CHAPTER 17 WORK, HEAT, & FIRST LAW OF THERMODYNAMICS
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1 CHAPTER 17 WORK, HEAT, and the FIRST LAW OF THERMODYNAMICS In this chapter, we will examine various thermal properties of matter, as well as several mechanisms by which energy can be transferred to and from a system without the use of mechanical work. Slide 1
2 17.1 It s All About Energy Most of chapter 17 is concerned with tracking changes in the total energy of a system. From PHYS 211, we know that E sys = K + U + E th = W ext (in chapter 15, we omitted the E th term why?) We also know that K + U defines the change in mechanical energy E mec of the system, and thus E sys = E mec + E th = W ext In most thermodynamic processes of interest,w ext is not equal to zero that is, the system is not isolated. We perform external work on it by heating, cooling, pressurizing, etc. We wish to determine how energy is transferred between the system and its environment. Slide 2
3 17.1 It s All About Energy Two examples from PHYS 211 are shown here. We will discuss them in class. Slide 3
4 17.1 It s All About Energy Now let s examine a situation that was not covered in PHYS 211. A pot of water is placed on the stove, and the burner is turned on. The water s temperature increases ( E th > 0), but its mechanical energy remains the same ( E mec = 0). Furthermore, no external work is done on the water (since there s neither a force nor a displacement). We must conclude that there s a problem with the last equation on slide 10. As it turns out, there is more than one way for a system to exchange energy with its environment (other than through the mechanical interaction W ext ). There can also be a thermal interaction. The energy transferred by a thermal interaction is called heat, and has the symbol Q: E sys = E mec + E th = W ext + Q Slide 4
5 17.2 Work in Ideal-Gas Processes Again, let s step back to PHYS 211 to remind ourselves how to calculate mechanical work. If a force F acts on an object through a displacement s = s f s i, the work done on the object is s ff W = ds (in most cases in PHYS 211, F was constant or a trivial function of position, so you didn t actually have to perform an integral). s i Slide 5
6 17.2 Work in Ideal-Gas Processes Consider a gas as it is compressed or expanded by a moveable piston of area A. In the top figure, the piston is held in placed by F ext, which must balance the force F gas arising due to the gas pressure: F ext = F gas = pa If the piston moves outward by a small distance dx, the external force will do an amount of work equal to dw = F ext dx = pa dx If dx is positive (i.e. the gas is expanding), then dw is negative, since the force and the displacement are in opposite directions. Slide 6
7 17.2 Work in Ideal-Gas Processes Now, as the piston moves a distance dx, the volume of the gas changes by dv = A dx. From the last equation on the previous slide, we then have dw = pdv Finally, as the volume changes from V i to V f, the amount of work done on the gas is W = V i V fp dv Note that as the volume changes, p generally changes as well. It is not a constant, and can not be brought out of the integral. Slide 7
8 17.2 Work in Ideal-Gas Processes If we have already drawn a pv diagram for a particular process, we can find the work done by examining the area under the curve. This will be discussed in class. Slide 8
9 17.2 Work in Ideal-Gas Processes For an isochoric process, the volume doesn t change, so W = 0 For an isobaric process, the pressure doesn t change this makes for an easy integral! W = p V Slide 9
10 17.2 Work in Ideal-Gas Processes For an isothermal process, we have to do a bit of work (pun intended). The ideal-gas law helps here. We know that p = nrt, where n and R are constants. V Also, in this case, T is a constant. Thus, W = V i v f p dv = V i v f nrt V dv = nrt V i v f dv V = nrt ln V f V i Finally, since nrt = p i V i = p f V f for an isothermal process, we can write W = p i V i ln V f V i = p f V f ln V f V i Slide 10
11 17.2 Work in Ideal-Gas Processes Path Dependence The area-under-a-curve interpretation of the work done on a gas during an ideal-gas process should convince you that W depends not just on the initial and final states, but also on how the gas transitioned between these states (i.e. on the path between i and f). Slide 11
12 Problem #1: Path Dependence RDK STT 17.2 Two processes take an ideal gas from state 1 to state 3. Compare the work done on the gas by process A to that done by process B. A W A = W B = 0 B W A = W B, but neither is zero C W A > W B D W A < W B Slide 12
