Thermodynamics. Slide 1 / 93. Slide 2 / 93. Slide 3 / 93. Chemical Thermodynamics

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1 Slide 1 / 93 Slide 2 / 93 Thermodynamics hemical Thermodynamics Slide 3 / 93 The Golden Gate Bridge is painted regularly to slow down the inevitable rusting of the iron on the bridge. In this unit you will study how heat and temperature relate to work and energy and apply principles of thermodyamics to predict when chemical reactions will occur.

2 First Law of Thermodynamics Slide 4 / 93 Recall a system is a portion of the universe that has been chosen for studying the changes that take place within it in response to varying conditions. A system can be relatively simple, like a glass of water, or it can be complex, like a planet, or the entire Universe can be considered a system. First Law of Thermodynamics Within every system exits a property called energy. Slide 5 / 93 In physics we learned about kinetic and potential energy. #E = W This year, we extend that by adding another way to change the energy of a system; by the flow of Heat (q). When two objects of different temperature are in contact, heat flow results in an increase of the energy of the cooler object and an identical decrease of the energy of the hotter object. T = 20 A B T = 10 heat flow #E = w + q The First Law of Thermodynamics #E = w + q The First Law of Thermodynamics tells us that energy cannot be created or destroyed. In other words the total energy of the universe is a constant. The same is true of any closed system. The First Law allows any process in which the total energy is conserved, including those where energy changes forms. Slide 6 / 93 Internal energy, E Initial state E < E 0 #E < 0 (-) Final state E E of system decreases E 0 Energy lost to surroundings. Internal energy, E Final state E > E 0 #E > 0 (+) Initial state E E 0 E of system increases Energy gained from surroundings

3 First Law of Thermodynamics Slide 7 / 93 Most of the processes in the natural world that involve transfer of energy from one form to another don't just happen naturally. For example, gold does not rust in the same way iron does. 4Au(s) + 3O 2(g) --> 2Au 2O 3(s) doesn't happen 4Fe(s) + 3O 2(g) --> 2Fe 2O 3(s) does happen As reserves of fossil fuels run low, people say we have an energy crisis. But if the First law of Thermodynamics is true, energy cannot be created or destroyed, so we're not actually running out of energy. What do people really mean? The First Law of Thermodynamics The First Law of Thermodynamics applies to any closed system. If our system is a cup, resting on a ledge at a certain height, we know the cup has potential energy and if it falls that energy is transfered to kinetic energy and thermal energy. Slide 8 / 93 E 0 + W = E f In this process as our system is definited, total energy remains conserved. If the initial and final energy of the system are equal to each other, why can't the process happen in reverse? Why don't we ever see a broken cup reassemble and return back to its initial position on the ledge? The Second Law of Thermodynamics Slide 9 / 93 The Second Law is a statement about which processes occur and which do not. There are many ways to state the second law: Heat can flow spontaneously from a hot object to a cold object; but not from a cold object to a hot object. It is impossible to build a perpetual motion machine. The universe always gets more disordered with time. Your bedroom will get increasingly messy unless you keep cleaning it up.

4 2nd Law: Order to isorder Slide 10 / 93 Natural processes tend to move toward a state of greater disorder. Stir sugar into coffee and you get coffee that is uniformly sweet. No amount of stirring will get the sugar back out. When a tornado hits a building, there is major damage. You never see a tornado pass through a pile of rubble and leave a building behind. You never walk past a lake on a summer day and see a puff of steam rise up, leaving a frozen lake behind. The First Law of Thermodynamics maintains that the above scenarios are possible. The Second Law maintains that they won't naturally occur. 2nd Law: Order to isorder Slide 11 / 93 The Second Law tell us which processes are naturally favorable - that is they can occur without more energy being put in than is released. Favorable doesn't mean fast, it just means that it will naturally occur if a system is left on its own. Thermodynamically Favorable Slide 12 / 93 Once the valve is opened, the gas in vessel B will effuse into vessel A and vice versa, but once the the gases are mixed, they will not spontaneously unmix. The mixing of these gases is favorable because there is much higher probability of the gases being mixed than unmixed. A thermodynamically favorable process is not reversible.

