2 Chapter 20 Second Law of Thermodynamics Chapter 20 opener. There are many uses for a heat engine, such as old steam trains and modern coal-burning power plants. Steam engines produce steam whichdoes work: on turbines to generate electricity, and on a piston that moves linkage to turn locomotive wheels. The efficiency of any engine—no matter how carefully engineered—is limited by nature as described in the second law of thermodynamics. This great law is best stated in terms of a quantity called entropy, which is unlike any other. Entropy is not conserved, but instead is constrained always to increase in any real process. Entropy is a measure of disorder. The second law of thermodynamics tells us that as time moves forward, the disorder in the universe increases. We discuss many practical matters including heat engines, heat pumps, and refrigeration.
3 Units of Chapter 20 The Second Law of Thermodynamics—Introduction Heat EnginesReversible and Irreversible Processes; the Carnot EngineRefrigerators, Air Conditioners, and Heat PumpsEntropyEntropy and the Second Law of Thermodynamics
4 Units of Chapter 20 Order to Disorder Unavailability of Energy; Heat DeathStatistical Interpretation of Entropy and the Second LawThermodynamic Temperature; Third Law of ThermodynamicsThermal Pollution, Global Warming, and Energy Resources
5 20-1 The Second Law of Thermodynamics—Introduction 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!Figure Have you ever observed this process, a broken cup spontaneously reassembling and rising up onto a table? This process could conserve energy and other laws of mechanics.
6 20-1 The Second Law of Thermodynamics—Introduction 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.
7 20-2 Heat EnginesIt 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.Figure Schematic diagram of energy transfers for a heat engine.
8 20-2 Heat EnginesWe 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.
9 20-2 Heat Engines A steam engine is one type of heat engine. Figure Steam engines.
10 20-2 Heat EnginesThe internal combustion engine is a type of heat engine as well.Figure Four-stroke-cycle internal combustion engine: (a) the gasoline–air mixture flows into the cylinder as the piston moves down; (b) the piston moves upward and compresses the gas; (c) the brief instant when firing of the spark plug ignites the highly compressed gasoline–air mixture, raising it to a high temperature; (d) the gases, now at high temperature and pressure, expand against the piston in this, the power stroke; (e) the burned gases are pushed out to the exhaust pipe; when the piston reaches the top, the exhaust valve closes and the intake valve opens, and the whole cycle repeats. (a), (b), (d), and (e) are the four strokes of the cycle.
11 20-2 Heat EnginesWhy 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.
12 20-2 Heat EnginesThe 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:
13 20-2 Heat Engines Example 20-1: 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.How much heat input is required, and(b) How much heat is discharged as waste heat from this engine, per second?Solution: a. We know W and e; solving for QH gives QH = 115 kJ.b. Now solve for QL: QL = 92 kJ.
14 20-2 Heat EnginesNo 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.Figure Schematic diagram of a hypothetical perfect heat engine in which all the heat input would be used to do work.
15 20-3 Reversible and Irreversible Processes; the Carnot Engine 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 expansionAdiabatic expansionIsothermal compressionAdiabatic compressionFigure The Carnot cycle. Heat engines work in a cycle, and the cycle for the Carnot engine begins at point a on this PV diagram. (1) The gas is first expanded isothermally, with the addition of heat QH, along the path ab at temperature TH. (2) Next the gas expands adiabatically from b to c—no heat is exchanged, but the temperature drops to TL. (3) The gas is then compressed at constant temperature TL, path cd, and heat QL flows out. (4) Finally, the gas is compressed adiabatically, path da, back to its original state. No Carnot engine actually exists, but as a theoretical idea it played an important role in the development of thermodynamics.
16 20-3 Reversible and Irreversible Processes; the Carnot Engine For an ideal reversible engine, the efficiency can be written in terms of the temperature: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.
17 20-3 Reversible and Irreversible Processes; the Carnot Engine Example 20-2: 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?Solution: The claimed efficiency of the engine is 56%; however, the theoretical maximum efficiency of an ideal heat engine operating between these temperatures is 34%. This engine would violate the 2nd law of thermodynamics if it operated as claimed (which it won’t).
18 20-3 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.Figure The Otto cycle.
19 20-3 Reversible and Irreversible Processes; the Carnot Engine Example 20-3: The Otto cycle.Show that for an ideal gas as working substance, the efficiency of an Otto cycle engine ise = 1 – (Va/Vb)1-γwhere γ is the ratio of specific heats (γ = CP/CV) and Va/Vb is the compression ratio.(b) Calculate the efficiency for a compression ratio Va/Vb = 8.0 assuming a diatomic gas like O2 and N2.Solution: a. See text for complete derivation.b. For a diatomic molecule, γ = 1.4, giving e = 0.56.
20 20-4 Refrigerators, Air Conditioners, and Heat Pumps 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.Figure Schematic diagram of energy transfers for a refrigerator or air conditioner.
