Chapter 19 Heat and the First Law of Thermodynamics Copyright © 2009 Pearson Education, Inc. Units of Chapter 19 • Heat as Energy Transfer • Internal Energy • Specific Heat • Calorimetry—Solving Problems • Latent Heat • The First Law of Thermodynamics • The First Law of Thermodynamics Applied; Calculating the Work Copyright © 2009 Pearson Education, Inc. Units of Chapter 19 • Molar Specific Heats for Gases, and the Equipartition of Energy • Adiabatic Expansion of a Gas • Heat Transfer: Conduction, Convection, Radiation Copyright © 2009 Pearson Education, Inc. 19-1 Heat as Energy Transfer 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. Copyright © 2009 Pearson Education, Inc. 19-1 Heat as Energy Transfer 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: 4.186 J = 1 cal 4.186 kJ = 1 kcal Copyright © 2009 Pearson Education, Inc. 19-1 Heat as Energy Transfer 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. Copyright © 2009 Pearson Education, Inc. 19-1 Heat as Energy Transfer Example 19-1: 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. 19-2 Internal Energy 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 Copyright © 2009 Pearson Education, Inc. 19-2 Internal Energy Internal energy of an ideal (atomic) gas: But since we know the average kinetic energy in terms of the temperature, we can write: Copyright © 2009 Pearson Education, Inc. 19-2 Internal Energy If the gas is molecular rather than atomic, rotational and vibrational kinetic energy need to be taken into account as well. Copyright © 2009 Pearson Education, Inc. 19-3 Specific Heat 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. Copyright © 2009 Pearson Education, Inc. 19-3 Specific Heat Example 19-2: 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? Copyright © 2009 Pearson Education, Inc. 19-4 Calorimetry—Solving Problems 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: Copyright © 2009 Pearson Education, Inc. heat lost = heat gained. 19-4 Calorimetry—Solving Problems Example 19-3: 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. 19-4 Calorimetry—Solving Problems 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. Copyright © 2009 Pearson Education, Inc. 19-4 Calorimetry—Solving Problems Example 19-4: 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. Copyright © 2009 Pearson Education, Inc. 19-5 Latent Heat Energy is required for a material to change phase, even though its temperature is not changing. Copyright © 2009 Pearson Education, Inc. 19-5 Latent Heat 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 Copyright © 2009 Pearson Education, Inc. 19-5 Latent Heat The total heat required for a phase change depends on the total mass and the latent heat: Example 19-5: 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. 19-5 Latent Heat Problem Solving: Calorimetry 1. Is the system isolated? Are all significant sources of energy transfer known or calculable? 2. Apply conservation of energy. 3. If no phase changes occur, the heat transferred will depend on the mass, specific heat, and temperature change. (continued) Copyright © 2009 Pearson Education, Inc. 19-5 Latent Heat 4. If there are, or may be, phase changes, terms that depend on the mass and the latent heat may also be present. Determine or estimate what phase the final system will be in. 5. Make sure that each term is in the right place and that all the temperature changes are positive. 6. There is only one final temperature when the system reaches equilibrium. 7. Solve. Copyright © 2009 Pearson Education, Inc. 19-5 Latent Heat Example 19-6: 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. 19-5 Latent Heat 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. Copyright © 2009 Pearson Education, Inc. 19-6 The First Law of Thermodynamics 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. Copyright © 2009 Pearson Education, Inc. 19-6 The First Law of Thermodynamics Example 19-7: 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. 19-6 The First Law of Thermodynamics The first law can be extended to include changes in mechanical energy—kinetic energy and potential energy: Example 19-8: 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. 19-7 The First Law of Thermodynamics Applied; Calculating the Work An isothermal process is one in which the temperature does not change. Copyright © 2009 Pearson Education, Inc. 19-7 The First Law of Thermodynamics Applied; Calculating the Work 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. Copyright © 2009 Pearson Education, Inc. 19-7 The First Law of Thermodynamics Applied; Calculating the Work An adiabatic process is one in which there is no heat flow into or out of the system. Copyright © 2009 Pearson Education, Inc. 19-7 The First Law of Thermodynamics Applied; Calculating the Work An isobaric process (a) occurs at constant pressure; an isovolumetric one (b) occurs at constant volume. Copyright © 2009 Pearson Education, Inc. 19-7 The First Law of Thermodynamics Applied; Calculating the Work The work done in moving a piston by an infinitesimal displacement is: Copyright © 2009 Pearson Education, Inc. 19-7 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. 19-7 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. 19-7 The First Law of Thermodynamics Applied; Calculating the Work Conceptual Example 19-9: 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. 19-7 The First Law of Thermodynamics Applied; Calculating the Work Example 19-10: 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. 19-7 The First Law of Thermodynamics Applied; Calculating the Work Example 19-11: 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. 19-7 The First Law of Thermodynamics Applied; Calculating the Work The following is a simple summary of the various thermodynamic processes. Copyright © 2009 Pearson Education, Inc. 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy For gases, the specific heat depends on the process—the isothermal specific heat is different from the isovolumetric one. Copyright © 2009 Pearson Education, Inc. 