5 Favorable Processes Slide 13 / 93 Processes that are favorable at one temperature may be not favorable at other temperatures. FOR EXAMPLE favorable at T > 0 favorable at T < 0 1 A reaction that is thermodynamically favorable. Slide 14 / 93 A B E is very rapid will proceed without a net increase in energy is also spontaneous in the reverse direction has an equilibrium position that lies far to the left is very slow 2 Which of the following statements is true? Slide 15 / 93 A B Processes that are favorable in one direction are not favorable in the opposite direction. Processes are favorable because they occur at an observable rate. Favorability can depend on the temperature. A and are true

6 Reversible Processes Slide 16 / 93 In a reversible process the system changes in such a way that the system and surroundings can be put back in their original states by exactly reversing the process. Surroundings System T-#T T Surroundings System T+#T T +q Heat Heat -q Endothermic Exothermic Irreversible Processes Slide 17 / 93 Piston Movable partition Vacuum Gas work Irreversible processes cannot be undone by exactly reversing the change to the system. Thermodynamically favorable processes are irreversible. 3 A reversible process is one that. Slide 18 / 93 A B E can be reversed with no net change in either system or surroundings is thermodynamically favorable is thermodynamically unfavorable must be carried out at low temperature must be carried out at high temperature

7 Entropy Slide 19 / 93 Entropy (S) is a term coined by Rudolph lausius in the 19th century. Entropy refers to the ratio of heat to the temperature at which the heat is delivered: S = q T Entropy Slide 20 / 93 Entropy can be thought of as a measure of the randomness of a system, or as a measure of the number of ways of arranging particles. It is related to the various modes of motion in molecules. Like total energy, E, and enthalpy, H, entropy is a state function. As a result, we are interested in measuring the change in entropy #S, as opposed to the absolute entropy, S #S = S final - S initial Entropy Slide 21 / 93 For a process occurring at constant temperature, the change in entropy is equal to the heat that would be transferred if the process were reversible divided by the temperature: #S = q rev T Isothermal process

8 Slide 22 / 93 Second Law of Thermodynamics The entropy of the universe increases for thermodynamically favorable processes and The entropy of the universe does not change for reversible processes. Second Law of Thermodynamics Slide 23 / 93 In other words: For reversible processes: S= #S system + #S surroundings = 0 For irreversible processes: #S= #S system + #S surroundings > 0 This means that the entropy of the universe constantly increases. 4 The thermodynamic quantity that expresses the degree of disorder in a system is. Slide 24 / 93 A B E enthalpy internal energy bond energy entropy heat flow

9 5 For an isothermal (constant temperature) process, #S =. Slide 25 / 93 A B E q q rev / T q rev Tq rev q + w 6 Which one of the following is always positive when a thermodynamically favorable process occurs? Slide 26 / 93 A B #S system #S surroundings #S universe E #H universe #H surroundings 7 The entropy of the universe is. Slide 27 / 93 A B E constant continually decreasing continually increasing zero the same as the energy, E

10 Entropy on the Molecular Scale Slide 28 / 93 Ludwig Boltzmann described the concept of entropy on the molecular level by using statistical analysis ** Statistical Interpretation of Entropy and the Second Law Slide 29 / 93 A macrostate of a system is specified by giving its macroscopic properties temperature, pressure, and so on. T = 16 P = 1 atm A microstate of a system describes the position and velocity of every particle. For every macrostate, there are one or more microstates. ** Statistical Interpretation of Entropy and the Second Law Slide 30 / 93 A simple example: tossing four coins. The macrostates describe how many heads and tails there are; the microstates list the different ways of achieving that macrostate.

11 ** Statistical Interpretation of Entropy and the Second Law Assume each microstate is equally probable; the probability of each macrostate then depends on how many microstates are in it. The number of microstates quickly becomes very large if we have even 100 coins instead of four. Slide 31 / 93 ** Statistical Interpretation of Entropy and the Second Law probabilities of various macrostates for 100 coin tosses Slide 32 / 93 Macrostate heads tails # of microstates Probability x x x x x x x x x x x x x x x10-31 This table lists some of the possible outcomes (macrostates) for 100 coin tosses, how many microstates they have, and the relative probability that each macrostate will occur. Note that the probability of getting fewer than 20 heads or tails is extremely small. ** Statistical Interpretation of Entropy and the Second Law Slide 33 / 93 The second law does not forbid certain processes; all microstates can occur with equal probability. However, some processes are extremely unlikely a lake freezing on a hot summer day, broken mug re-assembling itself; all the air in a room moving into a single corner. Tossing a coin 100 times led to some macrostates being so unlikely that they will probably not ever occur...now think of the odds with even one mole of matter (6 x particles)...some macrostates become so rare as to be effectively impossible in the lifetime of the universe.

