 18-1: Molecular kinetic energy. What is the average translational kinetic energy of molecules in an ideal gas at 37° C? The Ideal Gas Law and the Molecular Interpretation of Temperature Copyright © 2009 Pearson Education, Inc. We can now calculate the average speed of molecules in a gas as a function of temperature: The Ideal Gas Law and the Molecular Interpretation of Temperature Copyright © 2009 Pearson Education, Inc. The molecules in a gas will not all have the same speed; their distribution of speeds is called the Maxwell distribution: Distribution of Molecular Speeds Copyright © 2009 Pearson Education, Inc. The Maxwell distribution depends only on the absolute temperature. This figure shows distributions for two different temperatures; at the higher temperature, the whole curve is shifted to the right. Distribution of Molecular Speeds
Calorimetry Quiz:
It is a well-known fact that water has a higher specific heat capacity than iron. Now, consider equal masses of water and iron that are initially in thermal equilibrium. The same amount of heat, 30 calories, is added to each. Which statement is true? They remain in thermal equilibrium. They are no longer in thermal equilibrium; the iron is warmer. They are no longer in thermal equilibrium; the water is warmer. d. It is impossible to say without knowing the exact mass involved. C e. It is impossible to say without knowing the exact specific heatcapacities.
2 The heatrequired tochangeasubstancefrom the solid tothe liquid stateis referred to as the heat of fusion. b heatofvaporization. a. C. heat of melting. d. heat of freezing. e. heat of condensation.
3. If 40 kcal of heat is added to 2.0 kg of water, what is the resultingtemperaturechange?
4. A camper is about to drink his morning coffee. He pours 400 grams of coffee, initially at 75.0°C, into a 250-g aluminum cup, initially at 16.0°C. What is the equilibrium temperature of the coffee-cup system, assuming no heat is lost to the surroundings? The specific heat of aluminum is 900J/(kg K). Assume that the specific heat of coffee is the same as the specific heat of water.
Heat and Calorimetry Slides: Copyright © 2009 Pearson Education, Inc. We often speak of heat as though it were a material that flows from one object to another; it is not. Rather, it is a form of energy. Unit of heat: calorie (cal) 1 cal is the amount of heat necessary to raise the temperature of 1 g of water by 1 Celsius degree. Don’t be fooled—the calories on our food labels are really kilocalories (kcal or Calories), the heat necessary to raise 1 kg of water by 1 Celsius degree. Heat as Energy Transfer Copyright © 2009 Pearson Education, Inc. If heat is a form of energy, it ought to be possible to equate it to other forms. The experiment below found the mechanical equivalent of heat by using the falling weight to heat the water: Heat as Energy Transfer 4.186 J = 1 cal 4.186 kJ = 1 kcal Copyright © 2009 Pearson Education, Inc. Definition of heat: Heat is energy transferred from one object to another because of a difference in temperature. • Remember that the temperature of a gas is a measure of the kinetic energy of its molecules. Heat as Energy Transfer Copyright © 2009 Pearson Education, Inc. Heat as Energy Transfer Example: Working off the extra calories. Suppose you throw caution to the wind and eat too much ice cream and cake on the order of 500 Calories. To compensate, you want to do an equivalent amount of work climbing stairs or a mountain. How much total height must you climb? Copyright © 2009 Pearson Education, Inc. The sum total of all the energy of all the molecules in a substance is its internal (or thermal) energy. Temperature: measures molecules’ average kinetic energy Internal energy: total energy of all molecules Heat: transfer of energy due to difference in temperature Internal Energy Copyright © 2009 Pearson Education, Inc. The amount of heat required to change the temperature