Basic Laws of Thermodynamics

Transcription

1 Basic Laws of Thermodynamics 1

2 First Law of Thermodynamics Conservation of Energy for Thermal Systems One form of work may be converted into another, or, work may be converted to heat, or, heat may be converted to work, but, final energy = initial energy The first law of thermodynamics is simply an expression of the conservation of energy principle, and it asserts that energy is a thermodynamic property. 2

3 Forms of internal energy 3

4 THE FIRST LAW OF THERMODYNAMICS The internal energy of a system changes from an initial value U i to a final value of U f due to heat Q and work. Qis positive when the system gains heat and negative when it loses heat. Wis positive when work is done by the system and negative when work is done on the system. For an isolated system there is no heat or work transferred with the surroundings, thus, by definition W = Q = 0 and therefore ΔU = 0. 4

5 1 st Law of TD Some conventions: For the gases perspective: heat added is positive, heat removed is negative. Work done on the gas is positive, work done by the gas is negative. Temperature increase means internal energy change is positive. 5

6 Positive and Negative Work 6

7 The figure illustrates a system and its surroundings. In part a, the system gains 1500 J of heat from its surroundings, and 2200 J of work is done bythe system on the surroundings. In part b, the system also gains 1500 J of heat, but 2200 J of work is done onthe system by the surroundings. In each case, determine the change in the internal energy of the system. (a) (b) 7

8 For a differential change, Equation 1 becomes: For a cyclic process, A B A, when the system returns to state A, it has the sameu, thus: In thermodynamics, the term heat simply means heat transfer. A process during which there is no heat transfer is called an adiabatic process. 8

9 Heat Capacity Before discussing isothermal or adiabatic processes, a new term is needed to make the calculations easier. Heat Capacity, C is equal to the ratio of the heat absorbed or withdrawn from the system to the resultant change in temperature. q C = T

10 Heat Capacity: The heat capacity is the amount of heat required to raise the temperature of a mass of a system by 1 C. It is denoted by C. 10

11 Joule Equivalent of Heat James Joule showed that mechanical energy could be converted to heat and arrived at the conclusion that heat was another form of energy. He showed that 1 calorie of heat was equivalent to J of work. 1 cal = J Called: Mechanical equivalent of heat 11

12 Energy Mechanical Energy: KE, PE, E Work is done by energy transfer. Heat is another form of energy. Need to expand the conservation of energy principle to accommodate thermal systems. 12

13 Modes of Heat transfer Heat can be transferred in three different ways: conduction, convection, and radiation. All modes of heat transfer require the existence of a temperature difference, and all modes of heat transfer are from the high-temperature medium to a lower-temperature one. Work Work, like heat, is an energy interaction between a system and its surroundings. As mentioned earlier, energy can cross the boundary of a closed system in the form of heat or work. Therefore, if the energy crossing theboundary of a closed system is not heat, it must be work. 13

14 There are two ways a process can be adiabatic: Either the system is well insulated so that only a negligible amount of heat can pass through the boundary, or both the system and the surroundings are at the same temperature and therefore there is no driving force (temperature difference) for heat transfer. An adiabatic process should not be confused with an isothermal process. Even though there is no heat transfer during an adiabatic process, the energy content and thus the temperature of a system can still be changed by other means such as work. 14

15 Thermal Processes quasi-staticmeans that it occurs slowly enough that a uniform pressure and temperature exist throughout all regions of the system at all times. An isobaric process is one that occurs at constant pressure. 15

16 For an isobaric process, a pressure-versus-volume plot is a horizontal straight line, and the work done [W = P(V f V i )] is the colored rectangular area under the graph. 16

17 (a) The substance in the chamber is being heated isochorically because the rigid chamber keeps the volume constant. (b) The pressure-volume plot for an isochoric process is a vertical straight line. The area under the graph is zero, indicating that no work is done. isochoric process, one that occurs at constant volume. 17

18 isothermal process, one that takes place at constant temperature.(when the system is an ideal gas.) There isadiabatic process, one that occurs without the transfer of heat. Since there is no heat transfer, Qequals zero, and the first law indicates that U= Q W= W. Thus, when work is done by a system adiabatically, Wis positive and the internal energy of the system decreases by exactly the amount of the work done. When work is done on a system adiabatically, Wis negative and the internal energy increases correspondingly. 18

19 The area under a pressurevolume graph is the work for any kind of process. The colored area gives the work done by the gas for the process from Xto Y. 19