12 ** Entropy on the Molecular Scale egrees of Freedom: Some molecules exhibit several types of motion Translational: Movement of the entire molecule from one place to another. Slide 34 / 93 Vibrational: Periodic motion of atoms within a molecule. Rotational: Rotation of the molecule on about an axis or rotation about s bonds. ** Entropy on the Molecular Scale Slide 35 / 93 Boltzmann envisioned the motions of a sample of molecules at a particular instant in time. This would be akin to taking a snapshot of all the molecules. Each thermodynamic state has a specific number of microstates, W, associated with it. Entropy can then be defined as: S = k lnw As the microstates increases, entropy (S) increases where k is the Boltzmann constant, 1.38 # 10 # 23 J/K. ** Entropy on the Molecular Scale Slide 36 / 93 The change in entropy for a process, then, is #S = k lnw final # k lnw initial #S = k ln W final W initial Entropy increases with the number of possible microstates.

13 ** Entropy on the Molecular Scale Slide 37 / 93 The number of microstates and, therefore, the entropy tends to increase with increases in: Temperature Volume The number of independently moving molecules The number of degrees of freedom of each molecule 8 Which of the following has the largest entropy? Slide 38 / 93 A B E They are all the same. Entropy is independent of temperature. ** Entropy and Physical States Slide 39 / 93 Entropy increases with the freedom of motion of molecules. S(g) > S(l) > S(s)

14 9 Which of the following phase changes would result in an increase in entropy? Slide 40 / 93 A L --> g B g --> s L --> s A and B E B and Entropy Slide 41 / _ 2+ _2+ _ 2+ _ 2+ When substances are dissolved in one another, such as when a solid is dissolved in a solvent, entropy increases. 10 In which of the following instances would entropy increase Slide 42 / 93 A Gases are formed from liquids and solids B Liquids or solutions are formed from solids The number of gas molecules increases The number of moles increases E All of the above

15 Third Law of Thermodynamics Slide 43 / 93 The entropy of a pure crystalline substance at absolute zero is 0. Solid crystal ordered arrangement Liquid less ordered more freedom an entropy ever equal 0? Standard Entropies Entropy values tend to increase with increasing molar mass as the molecules have more degrees of freedom. Entropy values are small compared to enthalpy. Therefore entropy values are expressed in J/mol-K. Although entropy values are always positive, the change in entropy, #S, can be either positive or negative since Substance H2(g) N2(g) O2(g) H2O(g) NH3(g) H3OH(g) 6H6(g) H2O(l) H3OH(l) 6H6(l) Li(s) Na(s) K(s) Fe(s) Fel3(s) Nal(s) S 0, J/mol-K Slide 44 / 93 Standard Entropies Slide 45 / 93 Larger and more complex molecules have greater entropies. H H H H H H H H H H H H Methane, H 4 #S = J/mol K Propane, 3H 8 #S = J/mol K

16 Predicting Entropy hanges Slide 46 / 93 To determine whether entropy increases or decreases in a reaction: FIRST, look at how the phases of the reactants and products are changing: Increase in S ecrease in S (l) to (g) vaporization (s) to (g) sublimation (s) to (l) melting (g) to (l) condensation (g) to (s) deposition (l) to (s) freezing * Predicting Entropy hanges Second, look at whether the number of gas moles is increasing or decreasing: Slide 47 / 93 Increase in S 2SO 3(g) --> 2S(s,rhombic) + 3O 2 ecrease in S 2H 2(g) + O 2(g) --> 2H 2O Negligible change in S H 2(g) + l 2(g) --> 2Hl(g) Lastly, check to see whether the number of moles increases or decreases. 11 Which of the following statements is false? A B The change in entropy in a system depends on the initial and final states of the system and the path taken from one state to the other. Any irreversible process results in an overall increase in entropy. The total entropy of the universe increases in any spontaneous process. Entropy increases with the number of microstates of the system. Slide 48 / 93