of a material is proportional to the mass and to the temperature change: The specific heat, c, is characteristic of the material. Some values are listed at left. Specific Heat Copyright © 2009 Pearson Education, Inc. Example: How heat transferred depends on specific heat. (a) How much heat input is needed to raise the temperature of an empty 20-kg vat made of iron from 10°C to 90°C? (b) What if the vat is filled with 20 kg of water? Specific Heat Copyright © 2009 Pearson Education, Inc. Closed system: no mass enters or leaves, but energy may be exchanged Open system: mass may transfer as well Isolated system: closed system in which no energy in any form is transferred For an isolated system, energy out of one part = energy into another part, or: heat lost = heat gained. Calorimetry—Solving Problems Copyright © 2009 Pearson Education, Inc. Calorimetry—Solving Problems Example: The cup cools the tea. If 200 cm3 of tea at 95°C is poured into a 150-g glass cup initially at 25°C, what will be the common final temperature T of the tea and cup when equilibrium is reached, assuming no heat flows to the surroundings? Copyright © 2009 Pearson Education, Inc. The instrument to the left is a calorimeter, which makes quantitative measurements of heat exchange. A sample is heated to a well-measured high temperature and plunged into the water, and the equilibrium temperature is measured. This gives the specific heat of the sample. Calorimetry—Solving Problems Copyright © 2009 Pearson Education, Inc. Calorimetry—Solving Problems Example: Unknown specific heat determined by calorimetry. An engineer wishes to determine the specific heat of a new metal alloy. A 0.150-kg sample of the alloy is heated to 540°C. It is then quickly placed in 0.400 kg of water at 10.0°C, which is contained in a 0.200-kg aluminum calorimeter cup. (We do not need to know the mass of the insulating jacket since we assume the air space between it and the cup insulates it well, so that its temperature does not change significantly.) The final temperature of the system is 30.5°C. Calculate the specific heat of the alloy.
Latent Heat Quiz:
1. A themmally isolated system is made up ofa hot piece of aluminum and a cold piece of copper, with the aluminum and the copper in thermal contact. The specific heat capacity of aluminum is more than double that of copper. Which object experiences the greater magnitude gain or loss of heat during the time the system takes to reach thermal equilibrium? a. the aluminum b. the copper (C) Neither one; both of them experience the same size gain or loss of heat. d. It is impossible to tell without knowing themasses. e. It is impossible to tell without knowing the volumes.
2. The figure shows a graph of the temperature of a pure substance as a function of time as heat is added to it at a constant rate in a closed container. If LF is the latent heat of fusion of this substanceand LV is its latent heat of vaporization, what is the value of the ratio LV/LF? a. 5.0 b. 4.5 7.2 )3.5 e. 1.5
3. How much heat must be removed from 456 g of water at 25.0°C to change it into ice at -10.0°C? The specific heat of ice is 2090 J/kg K, the latentheat of fusion of water is 33.5 x 10ʻ J/kg, and the specific heat of water is 4186 J/kg K.
4. Two experimental runs are performed to determine the calorimetric properties of an alcohol which has a melting point of -10.0° C. In the first run, a 200-g cube of frozen alcohol, at the melting point, is added to 300 gof water at 20.0°C in a styrofoam container. When thermal equilibrium is reached, the alcohol-water solution is at a temperature of 5.0°C. In thesecond run, an identical cube of alcohol is added to 500 g of water at 20.0°C and the temperature at thermal equilibrium is 10.0°C. The specific heat capacity of water is 4190 J/kg K. Assumeno heat isexchangedwith the styrofoam container and the surroundings. What is the heat of fusion of the alcohol?