13 Problem #2: Isobaric Process 80 J of work are done on the gas in the process shown in the figure. What is V 1 in units of cm 3? Solution: in class RDK EX 17.3 Slide 13
14 17.3 Heat Heat is not a substance. It s energy. Specifically, it s the energy transferred between a system and its environment as a consequence of a temperature difference between them. For example, we can increase the temperature of a container of water by placing it over a flame, or by rapidly spinning a paddle in it. The former achieves a transfer of energy via heat, in what is termed a thermal interaction, while the latter achieves it by mechanical work (which increases the kinetic energy of the water molecules, bringing us back to PHYS 211). The details of the heat transfer mechanisms will be discussed in chapter 18. Slide 14
15 17.3 Heat The sign of heat (Q) depends on the direction of energy flow, which is from hot to cold. If a system is colder than its environment, heat flows into the system, and Q > 0. If the system is hotter than its environment, heat flows out of the system, and Q < 0. In thermal equilibrium, the system and its environment are at the same temperature. Here, Q = 0 (there is no energy flow). Since heat is energy, it must have units of Joules. In some branches of science, it is measured in calories (1 cal = J). For added confusion, the calorie that we know of in relation to food is actually a kilocalorie (4186 J). Slide 15
16 17.3 Heat We must distinguish among heat, temperature, and thermal energy: Thermal energy E th is a quantity of energy of a system due to the motion of its atoms. It is a state variable, so we are sometimes interested in howe th changes during a process. Heat Q is energy transferred between a system and its environment as they interact. It is not a property of the system itself, and therefore is not a state variable. If a system does not interact thermally with its environment, then Q = 0. If it does interact with its environment, then it is possible that Q 0. Furthermore, it is possible that heat may causee th to change. Temperature is a state variable that quantifies the hotness of a system. It is related to the thermal energy per molecule. A temperature difference between system and environment causes Q 0. Slide 16
17 Problem #3: Heat or Not Heat? RDK STT 17.3 Which of the following processes involves heat? A The brakes in your car get hot when you stop B A steel block is held over a candle C You push a rigid cylinder of gas across a frictionless surface D You push a piston into a cylinder of gas, increasing the temperature of the gas E You place a cylinder of gas in hot water. The gas expands, causing a piston to rise and lift a weight. The temperature of the gas does not change Slide 17
18 17.4 The First Law of Thermodynamics As we learned earlier in this chapter, a complete statement of the law of conservation of energy requires a heat term: E sys = E mech + E th = W + Q For our study of thermodynamics, we are not interested in systems that have any macroscopic degree of motion. That is, we will assume E mech = 0, and therefore claim that E th = W + Q This is a statement of the first law of thermodynamics. It s illustrated by the figure to the right, which we will discuss in class. Slide 18
19 17.4 The First Law of Thermodynamics Three Special Ideal-Gas Processes It is possible to define processes in which one of the three terms in the first law ( E th, W, or Q) is zero. The setup shown here has three properties: 1. A locking pin can hold the piston in place, keeping volume constant. 2. Adding or removing masses on top of the piston can change the pressure. Work is done on the masses as they move. 3. The gas can be warmed or cooled through the bottom surface. Other surfaces are insulated. Slide 19
20 17.4 The First Law of Thermodynamics When the piston is unlocked but stationary, the gas pressure is related to the mass M as p gas = p 0 + Mg A where A is the area of the piston. When the piston is locked, there is no such relation. We are now ready to examine the three aforementioned processes Slide 20
21 17.4 The First Law of Thermodynamics Isochoric Cooling As we learned previously, no work is done in an isochoric process, since the volume doesn t change. This is achieved as follows: the locking pin is inserted the cylinder is placed on a block of ice this produces a Q < 0 the gas temperature and pressure fall the cylinder is removed from the ice when the desired pressure is reached masses are removed from the piston until the desired pressure is reached the locking pin is removed Overall: W = 0, Q < 0, E th < 0. Slide 21