17 12 Of the following, the entropy of is the largest. A B E Hl (l) Hl (s) Hl (g) HBr (g) HI (g) Slide 49 / Of the following, the entropy of gaseous is the largest at 25 o and 1 atm. Slide 50 / 93 A H 2 B 2H 6 2H 2 H 4 E 2H 4 14 The entropy of a pure crystalline substance at 0 o is zero. Slide 51 / 93 True False

18 15 Which one of the following processes produces a decrease in the entropy of the system? Slide 52 / 93 A B E boiling water to form steam dissolution of solid Kl in water mixing of two gases into one container freezing water to form ice melting ice to form water 16 #S is positive for. Slide 53 / 93 A B E 2H 2(g) + O 2(g) --> 2H 2O(g) 2NO 2(g) --> N 2O 4(g) O 2(g) --> O 2(s) BaF 2(s) --> Ba 2+ (aq) + 2F - (aq) 2Hg(l) + O 2(g) --> 2HgO(s) 17 For which of the following processes would #S be negative? Slide 54 / 93 A H 2O(l) --> H 2O (g) B ao 3(s) --> ao(s) + O 2(g) a 2+ (aq) + O 2-3 (aq) --> ao 3(s) 2NH 3(g) --> N 2(g) + 3H 2(g) E A and

19 alculating Entropy hanges Slide 55 / 93 The entropy change, #S# #for a reaction can be estimated in the same way we estimated enthalpy change, #H: S# = # #ns#(products)) - # #ms#(reactants)) where n and m are the coefficients in the balanced chemical equation. 18 The value of #S o for the catalytic hydrogenation of acetylene to ethane is J/K mol. A B E H 2(g) + H 2(g) # 2H 4(g) Slide 56 / The value of #S o for the oxidation of carbon to carbon dioxide, is J/K. A B E +2.9 (s, graphite) + O 2(g) # O 2(g) Slide 57 / 93

20 Entropy hanges in Surroundings Slide 58 / 93 Heat that flows into or out of the system changes the entropy of the surroundings. For an isothermal process: #S surr = - qsys T At constant pressure, q sys is simply #H o for the system, so #S surr = - #Hsys T Entropy hange in the Universe Slide 59 / 93 The universe is composed of the system and the surroundings. Therefore, #S universe = #S system + #S surroundings For thermodynamically favorable processes... #S universe > 0 Gibbs Free Energy Slide 60 / 93 #S universe = #S system + #S surroundings #S surr = - #Hsys T #S universe = #S system + - #Hsys T Multiply both sides by (-T) -T#S universe = #H system - T#S system -T#S universe = #G #G = #H system - T#S system

21 Entropy hange in the Universe Slide 61 / 93 #G refers to the change in Gibbs free energy This is the energy available to do work. #G = #H - T#S This is called the Gibbs equation. # Gibbs Free Energy -T S universe = #G Slide 62 / 93 When #S universe is positive, #G is negative. If #S universe is positive this means the reaction is irreversible. Irreversible reactions/processes are spontaneous. When #G is negative, a process is thermodynamically favorable. # Gibbs Free Energy Slide 63 / 93 Free energy N2 + 3H2 <--> 2NH3 spontaneous spontaneous Equilibrium ourse of reaction If #G is negative, the reaction is favorable. If #G is 0, the system is at equilibrium. If #G is positive, the reaction is unfavorable. However, the reverse reaction would be favorable. NN NN NN HH NN N HH HH HH N HH HH HH HH HH pure N 2 +H 2 HH HH HH HH HH N HH N NN HH N HH N HH HH N NN HH N Equilibrium mixture ΔG=0 HH N NHH HH N HH N NHH NHH HH N NHH HH N NHH NHH NHH NHH NHH HH N pure NH3

22 20 Which of the following MUST be true for a reaction to be thermodynamically favorable? Slide 64 / 93 A #G is positive B #G is negative #S universe is positive #S system is positive E B and # Standard Free Energy hanges Slide 65 / 93 Analogous to standard enthalpies of formation are standard free energies of formation, G f#. #G# = # #n#g# f(products) - #(m#g# f (reactants)) where n and m are the stoichiometric coefficients As it was for ΔH f 0 (but not ΔS f0 ), the ΔG f 0 for elements in their standard state is zero # Slide 66 / The standard Gibbs free energy of formation of is zero. (I) Al (s) (II) Br 2 (l) (III) Hg (l) A B E I only II only III only II and III I, II, and III