Latent Heat and Heat Transfer Slides: Copyright © 2009 Pearson Education, Inc. Energy is required for a material to change phase, even though its temperature is not changing. Latent Heat Copyright © 2009 Pearson Education, Inc. Heat of fusion, LF : heat required to change 1.0 kg of material from solid to liquid Heat of vaporization, LV : heat required to change 1.0 kg of material from liquid to vapor Latent Heat Copyright © 2009 Pearson Education, Inc. The total heat required for a phase change depends on the total mass and the latent heat: Latent Heat Example: Will all the ice melt? A 0.50-kg chunk of ice at -10°C is placed in 3.0 kg of “iced” tea at 20°C. At what temperature and in what phase will the final mixture be? The tea can be considered as water. Ignore any heat flow to the surroundings, including the container. Copyright © 2009 Pearson Education, Inc. The latent heat of vaporization is relevant for evaporation as well as boiling. The heat of vaporization of water rises slightly as the temperature decreases. On a molecular level, the heat added during a change of state does not go to increasing the kinetic energy of individual molecules, but rather to breaking the close bonds between them so the next phase can occur. Latent Heat Copyright © 2009 Pearson Education, Inc. Latent Heat Example: Determining a latent heat. The specific heat of liquid mercury is 140 J/kg·°C. When 1.0 kg of solid mercury at its melting point of -39°C is placed in a 0.50-kg aluminum calorimeter filled with 1.2 kg of water at 20.0°C, the mercury melts and the final temperature of the combination is found to be 16.5°C. What is the heat of fusion of mercury in J/kg? Copyright © 2009 Pearson Education, Inc. Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. Heat conduction can be visualized as occurring through molecular collisions. The heat flow per unit time is given by: Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. Conduction Copyright © 2009 Pearson Education, Inc. The constant k is called the thermal conductivity. Materials with large k are called conductors; those with small k are called insulators. Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. Heat Transfer: Conduction, Convection, Radiation Example: Heat loss through windows. A major source of heat loss from a house is through the windows. Calculate the rate of heat flow through a glass window 2.0 m x 1.5 m in area and 3.2 mm thick, if the temperatures at the inner and outer surfaces are 15.0°C and 14.0°C, respectively. Copyright © 2009 Pearson Education, Inc. Convection occurs when heat flows by the mass movement of molecules from one place to another. It may be natural or forced; both these examples are natural convection. Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. Convection Copyright © 2009 Pearson Education, Inc. Convection Copyright © 2009 Pearson Education, Inc. Radiation is the form of energy transfer we receive from the Sun; if you stand close to a fire, most of the heat you feel is radiated as well. The energy radiated has been found to be proportional to the fourth power of the temperature: Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. Heat Transfer: Conduction, Convection, Radiation The constant σ is called the Stefan-Boltzmann constant: The emissivity ε is a number between 0 and 1 characterizing the surface; black objects have an emissivity near 1, while shiny ones have an emissivity near 0. It is the same for absorption; a good emitter is also a good absorber. Copyright © 2009 Pearson Education, Inc. Heat Transfer: Conduction, Convection, Radiation Example: Cooling by radiation. An athlete is sitting unclothed in a locker room whose dark walls are at a temperature of 15°C. Estimate his rate of heat loss by radiation, assuming a skin temperature of 34°C and ε = 0.70. Take the surface area of the body not in contact with the chair to be 1.5 m2 . Copyright © 2009 Pearson Education, Inc. Radiation Copyright © 2009 Pearson Education, Inc. If you are in the sunlight, the Sun’s radiation will warm you. In general, you will not be perfectly perpendicular to the Sun’s rays, and will absorb energy at the rate: Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. This cos θ effect is also responsible for the seasons. Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. Thermography—the detailed measurement of radiation from the body—can be used in medical imaging. Warmer areas may be a sign of tumors or infection; cooler areas on the skin may be a sign of poor circulation. Heat Transfer: Conduction, Convection, Radiation
Thermodynamics Quiz:
An engine manufacturer makes the claim that the engine they have developed will, on each cycle, take 100 J of heat out of boiling water at 100°C, do mechanical work of 80 J, and exhaust 20 J of heat at 10°C. What, if anything, is wrong with this claim? a. The heat exhausted must always be greater than the work done according to the second law ofthermodynamics. b. This engineviolatesthe first law ofthermodynamicsbecause 100J+ 20J # 80 JX c. An engine would operate by taking in heat at the lower temperature and e. exhausting heat at the higher temperature. The efficiency of this engine is greater than the ideal Carnot cycle efficiency. There is nothing wrong with this claim because 100 J= 20 J + 80 J.