20 Thermal Processes Using an Ideal Gas 20

21 Example: Isobaric Expansion of Water One gram of water is placed in the cylinder and the pressure is maintained at Pa. The temperature of the water is raised by 31 C. In one case, the water is in the liquid phase and expands by the small amount of m 3. In another case, the water is in the gas phase and expands by the much greater amount of m 3. For the water in each case, find (a) the work done and (b) the change in the internal energy. c = 4186 J/(kg C ) c P = 2020 J/(kg C ). 21

22 (a) (b) 22

23 ISOTHERMAL EXPANSION OR COMPRESSION P = nrt/v W = P V = P(V f V i ) 23

24 Example Isothermal Expansion of an Ideal Gas Two moles of the monatomic gas argon expand isothermally at 298 K, from an initial volume of V i = m 3 to a final volume of V f = m 3. Assuming that argon is an ideal gas, find (a) the work done by the gas, (b) the change in the internal energy of the gas, and (c) the heat supplied to the gas. (a). (b) (c) 24

25 ADIABATIC EXPANSION OR COMPRESSION 25

26 [T i = P i V i /(nr)] [T f = P f V f /(nr)]. 26

27 Type of Thermal Process Work Done First Law of Thermodynamics ( U = Q W) Isobaric (constant pressure) W = P(V f V i ) Isochoric (constant volume) W = 0 J Isothermal (constant temperature) (for an ideal gas) Adiabatic (no heat flow) (for a monatomic ideal gas) 27

28 1 st Law of TD Example: 25 L of gas is enclosed in a cylinder/piston apparatus at 2 atm of pressure and 300 K. If 100 kg of mass is placed on the piston causing the gas to compress to 20 L at constant pressure. This is done by allowing heat to flow out of the gas. What is the work done on the gas? What is the change in internal energy of the gas? How much heat flowed out of the gas? 28

29 P o = 202,600 Pa, V o = m 3, T o = 300 K, P f = 202,600 Pa, V f =0.020 m 3, T f = n = PV/RT. W = -P V U = 3/2 nr T Q = W + U W =-P V = -202,600 Pa ( )m 3 =1013 J energy added to the gas. U =3/2 nr T=1.5(2.03)(8.31)(-60)=-1518 J Q = W + U = = -505 J heat out 29

30 Specific Heat Capacities where the capital letter C refers to the molar specific heat capacity in units of J/(mol K). 30

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33 The Second Law of Thermodynamics THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENT Heat flows spontaneously from a substance at a higher temperature to a substance at a lower temperature and does not flow spontaneously in the reverse direction. 33

34 Heat Engines A heat engineis any device that uses heat to perform work. It has three essential features: 1. Heat is supplied to the engine at a relatively high input temperature from a place called the hot reservoir. 2. Part of the input heat is used to perform work by the working substanceof the engine, which is the material within the engine that actually does the work (e.g., the gasoline-air mixture in an automobile engine). 3. The remainder of the input heat is rejected to a place called the cold reservoir,which has a temperature lower than the input temperature. 34

35 These three symbols refer to magnitudes only, without reference to algebraic signs. Therefore, when these symbols appear in an equation, they do not have negative values assigned to them. 35

36 Efficiencies are often quoted as percentages obtained by multiplying the ratio W/QHby a factor of

37 Example: An Automobile Engine An automobile engine has an efficiency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine? 37

38 Carnot's Principle and the Carnot Engine A reversible process is one in which both the system and its environment can be returned to exactly the states they were in before the process occurred. CARNOT S PRINCIPLE: AN ALTERNATIVE STATEMENT OF THE SECOND LAW OF THERMODYNAMICS No irreversible engine operating between two reservoirs at constant temperatures can have a greater efficiency than a reversible engine operating between the same temperatures. Furthermore, all reversible engines operating between the same temperatures have the same efficiency. 38

39 A Carnot engineis a reversible engine in which all input heat Q H originates from a hot reservoir at a single temperature TH, and all rejected heat Q C goes into a cold reservoir at a single temperature T C. The work done by the engine is W. 39

40 where the temperatures T C and T H must be expressed in Kelvins. 40

41 Example: A Tropical Ocean as a Heat Engine Water near the surface of a tropical ocean has a temperature of K (25.0 C), whereas water 700 m beneath the surface has a temperature of K (7.0 C). It has been proposed that the warm water be used as the hot reservoir and the cool water as the cold reservoir of a heat engine. Find the maximum possible efficiency for such an engine. T H = K and T C = K 41