23 * 22 The value of ΔG o at 25 o for the decomposition of gaseous sulfur trioxide to solid elemental sulfur and gaseous oxygen, as shown below, is kj/ mol. A B E SO 3(g) à 2S(s, rhombic) + 3O 2(g) Sulfur S (s, rhombic) SO 2(g) SO 3(g) Oxygen #H 0 f #G 0 f S (kj/mol) (kj/mol) (J/k-mol) O 2(g) Slide 67 / 93 # Slide 68 / The standard Gibbs free energy of formation of is zero. A B E (I) H 2O(l) (II) Na(s) (III) H 2(g) I only II only III only II and III I, II, and III * P 2(g) + 3l 2(g) à 2Pl 3(g) 24 The value of ΔG o at 25 o for the formation of phosphorous trichloride from its constituent elements, as shown below, is kj/mol. Slide 69 / 93 A B E #H 0 f #G 0 f S (kj/mol) (kj/mol) (J/k-mol) hlorine l 2 (g) % l - (aq) Phosphorous P 2 (g) Pl 3(g) POl 3(g)

24 Free Energy hanges Slide 70 / 93 At temperatures other than 25 o, #G# = #H# - T #S# How does #G# change with temperature? Free Energy and Temperature Slide 71 / 93 There are two parts to the free energy equation: #H# is the enthalpy term T#S# is the entropy term Note that the temperature dependence of free energy comes from the entropy term. #G = #H# -T#S# When you use the Gibbs equation above, be sure to have both enthalpy and entropy terms in the same units. Generally, it is easiest to divide #S by 1000 in order for it to be in kj/mol-k. S(s, rhombic) + O 2(g) à SO 2(g) 25 The value of ΔG o at 373 K for the oxidation of solid elemental sulfur to gaseous sulfur dioxide, as shown below, is kj/mol. The ΔH o for this reaction is kj/mol, and ΔS o is J/K. Slide 72 / 93 A B ,597 #G = #H -T#S #G = # #n#g# f(products) - #(m#g# f (reactants)) +4,597 E

25 #G and Thermodynamic Favorability Slide 73 / 93 As we have seen, the enthalpy, entropy, and the temperature influence both the sign and magnitude of #G for a given process. Let's examine four different processes and determine the qualitative impact of these factors on #G. #G and Thermodynamic Favorability ase 1: Freezing of 1 atm: H 2O(s) --> H 2O(l) Slide 74 / 93 #H = - (energy is released as bonds form) #S = - (disorder decreases as the crystal forms) At what temperatures will this reaction be thermodynamically favorable? #G = #H -T#S #G = (-) + (-T(-)) = (-) + (T(+)) To be thermodynamically favorable, #G must be negative, so a relatively small T is required. #G and Thermodynamic Favorability ase 2: ombustion of 1 atm: 2 8H 18(g) + 25O 2(g) --> 16 O 2(g) + 18H 2O(g) Slide 75 / 93 #H = - (energy is released as new bonds form) #S = + (few moles are converted to many) At what temperatures will this reaction be thermodynamically favorable? #G = #H -T#S G = (-) + (-T(+)) = (-) + (T(-)) This reaction is favorable at all temperatures as both the enthalpy and entropy terms are negative.

26 #G and Thermodynamic Favorability Slide 76 / 93 ase 3: Photosynthesis 6O 2(g) + 6H 2O(g) --> 6O 2(g) + 6H 12O 6(s) #H = + (energy is absorbed as to break existing bonds) #S = - (many moles are converted to few) At what temperatures will this reaction be thermodynamically favorable? #G = #H -T#S #G = (+) + (-T(-)) = (+) + (T(+)) move for answer This reaction is unfavorable at all temperatures as both the enthalpy and entropy terms are positive. Plants need energy from the sun for this reaction to occur. #G and Thermodynamic Favorability Slide 77 / 93 ase 4: issolving of ammonium chloride in water NH 4l(s) --> NH 4+ (aq) + l - (aq) #H = + (energy is absorbed as to break existing bonds) #S = + (increase in disorder as ions mix with water) At what temperatures will this reaction be thermodynamically favorable? #G = #H -T#S #G = (+) + (-T(+)) move for answer = (+) + (T(-)) This reaction is favorable only at relatively high temperatures. #G and Thermodynamic Favorability Fill in the following summary chart #G = #H -T#S Slide 78 / 93 #H #S T #G all high low high low all