The second law of thermodynamics leads us to conclude that a. the total energy of the universe is constant. (6) disorder in the universe is increasing with thepassage of time. c. it is theoretically possible to convert heat into work with 100% efficiency. d. the average temperature of the universe is increasing with the passage of time. e. the average temperature of the universe is decreasing with the passage of time.
During each cycle of operation, a refigerator absorbs 230 Jof heat from the freezer and expels 356 J of heat to the room. How much work input is required in each cycle?
A Carnot cycle engine operates between a low temperature reservoir at 20°C and a high temperature reservoir at 800°C. If the engine is required to output 20.0 kJ of work per cycle, how much heat must the high temperature reservoir transfer to the engine during cach cycle?
The First Law of Thermodynamics Slides: The change in internal energy of a closed system will be equal to the energy added to the system minus the work done by the system on its surroundings. This is the law of conservation of energy, written in a form useful to systems involving heat transfer. © 2009 Pearson Education, Inc. The First Law of Thermodynamics Example: Using the first law. 2500 J of heat is added to a system, and 1800 J of work is done on the system. What is the change in internal energy of the system? Copyright © 2009 Pearson Education, Inc. The First Law of Thermodynamics The first law can be extended to include changes in mechanical energy—kinetic energy and potential energy: Example: Kinetic energy transformed to thermal energy. A 3.0-g bullet traveling at a speed of 400 m/s enters a tree and exits the other side with a speed of 200 m/s. Where did the bullet’s lost kinetic energy go, and what was the energy transferred? Copyright © 2009 Pearson Education, Inc. An isothermal process is one in which the temperature does not change. The First Law of Thermodynamics Applied; Calculating the Work Copyright © 2009 Pearson Education, Inc. In order for an isothermal process to take place, we assume the system is in contact with a heat reservoir. In general, we assume that the system remains in equilibrium throughout all processes. The First Law of Thermodynamics Applied; Calculating the Work Copyright © 2009 Pearson Education, Inc. An adiabatic process is one in which there is no heat flow into or out of the system. The First Law of Thermodynamics Applied; Calculating the Work Copyright © 2009 Pearson Education, Inc. An isobaric process (a) occurs at constant pressure; an isovolumetric one (b) occurs at constant volume. The First Law of Thermodynamics Applied; Calculating the Work Copyright © 2009 Pearson Education, Inc. The work done in moving a piston by an infinitesimal displacement is: The First Law of Thermodynamics Applied; Calculating the Work Copyright © 2009 Pearson Education, Inc. The First Law of Thermodynamics Applied; Calculating the Work For an isothermal process, P = nRT/V. Integrating to find the work done in taking the gas from point A to point B gives: Copyright © 2009 Pearson Education, Inc. The First Law of Thermodynamics Applied; Calculating the Work A different path takes the gas first from A to D in an isovolumetric process; because the volume does not change, no work is done. Then the gas goes from D to B at constant pressure; with constant pressure no integration is needed, and W = PΔV. Copyright © 2009 Pearson Education, Inc. The First Law of Thermodynamics Applied; Calculating the Work Conceptual Example: Work in isothermal and adiabatic processes. Reproduced here is the PV diagram for a gas expanding in two ways, isothermally and adiabatically. The initial volume VA was the same in each case, and the final volumes were the same (VB = VC ). In which process was more work done by the gas? Copyright © 2009 Pearson Education, Inc. The First Law of Thermodynamics Applied; Calculating the Work Example: First law in isobaric and isovolumetric processes. An ideal gas is slowly compressed at a constant pressure of 2.0 atm from 10.0 L to 2.0 L. (In this process, some heat flows out of the gas and the temperature drops.) Heat is then added to the gas, holding the volume constant, and the pressure and temperature are allowed to rise (line DA) until the temperature reaches its original value (TA = TB ). Calculate (a) the total work done by the gas in the process BDA, and (b) the total heat flow into the gas. Copyright © 2009 Pearson Education, Inc. The First Law of Thermodynamics Applied; Calculating the Work Example: Work done in an engine. In an engine, 0.25 mol of an ideal monatomic gas in the cylinder expands rapidly and adiabatically against the piston. In the process, the temperature of the gas drops from 1150 K to 400 K. How much work does the gas do? Copyright © 2009 Pearson Education, Inc. The First Law of Thermodynamics Applied; Calculating the Work The following is a simple summary of the various thermodynamic processes.