22 17.4 The First Law of Thermodynamics Isothermal Expansion This is achieved as follows: the cylinder is placed over the flame this produces a Q > 0 the gas will begin to expand since pv must be constant for an isothermal process, remove masses as the gas expands in order to reduce the gas pressure the temperature remains constant, since heat energy balances the negative work done on the expanding gas when the desired volume is reached, remove the cylinder from the flame. Overall: W < 0, Q > 0, E th = 0. Slide 22
23 17.4 The First Law of Thermodynamics Adiabatic Compression An adiabatic process is one in which no heat energy is transferred between the system and its environment. add insulation beneath the cylinder add masses to the piston this will increase the pressure the piston will descend, decreasing the volume stop adding masses when the desired volume is reached Overall: W > 0, Q = 0, E th > 0. Slide 23
24 Problem #4: First-Law Processes RDK STT 17.4 Which first-law bar chart describes the process shown in this pv diagram? A B C D Slide 24
25 Problem #5: Thermal Processes A gas is compressed from 600 cm 3 to 200 cm 3 at a constant pressure of 400 kpa. At the same time, 100 J of heat energy is transferred out of the gas. What is the change in thermal energy of the gas during this process? Solution: in class RDK EX 17.9 Slide 25
26 17.5 Thermal Properties of Matter We re now aware that the thermal energy of a system can increase or decrease. This can have two effects it can change the temperature of the system, or it can change its phase (i.e. the system can melt, boil, etc.) In the first case, the natural question is: what quantity of thermal energy increase corresponds to a given temperature increase? It s not hard to predict that the temperature change depends on the amount of the substance that is, if a certain E th produces a 1 K temperature increase in 1 liter of water, then it should produce a 2 K temperature increase in ½ liter of water (since the thermal energy per molecule) is twice as great. However, it turns out that the relation between temperature change and thermal energy also depends entirely on which material is being heated or cooled. This leads to the concept of specific heat. Slide 26
27 17.5 Thermal Properties of Matter The specific heat of a substance is the amount of energy required to raise the temperature of 1 kg of the substance by 1 K. It has the symbol c, and SI units of J/kg K. Based on this description, we can write the relation between thermal energy change and temperature change as E th = Mc T We will examine the specific heat of gases in section calorie / g K Slide 27
28 17.5 Thermal Properties of Matter From the First Law, we can write W + Q = Mc T For liquids and solids, we re almost always changing the thermal energy by heating, in which case W = 0 (since liquids and solids are incompressible, we can t really perform work on them by changing their volume). In this case, we can write Q = Mc T The molar specific heat is the thermal energy required to change the temperature of 1 mole of a substance (rather than 1 kg) by 1 K. It has a symbol C and units of J/mol K. The relevant relation is Q = nc T According to the table, for elemental solids, C is always very close to 25 J/mol K. We ll find out why this is in the next chapter. Slide 28
29 17.5 Thermal Properties of Matter Phase Change and Heat of Transformation From the previous slide, it should be clear that if we plot T vs. Q for any substance, the graph should appear as a straight line with slope 1/Mc. However, plotting actual experimental data results in the figure shown here. The first interesting property of this plot is that the solid, liquid, and gas phases of a substance have a different specific heat (notice that the slopes are not identical). Even more interesting is that there are regions of zero slope, corresponding to the melting and boiling temperatures, T m and T c. Slide 29
30 17.5 Thermal Properties of Matter Phase Change and Heat of Transformation cont Physically, what is happening in these latter regions is that the additional heat energy does not raise the temperature of the material; rather, it is used to break molecular bonds in order to produce the phase change. The heat of transformation (L, units of J/kg) is the amount of heat energy required to cause a phase change in 1 kg of material. A subscript indicates if the phase change is freezing/melting (f, heat of fusion ) or boiling/condensing (v, heat of vaporization ). The heat needed to cause a phase change in a mass M of a substance is therefore Q = ±ML f (melt / freeze) Q = ±ML v (boil / condense) (the negative sign indicates that heat must be removed from the system in order to freeze or condense the substance). Slide 30