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy In this table, we see that the specific heats for gases with the same number of molecules are almost the same, and that the difference CP – CV is almost exactly equal to 2 in all cases. Copyright © 2009 Pearson Education, Inc. 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy For a gas in a constant-volume process, no work is done, so QV = ΔEint. For a gas at constant pressure, QP = ΔEint + PΔV. Comparing these two processes for a monatomic gas when the temperature change is the same gives which is consistent with the measured values. Copyright © 2009 Pearson Education, Inc. 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy In addition, since we expect that This is also in agreement with measurements. Copyright © 2009 Pearson Education, Inc. 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy For a gas consisting of more complex molecules (diatomic or more), the molar specific heats increase. This is due to the extra forms of internal energy that are possible (rotational, vibrational). Copyright © 2009 Pearson Education, Inc. 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy Each mode of vibration or rotation is called a degree of freedom. The equipartition theorem states that the total internal energy is shared equally among the active degrees of freedom, each accounting for ½ kT. The actual measurements show a more complicated situation. Copyright © 2009 Pearson Education, Inc. 19-8 Molar Specific Heats for Gases, and the Equipartition of Energy For solids at high temperatures, CV is approximately 3R, corresponding to six degrees of freedom (three kinetic energy and three vibrational potential energy) for each atom. Copyright © 2009 Pearson Education, Inc. 19-9 Adiabatic Expansion of a Gas For an adiabatic expansion, dEint = -PdV, since there is no heat transfer. From the relationship between the change in internal energy and the molar heat capacity, dEint = nCVdT. From the ideal gas law, PdV + VdP = nRdT. Combining and rearranging gives (CP/CV)PdV + VdP = 0. Copyright © 2009 Pearson Education, Inc. 19-9 Adiabatic Expansion of a Gas Define: Integration then gives the result: Copyright © 2009 Pearson Education, Inc. 19-9 Adiabatic Expansion of a Gas Example 19-12: Compressing an ideal gas. An ideal monatomic gas is compressed starting at point A, where PA = 100 kPa, VA = 1.00 m3, and TA = 300 K. The gas is first compressed adiabatically to state B (PB = 200 kPa). The gas is then further compressed from point B to point C (VC = 0.50 m3) in an isothermal process. (a) Determine VB. (b) Calculate the work done on the gas for the whole process. Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation Heat conduction can be visualized as occurring through molecular collisions. The heat flow per unit time is given by: Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation The constant k is called the thermal conductivity. Materials with large k are called conductors; those with small k are called insulators. Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation Example 19-13: 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. 19-10 Heat Transfer: Conduction, Convection, Radiation Building materials are measured using Rvalues rather than thermal conductivity: Here, l is the thickness of the material. Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation 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. Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation 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: Copyright © 2009 Pearson Education, Inc. 19-10 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. 19-10 Heat Transfer: Conduction, Convection, Radiation Example 19-14: 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. 19-10 Heat Transfer: Conduction, Convection, Radiation 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: Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation This cos θ effect is also responsible for the seasons. Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation 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. Copyright © 2009 Pearson Education, Inc. 19-10 Heat Transfer: Conduction, Convection, Radiation Example 19-15: Star radius. The giant star Betelgeuse emits radiant energy at a rate 104 times greater than our Sun, whereas its surface temperature is only half (2900 K) that of our Sun. Estimate the radius of Betelgeuse, assuming ε = 1 for both. The Sun’s radius is rS = 7 x 108 m. Copyright © 2009 Pearson Education, Inc. Summary of Chapter 19 • Internal energy, Eint, refers to the total energy of all molecules in an object. For an ideal monatomic gas, • Heat is the transfer of energy from one object to another due to a temperature difference. Heat can be measured in joules or in calories. • Specific heat of a substance is the energy required to change the temperature of a fixed amount of matter by 1°C. Copyright © 2009 Pearson Education, Inc. Summary of Chapter 19 • In an isolated system, heat gained by one part of the system must be lost by another. • Calorimetry measures heat exchange quantitatively. • Phase changes require energy even though the temperature does not change. • Heat of fusion: amount of energy required to melt 1 kg of material • Heat of vaporization: amount of energy required to change 1 kg of material from liquid to vapor Copyright © 2009 Pearson Education, Inc. Summary of Chapter 19 • The first law of thermodynamics: ΔEint = Q – W. • Thermodynamic processes: adiabatic (no heat transfer), isothermal (constant temperature), isobaric (constant pressure), isovolumetric (constant volume). • Work done: dW = PdV. • Molar specific heats: CP – CV = R. Copyright © 2009 Pearson Education, Inc. Summary of Chapter 19 • Heat transfer takes place by conduction, convection, and radiation. • In conduction, energy is transferred through the collisions of molecules in the substance. • In convection, bulk quantities of the substance flow to areas of different temperature. • Radiation is the transfer of energy by electromagnetic waves. Copyright © 2009 Pearson Education, Inc.