The Second Law of Thermodynamics Slides:
Copyright © 2009 Pearson Education, Inc. The first law of thermodynamics tells us that energy is conserved. However, the absence of the process illustrated above indicates that conservation of energy is not the whole story. If it were, movies run backwards would look perfectly normal to us! The Second Law of Thermodynamics Copyright © 2009 Pearson Education, Inc. The second law of thermodynamics is a statement about which processes occur and which do not. There are many ways to state the second law; here is one: Heat can flow spontaneously from a hot object to a cold object; it will not flow spontaneously from a cold object to a hot object. The Second Law of Thermodynamics Copyright © 2009 Pearson Education, Inc. It is easy to produce thermal energy using work, but how does one produce work using thermal energy? This is a heat engine; mechanical energy can be obtained from thermal energy only when heat can flow from a higher temperature to a lower temperature. Heat Engines Copyright © 2009 Pearson Education, Inc. We will discuss only engines that run in a repeating cycle; the change in internal energy over a cycle is zero, as the system returns to its initial state. The high-temperature reservoir transfers an amount of heat QH to the engine, where part of it is transformed into work W and the rest, QL , is exhausted to the lower temperature reservoir. Note that all three of these quantities are positive. Heat Engines Copyright © 2009 Pearson Education, Inc. A steam engine is one type of heat engine. Heat Engines Copyright © 2009 Pearson Education, Inc. The internal combustion engine is a type of heat engine as well. Heat Engines Copyright © 2009 Pearson Education, Inc. Why does a heat engine need a temperature difference? Otherwise the work done on the system in one part of the cycle would be equal to the work done by the system in another part, and the net work would be zero. Heat Engines Copyright © 2009 Pearson Education, Inc. The efficiency of the heat engine is the ratio of the work done to the heat input: Using conservation of energy to eliminate W, we find: Heat Engines Copyright © 2009 Pearson Education, Inc. Heat Engines Example: Car efficiency. An automobile engine has an efficiency of 20% and produces an average of 23,000 J of mechanical work per second during operation. (a) How much heat input is required, and (b) How much heat is discharged as waste heat from this engine, per second? Copyright © 2009 Pearson Education, Inc. Heat Engines No heat engine can have an efficiency of 100%. This is another way of writing the second law of thermodynamics: No device is possible whose sole effect is to transform a given amount of heat completely into work. Copyright © 2009 Pearson Education, Inc. The Carnot engine was created to examine the efficiency of a heat engine. It is idealized, as it has no friction. Each leg of its cycle is reversible. The Carnot cycle consists of: • Isothermal expansion • Adiabatic expansion • Isothermal compression • Adiabatic compression Reversible and Irreversible Processes; the Carnot Engine Copyright © 2009 Pearson Education, Inc. From this we see that 100% efficiency can be achieved only if the cold reservoir is at absolute zero, which is impossible. Real engines have some frictional losses; the best achieve 60–80% of the Carnot value of efficiency. For an ideal reversible engine, the efficiency can be written in terms of the temperature: Reversible and Irreversible Processes; the Carnot Engine Copyright © 2009 Pearson Education, Inc. Reversible and Irreversible Processes; the Carnot Engine Example: A phony claim? An engine manufacturer makes the following claims: An engine’s heat input per second is 9.0 kJ at 435 K. The heat output per second is 4.0 kJ at 285 K. Do you believe these claims? Copyright © 2009 Pearson Education, Inc. Reversible and Irreversible Processes; the Carnot Engine Automobiles run on the Otto cycle, shown here, which is two adiabatic paths alternating with two constant-volume paths. The gas enters the engine at point a and is ignited at point b. Curve cd is