21 20-4 Refrigerators, Air Conditioners, and Heat Pumps This figure shows more details of a typical refrigerator.Figure (a) Typical refrigerator system. The electric compressor motor forces a gas at high pressure through a heat exchanger (condenser) on the rear outside wall of the refrigerator where QH is given off and the gas cools to become liquid. The liquid passes from a high-pressure region, via a valve, to low-pressure tubes on the inside walls of the refrigerator; the liquid evaporates at this lower pressure and thus absorbs heat (QL) from the inside of the refrigerator. The fluid returns to the compressor where the cycle begins again.(b) Schematic diagram, like Fig. 20–9.
22 20-4 Refrigerators, Air Conditioners, and Heat Pumps Refrigerator performance is measured by the coefficient of performance (COP):Substituting:For an ideal refrigerator,
23 20-4 Refrigerators, Air Conditioners, and Heat Pumps Example 20-4: Making ice.A freezer has a COP of 3.8 and uses 200 W of power. How long would it take this otherwise empty freezer to freeze an ice-cube tray that contains 600 g of water at 0°C?Solution: The total amount of energy that needs to be removed from the water in order to freeze it is 2.0 x 105 J. The refrigerator does work at a rate of 200 J/s, and the amount of work it needs to do in order to remove 2.0 x 105 J of energy from the water is 2.0 x 105 J/COP. It can do this in a time of 260 seconds.
24 20-4 Refrigerators, Air Conditioners, and Heat Pumps A heat pump can heat a house in the winter:Figure A heat pump uses an electric motor to “pump” heat from the cold outside to the warm inside of a house.
25 20-4 Refrigerators, Air Conditioners, and Heat Pumps A heat pump is evaluated on a different type of “Coefficient of Performance” COP:COP (heat pump) = QH/Wvs.COP (refrigerator) = Qc/WFigure A heat pump uses an electric motor to “pump” heat from the cold outside to the warm inside of a house.
26 Example 20-5: Heat pump.A heat pump has a coefficient of performance of 3.0* and is rated to do work at 1500 W.How much heat can it add to a room per second?COP = 3.0 = Qh/W => Qh = 3.0 x W = 4500 W = 4500 Joule/Sec so… 4500 J per second, or ~4.3 BTUSolution: a. It can add 4500W (4500 J/s).b. COP = 2.0 (reversing the inside and the outside of the house)
27 Example 20-5: Heat pump.A heat pump has a coefficient of performance of 3.0* and is rated to do work at 1500 W.(b) If the heat pump were turned around to act as an air conditioner in the summer, what would you expect its coefficient of performance to be, assuming all else stays the same?Qc = Qh – W = 4500 J – 1500 J = 3000 JAnd COP (refrigerator) = Qc/W = 3000/1500 = 2Solution: a. It can add 4500W (4500 J/s).b. COP = 2.0 (reversing the inside and the outside of the house)
28 20-5 Entropy You can’t measure the “heat” of something! You CAN measure how much heat must be added to change something’s temperature.Consider isothermal process (T constant):How much heat must you add as the gas expands to keep it at T?PV
29 20-5 EntropyHow much heat must you add as the gas expands to keep it at T?DU = Qin – Wout = 0 (isothermal)Q in = W out = nRT*ln(Vf/Vi)Define Entropy ChangeDS = Q/TPFor this reversible isothermal process,DS = nR*ln(Vf/Vi)V
30 20-5 Entropy Entropy = “S” A *state* variable for a gas (P, V, T) A measure of “disorder” of a systemAn indicator of the likelihood (probability) of reaction directions (towards higher S)Measured only when it *changes*
31 20-5 EntropyDefinition of the change in entropy S when an amount of heat Q is added:if the process is reversible and the temperature is constant. Otherwise,
32 20-5 EntropyIF the temperature is NOT constant:
33 20-5 Entropy Consider Carnot cycles: TH/TL = QH/QL So… QH/TH = QL/TL Figure Any reversible cycle can be approximated as a series of Carnot cycles. (The dashed lines represent isotherms.)
34 20-5 EntropySince for any Carnot cycle QH/TH + QL/TL = 0, we see that the integral of dQ/T around a closed path is zero. This means that entropy is a state variable—the change in its value depends only on the initial and final states.Figure The integral of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b is the same for path I as for path II.
35 20-5 EntropyAny 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.Figure Any reversible cycle can be approximated as a series of Carnot cycles. (The dashed lines represent isotherms.)
36 20-5 EntropySince for any Carnot cycle QH/TH + QL/TL = 0, if we approximate any reversible cycle as an infinite sum of Carnot cycles, we see that the integral of dQ/T around a closed path is zero. This means that entropy is a state variable—the change in its value depends only on the initial and final states.Figure The integral of the entropy for a reversible cycle is zero. Hence the difference in entropy between states a and b is the same for path I as for path II.