31 17.5 Thermal Properties of Matter Phase Change and Heat of Transformation cont A few heats of transformation are shown here. Notice that L v is always a few times larger than L f. Slide 31
32 Problem #6: Boiling Booze What is the maximum mass of ethyl alcohol you could boil with 1000 J of heat, starting from 20 C? Solution: in class RDK EX Slide 32
33 17.6 Calorimetry Calorimetry is a fancy term that refers to the science of measuring the heat of physical or chemical reactions. The latter is done in most introductory (or high school) chemistry labs; we ll be concerned with the former. Experiments in calorimetry involve systems or collections of systems that are thermally isolated. This means that the net heat Q net is zero. There s really not much more to say it s best explained by working through a problem. Slide 33
34 Problem #7: Hot Pan! RDK EX A 750 g aluminum pan is removed from the stove and plunged into a sink filled with 10.0 L of water at 20.0 C. The water temperature quickly rises to 24.0 C. What was the initial temperature of the pan, in C? Useful data: c Al = 900 J/kg K, c H2 O = 4190 J/kg K Solution: in class Slide 34
35 17.7 The Specific Heats of Gases Our previous discussion of specific heat was concerned only with solids and liquids. This is because the relation between heat and temperature change of a gas is more complicated as we have seen, there are many mechanisms by which the temperature of a gas can change. As an example, the figure shows an isochoric process (A) and an isobaric process (B), both of which have the same initial and final temperatures. Since work is done in process B but not in process A, a different amount of heat is required in the two cases. Thus, we require a different version of specific heat for the two cases. Slide 35
36 17.7 The Specific Heats of Gases Since we usually deal with moles rather than mass when performing gas calculations, we will only be concerned with the molar specific heats. As before, the quantity of heat needed to raise the temperature of n moles of gas by T is isobaric isochoric R Q = nc V T (constant volume) Q = nc P T (constant pressure) For a general gas process, in which neither volume nor pressure is constant, there is no way to define specific heat. Slide 36
37 17.7 The Specific Heats of Gases Path Dependence of Heat We have already seen that the work done in changing the state of a gas depends on the path taken on the pv diagram. In fact, the heat required for the state change depends on the path as well. This can be shown using the First Law, E th = W + Q. Recall that E th is a state variable, and thus it depends on the state of the gas, and not the process by which the gas arrived at that state. Therefore, W A + Q A = W B + Q B. But we already know that W is path-dependent (W A W B ) Therefore, Q A Q B. The heat added or removed during an ideal-gas process depends on the path followed through the pv diagram. Slide 37
38 17.7 The Specific Heats of Gases Adiabatic Processes Recall that an adiabatic process is one for which Q = 0 (no heat energy is transferred). This can be achieved by thoroughly insulating a system from its environment. Alternatively, it can come about by expanding or compressing a gas rapidly enough that heat transfer while allowed simply doesn t have enough time to occur. The relationship of isothermal, isochoric, and adiabatic processes to the First Law is shown here: Slide 38
39 17.7 The Specific Heats of Gases Adiabatic Processes cont For any ideal-gas process, it can be shown (see p. 487) that E th = nc V T Since E th = W for an adiabatic process, it follows that W = nc V T (adiabatic process) This provides a fourth description of work, along with W = 0 (isochoric process) W = p V (isobaric process) W = p i V i ln V f V i = p f V f ln V f V i (isothermal process) Slide 39
40 17.7 The Specific Heats of Gases Adiabatic Processes cont Finally, we must examine the trajectory of an adiabatic process on a pv diagram. In order to do this, we start by defining the specific heat ratio: γ = C p = 1.67 monatomic gas C V 1.40 diatomic gas (these numbers represent the fractions 5/3 and 7/5, respectively there s some slick physics in there, but it s well beyond the scope of PHYS 212). Then, an adiabatic process is one for which pv γ is constant; i.e. p i V i γ = p f V f γ. The resulting curve is called an adiabat. There s a proof of this result on p. 490 of the text. Don t sweat it. Slide 40