the power stroke, and da is the exhaust. Copyright © 2009 Pearson Education, Inc. Reversible and Irreversible Processes; the Carnot Engine Example: The Otto cycle. (a) Show that for an ideal gas as working substance, the efficiency of an Otto cycle engine is e = 1 – (V a /Vb ) 1-γ where γ is the ratio of specific heats (γ = CP /CV ) and V a /Vb is the compression ratio. (b) Calculate the efficiency for a compression ratio V a /Vb = 8.0 assuming a diatomic gas like O2 and N2 . Copyright © 2009 Pearson Education, Inc. These appliances are essentially heat engines operating in reverse. By doing work, heat is extracted from the cold reservoir and exhausted to the hot reservoir. Refrigerators, Air Conditioners, and Heat Pumps Copyright © 2009 Pearson Education, Inc. Refrigerators, Air Conditioners, and Heat Pumps This figure shows more details of a typical refrigerator. Copyright © 2009 Pearson Education, Inc. A heat pump can heat a house in the winter: Refrigerators, Air Conditioners, and Heat Pumps Copyright © 2009 Pearson Education, Inc. Entropy Definition of the change in entropy S when an amount of heat Q is added: if the process is reversible and the temperature is constant. Copyright © 2009 Pearson Education, Inc. Entropy Any reversible cycle can be written as a succession of Carnot cycles; therefore, what is true for a Carnot cycle is true of all reversible cycles. Copyright © 2009 Pearson Education, Inc. Entropy and the Second Law of Thermodynamics Example: Entropy change when mixing water. A sample of 50.0 kg of water at 20.00°C is mixed with 50.0 kg of water at 24.00°C. Estimate the change in entropy. Copyright © 2009 Pearson Education, Inc. Entropy and the Second Law of Thermodynamics The total entropy always increases when heat flows from a warmer object to a colder one in an isolated two-body system. The heat transferred is the same, and the cooler object is at a lower average temperature than the warmer one, so the entropy gained by the cooler one is always more than the entropy lost by the warmer one. Copyright © 2009 Pearson Education, Inc. Entropy and the Second Law of Thermodynamics Example: Entropy changes in a free expansion. Consider the adiabatic free expansion of n moles of an ideal gas from volume V1 to volume V2 , where V2 > V1 . Calculate the change in entropy (a) of the gas and (b) of the surrounding environment. (c) Evaluate ΔS for 1.00 mole, with V2 = 2.00 V1 . Copyright © 2009 Pearson Education, Inc. Entropy and the Second Law of Thermodynamics The fact that after every interaction the entropy of the system plus the environment increases is another way of putting the second law of thermodynamics: The entropy of an isolated system never decreases. It either stays constant (reversible processes) or increases (irreversible processes). Copyright © 2009 Pearson Education, Inc. Entropy is a measure of the disorder of a system. This gives us yet another statement of the second law: Natural processes tend to move toward a state of greater disorder. Example: If you put milk and sugar in your coffee and stir it, you wind up with coffee that is uniformly milky and sweet. No amount of stirring will get the milk and sugar to come back out of solution. Order to Disorder Copyright © 2009 Pearson Education, Inc. Another consequence of the second law: In any natural process, some energy becomes unavailable to do useful work. If we look at the universe as a whole, it seems inevitable that, as more and more energy is converted to unavailable forms, the ability to do work anywhere will gradually vanish. This is called the heat death of the universe. Unavailability of Energy; Heat Death Copyright © 2009 Pearson Education, Inc. 20-11 Thermal Pollution, Global Warming, and Energy Resources Over 90% of the energy used in the U.S. is generated using heat engines to drive turbines and generators—even nuclear power plants use the energy generated from fission heat water for a steam engine. The thermal output QL of all these heat engines contributes to warming of the atmosphere and water. This is an inevitable consequence of the second law of thermodynamics.