42 Example: Limits on the Efficiency of a Heat Engine Consider a hypothetical engine that receives 1000 J of heat as input from a hot reservoir and delivers 1000 J of work, rejecting no heat to a cold reservoir whose temperature is above 0 K. Decide whether this engine violates the first or the second law of thermodynamics, or both. It is the second law, not the first law, that limits the efficiencies of heat engines to values less than 100%. 42

43 Refrigerators, Air Conditioners, and Heat Pumps In the refrigeration process, work Wis used to remove heat Q C from the cold reservoir and deposit heat Q H into the hot reservoir. 43

44 In a refrigerator, the interior of the unit is the cold reservoir, while the warmer exterior is the hot reservoir. 44

45 A window air conditioner removes heat from a room, which is the cold reservoir, and deposits heat outdoors, which is the hot reservoir. 45

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47 In a heat pump the cold reservoir is the wintry outdoors, and the hot reservoir is the inside of the house. 47

48 This conventional electric heating system is delivering 1000 J of heat to the living room. Q H = W + Q C and Q C /Q H = T C /T H 48

49 Example: A Heat Pump An ideal or Carnot heat pump is used to heat a house to a temperature of T H = 294 K (21 C). How much work must be done by the pump to deliver Q H = 3350 J of heat into the house when the outdoor temperature T C is (a) 273 K (0 C) and (b) 252 K ( 21 C)? (a) (b) 49

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51 Check Your Understanding Each drawing represents a hypothetical heat engine or a hypothetical heat pump and shows the corresponding heats and work. Only one is allowed in nature. Which is it? (c) 51

52 Entropy In general, irreversible processes cause us to lose some, but not necessarily all, of the ability to perform work. This partial loss can be expressed in terms of a concept called entropy. Reversible processes do not alter the total entropy of the universe. 52

53 Although the relation S = (Q/T)R applies to reversible processes, it can be used as part of an indirect procedure to find the entropy change for an irreversible process. 53

54 Example: The Entropy of the Universe Increases 1200 J of heat flow spontaneously through a copper rod from a hot reservoir at 650 K to a cold reservoir at 350 K. Determine the amount by which this irreversible process changes the entropy of the universe, assuming that no other changes occur. 54

55 Any irreversible process increases the entropy of the universe. 55

56 THE SECOND LAW OF THERMODYNAMICS STATED IN TERMS OF ENTROPY The total entropy of the universe does not change when a reversible process occurs ( S universe = 0 J/K) and does increases when an irreversible process occurs ( S universe > 0 J/K). 56

57 Example: Energy Unavailable for Doing Work 57

58 Suppose that 1200 J of heat is used as input for an engine under two different conditions. In Figure part athe heat is supplied by a hot reservoir whose temperature is 650 K. In part bof the drawing, the heat flows irreversibly through a copper rod into a second reservoir whose temperature is 350 K and then enters the engine. In either case, a 150-K reservoir is used as the cold reservoir. For each case, determine the maximum amount of work that can be obtained from the 1200 J of heat. 58

59 A block of ice is an example of an ordered system relative to a puddle of water. 59

60 Example: Order to Disorder Find the change in entropy that results when a 2.3kg block of ice melts slowly (reversibly) at 273 K (0 C). 60

61 The Third Law of Thermodynamics THE THIRD LAW OF THERMODYNAMICS It is not possible to lower the temperature of any system to absolute zero (T= 0 K) in a finite number of steps. 61

62 Concepts & Calculations Example: The Sublimation of Zinc The sublimation of zinc (mass per mole = kg/mol) occurs at a temperature of K, and the latent heat of sublimation is J/kg. The pressure remains constant during the sublimation. Assume that the zinc vapor can be treated as a monatomic ideal gas and that the volume of solid zinc is negligible compared to the corresponding vapor. What is the change in the internal energy of the zinc when 1.50 kg of zinc sublimates? Q = U + W, Q = ml s, W = nrt 62

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64 Example: The Work Energy Theorem Each of two Carnot engines uses the same cold reservoir at a temperature of 275 K for its exhaust heat. Each engine receives 1450 J of input heat. 64

65 The work from either of these engines is used to drive a pulley arrangement that uses a rope to accelerate a 125-kg crate from rest along a horizontal frictionless surface. With engine 1 the crate attains a speed of 2.00 m/s, while with engine 2 it attains a speed of 3.00 m/s. Find the temperature of the hot reservoir for each engine. 65

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