41 Problem #8: Adiabatic Expansion A gas cylinder holds 0.10 mol of O 2 at a temperature of 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halved. a) What is the final volume of the gas? b) What is the final temperature of the gas? c) How would these answers change if the gas was N 2? Solution: in class RDK EX Slide 41
42 17.8 Heat-Transfer Mechanisms Much of this chapter has involved the transfer of heat. In this last section of the notes, we will investigate the various physical mechanisms by which this transfer occurs conduction, convection, and radiation. A fourth mechanism evaporation was already described earlier in these notes (heat of vaporization). Slide 42
43 17.8 Heat-Transfer Mechanisms Conduction The figure shows a material sandwiched between a higher temperature T H and a lower temperature T L, resulting in a transfer of thermal energy from the hot side to the cold side. The microscopic mechanism of conduction will be described in class. An experimental observation of conduction reveals two properties: i. More heat is transferred if T = T H T L is larger ii. Materials with a larger cross-section A transfer more heat, while those with a larger length L transfer less heat Slide 43
44 17.8 Heat-Transfer Mechanisms Conduction cont If a heat Q is transferred in a time interval t, then the rate of heat transfer is Q/ t (units of J/s recall from PHYS 211 that a rate of energy transfer is a power, with units of Watts). Based on our experimental observations, we can write Q t = k A L T Here, k is the thermal conductivity of the material (SI units: W/m K); a large value of k means that the material is a good conductor of heat. Slide 44
45 Problem #9: Leaky Window A glass window of area 2.0 m 2 has a thickness of 4.0 mm. If the indoor temperature is 25 C and the outdoor temperature is 0 C, how much heat energy transfers through the window in 1 hour? Solution: in class Slide 45
46 17.8 Heat-Transfer Mechanisms Convection In this process, thermal energy is transferred by the physical motion of a fluid. For example, when water is heated in a pot, the water that contacts the bottom surface is heated by conduction, but then it rises to the surface via convection. Mathematically, convection is too complex to describe in any detail in PHYS 212. Just be aware that it exists! Aerogels (shown here) are extraordinarily good insulators, since they almost entirely prevent both conduction and convection. Slide 46
47 17.8 Heat-Transfer Mechanisms Radiation There s ample proof that another mechanism of heat transfer must exist in addition to conduction and convection. Although there is essentially no material between the sun and the earth with which to conduct or convect thermal energy, we can still feel the sun s heat. Radiation here refers to any process by which electromagnetic energy is generated at one location and beamed to another. It requires no intervening medium and in fact, it s more efficient if there s no medium at all. The properties of this electromagnetic energy depend on an object s temperature. This is useful for satellite imaging of ocean temperature, for example. Slide 47
48 17.8 Heat-Transfer Mechanisms Radiation cont The rate of energy transfer by radiation (i.e. the radiated power) depends on an object s temperature and on its surface area: Q t = eσat4, where σ = W/m 2 K 4 σ is known as the Stefan-Boltzmann constant. Here, e is the emissivity of the surface, a unitless number between 0 and 1. It is a measure of how effectively the surface radiates. When e = 1, the radiation is referred to as black-body radiation. Slide 48
49 17.8 Heat-Transfer Mechanisms Radiation cont All objects that emit radiation also absorb radiation from their surroundings. Therefore, we are more interested in the net rate of energy transfer, Q net t = eσa(t4 T 0 4 ) where T and T 0 are the temperatures of the object and its environment, respectively. This equation indicates that an object in thermal equilibrium with its surroundings undergoes no net thermal radiation, while an object that is colder than its surroundings has a negative value of Q net. Be careful about the sign convention here. Since radiation, by definition, refers to energy leaving an object, a positive Q net refers to a net energy transfer out of the object. Slide 49
50 Problem #10: Lead Sphere What maximum power can be radiated by a 10-cm-diameter solid lead sphere? Assume that e = 1. Useful information: the surface area of a sphere of radius r is A = 4πr 2 More useful information: you ll need some data from a table elsewhere in these notes Solution: in class RDK EX Slide 50
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