37 20-6 Entropy and the Second Law of Thermodynamics Example 20-6: 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.Solution: The final water temperature is 22.00°C. The heat flow out of the warmer water equals the heat flow into the cooler water. To calculate the change in entropy for each, use the average temperature 21.00°C for the cooler water and 23.00°C for the warmer. This gives J/K J/K = 10 J/K.
38 20-6 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.
39 20-6 Entropy and the Second Law of Thermodynamics Example 20-7: 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.Solution. a. ΔSgas = nR ln (V2/V1)b. ΔSenv = 0.c. ΔSgas = 5.76 J/K.
40 20-6 Entropy and the Second Law of Thermodynamics Example 20-8: Heat transfer.A red-hot 2.00-kg piece of iron at temperature T1 = 880 K is thrown into a huge lake whose temperature is T2 = 280 K. Assume the lake is so large that its temperature rise is insignificant. Determine the change in entropy(a) of the iron and(b) of the surrounding environment (the lake).Solution: a. ΔSiron = − mc ln (T1/T2) = J/K.b. ΔSenv = Q/T = 1930 J/K.
41 20-6 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).
42 20-7 Order to DisorderEntropy 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.
43 20-7 Order to DisorderAnother example: When a tornado hits a building, there is major damage. You never see a tornado approach a pile of rubble and leave a building behind when it passes.Thermal equilibrium is a similar process—the uniform final state has more disorder than the separate temperatures in the initial state.
44 20-8 Unavailability of Energy; Heat Death 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.
45 Entropy vs. EnthalpyEnthalpy = Internal Energy PLUS Energy in the surrounding system:The enthalpy of a system is defined as:H = U + pVAnd the change in enthalpy =DH = DU + D (pV)For constant pressure processes:DH = DU + pDV = DU +Work = Q!
46 Entropy vs. EnthalpyEnthalpy is sometimes described as the heat content of a system under a given pressure, Such a visualization assumes no energy exchange with the environment other than heat or expansion work. Given such restrictions, it can be shown that:The enthalpy is the total amount of energy which the system can emit through heat,Adding or removing energy through heat is the only way to change the enthalpy, andThe amount of change in enthalpy is equal to the amount of energy added through heat.
47 Entropy vs. EnthalpyENTHALPHY: Is the energy content of a process (chemical, thermodynamic, mechanical, etc) that can be recovered. It is also described as useful energy. ENTROPY: Is the energy content of a process (chemical, thermodynamic, mechanical, etc) that CANNOT be recovered. It is also described as chaos.
48 20-9 Statistical Interpretation of Entropy and the Second Law Microstate: a particular configuration of atomsMacrostate: a particular set of macroscopic variablesThis example uses coin tosses:
49 20-9 Statistical Interpretation of Entropy and the Second Law Similarly, the most probable distribution of velocities in a gas is Maxwellian:The most probable state is the one with the greatest disorder, or the greatest entropy. With k being Boltzmann’s constant and W the number of microstates,Figure (a) Most probable distribution of molecular speeds in a gas (Maxwellian, or random); (b) orderly, but highly unlikely, distribution of speeds in which allmolecules have nearly the same speed.
50 20-9 Statistical Interpretation of Entropy and the Second Law Example 20-9: Free expansion—statistical determination of entropy.Determine the change in entropy for the adiabatic free expansion of one mole of a gas as its volume doubles. Assume W, the number of microstates for each macrostate, is the number of possible positions.Solution: A doubling of the volume gives twice as many possibilities per molecule, so the number of microstates increases by a factor of 2NA. Then the change in entropy is k NA ln 2, or R ln 2.
51 20-9 Statistical Interpretation of Entropy and the Second Law In this form, the second law of thermodynamics does not forbid processes in which the total entropy decreases; it just makes them exceedingly unlikely.
52 20-10 Thermodynamic Temperature; Third Law of Thermodynamics Since the ratio of heats exchanged between the hot and cold reservoirs in a Carnot engine is equal to the ratio of temperatures, we can define a temperature scale using the triple point of water:T = (273.16K)(Q/Qtp).Here, Q and Qtp are the heats exchanged by a Carnot engine with reservoirs at temperatures T and Ttp.
53 20-10 Thermodynamic Temperature; Third Law of Thermodynamics Also, since the maximum efficiency of a heat engine isthere is no way to achieve a temperature of absolute zero. This is the third law of thermodynamics:It is not possible to reach absolute zero in any finite number of processes.
54 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.
55 Summary of Chapter 20 Heat engine changes heat into useful work Efficiency: work/heat inputMaximum efficiency: 1 – TL/THRefrigerators and air conditioners are heat engines, reversed; COP = heat removed/workHeat pump: COP = heat delivered/workSecond law of thermodynamics: Natural processes always tend to increase entropy
56 Summary of Chapter 20 Entropy change in reversible process: Change in entropy gives direction to “arrow of time”As time goes on, energy becomes degraded.Heat engines cause thermal pollution.