27 26 For a reaction to be thermodynamically favorable under standard conditions at all temperatures, the signs of ΔH o and ΔS o must be and, respectively. Slide 79 / 93 A +, + #G = #H -T#S B +, - -, + -, - E +, 0 27 For a reaction to be unfavorable under standard conditions at all temperatures, the signs of ΔH o and ΔS o must be and, respectively. Slide 80 / 93 A +, + #G = #H -T#S B +, - -, + -, - E +, 0 28 A reaction that is unfavorable at low temperature can become favorable at high temperature if ΔH is and ΔS is. Slide 81 / 93 A +, + B -, - +, - -, + #G = #H -T#S E +, 0

28 29 A reaction that is not favorable at one temperature can become favorable if the temperature is lowered. Therefore, ΔH is and ΔS is. Slide 82 / 93 A +, + #G = #H -T#S B -, - +, - -, + E +, 0 30 The values for #H and #S values are as follows for a reaction. #H = 137 kj/mol #G = #H -T#S #S = 120 J/K mol Slide 83 / 93 This reaction is. A favorable only at low temperatures. B favorable only at high temperature. favorable at all temperatures. Not enough information is provided. 31 When ammonium chloride dissolves in water the temperature of the solution is less than that of the original water sample. Thus, we know that ΔH is and that ΔS is. A - - B E - 0 #G = #H -T#S Slide 84 / 93

29 Free Energy and Temperature Slide 85 / 93 Occasionally, you are asked to calculate one of the following: 1) the specific temperature at which a reaction changes from being favorable to unfavorable or 2) the specific temperature at which a reaction changes from being unfavorable to favorable Free Energy and Temperature At what specific temperature will a reaction change from being favorable to unfavorable? Slide 86 / 93 In this situation, if #G < 0 at low temperatures and #G > 0 at high temperatures, then we can conclude that #H < 0 and #S < 0. You can calculate the specific temperature at which #G changes sign by setting #G = 0. #G = #H -T#S 0 = #H -T#S T#S = #H #T = #H/#S Below this temperature, the reaction will be favorable, while above it, the process is unfavorable. Remember to note that enthalpy values are usually in kj/mol while entropy values are usually given in J/mol-K. Therefore, divide entropy values by Free Energy and Temperature Slide 87 / 93 At what specific temperature will a reaction change from being unfavorable to favorable? In this situation, if #G > 0 at low temperatures and #G < 0 at high temperatures, then we can conclude that #H > 0 and #S > 0. #G = #H -T#S 0 = #H -T#S T#S = #H #T = #H/#S Below this temperature, the reaction will be unfavorable, while above it, the process is favorable. Remember to note that enthalpy values are usually in kj/mol while entropy values are usually given in J/mol-K. Therefore, divide entropy values by 1000.

30 32 For the below reaction, ΔH o = kj/mol and ΔS o = J/mol at 298 K. At temperatures greater than this reaction is spontaneous under standard conditions Slide 88 / 93 (s) + H 2O(g) --> O(g) + H 2(g) A 273 #G = #H -T#S B E 983 # # Unavailability of Energy: Heat eath Slide 89 / 93 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. # # Evolution and Growth: Time s Arrow Slide 90 / 93 Growth of an individual, and evolution of a species, are both processes of increasing order. o they violate the second law of thermodynamics? No! These are not isolated systems. Energy comes into them in the form of food, sunlight, and air, and energy also leaves them. The second law of thermodynamics is the one that defines the arrow of time processes will occur that are not reversible, and movies that run backward will look silly.

31 Thermal Pollution and limate hange Slide 91 / 93 Air pollution is also emitted by power plants, industries, and consumers. Some of this pollution results in a buildup of O 2 in the atmosphere, contributing to global warming. This can be minimized through careful choices of fuels and processes. Thermal pollution, however, is a consequence of the second law, and is unavoidable; it can be reduced only by reducing the amount of energy we use. Thermal Pollution and Alternative Power Slide 92 / 93 The generation of electricity using solar energy does not involve a heat engine, but fossil-fuel plants and nuclear plants do. However, any use of energy will always result in some loss of "heat death" as no system is 100% effective due to the second law of thermodynamics. Slide 93 / 93