Thermodynamics Heat & Work The First Law of Thermodynamics

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1 Thermodynamics Heat & Work The First Law of Thermodynamics Lana Sheridan De Anza College April 20, 2016

2 Last time applying the ideal gas equation thermal energy heat capacity phase changes

3 Overview latent heat more about phase changes work, heat, and the first law of thermodynamics P-V diagrams applying the first law in various cases

4 Phase Changes

5 Latent Heat latent heat of fusion, L f The amount of energy (heat) per unit mass required to change a solid to a liquid. Q = ml f where m is the mass of solid that is transformed into a liquid. latent heat of vaporization, L v The amount of energy (heat) per unit mass required to change a liquid to a gas. Q = ml v where m is the mass of liquid that is transformed into a gas.

6 Latent Heat 20 The First Law of Thermodynamics Table 20.2 Latent Heats of Fusion and Vaporization Latent Heat Melting of Fusion Boiling Latent Heat Substance Point ( C) (J/kg) Point ( C) of Vaporization ( J/kg) Helium a Oxygen Nitrogen Ethyl alcohol Water Sulfur Lead Aluminum Silver Gold Copper a Helium does not solidify at atmospheric pressure. The melting point given here corresponds to a pressure of 2.5 MPa. plate melts completely, the change in mass of the water is m f m, which is the mass of new water and is also equal to the initial mass of the ice cube. From the definition of latent heat, and again choosing heat as our energy transfer 1 mechanism, the energy required to change the phase of a pure substance is Table from Serway & Jewett, page 598.

7 Practice The specific heat capacity of ice is about 0.5 cal/g C. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1 g ice cube at absolute zero ( 273 C) to 1 g of boiling water. How does this number of calories required to change the same gram of 100 C boiling water to 100 C steam? Reminder: 1 cal is the heat required to raise the temperature of 1 g of water by 1 C. 1 Hewitt, Problem 2, page 314.

8 Practice The specific heat capacity of ice is about 0.5 cal/g C. Calculate the number of calories it would take to change a 1 g ice cube at absolute zero ( 273 C) to 1 g of boiling water. 1 Hewitt, Problem 2, page 314.

9 Practice The specific heat capacity of ice is about 0.5 cal/g C. Calculate the number of calories it would take to change a 1 g ice cube at absolute zero ( 273 C) to 1 g of boiling water. warming ice: melting: Q 1 = mc ice T = (1 g)(0.5 cal/g C)(273 C) = cal Q 2 = ml f = (1 g) warming water: ( ) ( ) J/kg 1 kg = cal J/cal 1000 g Q 3 = mc water T = (1 g)(1.0 cal/g C)(100 C) = 100 cal 1 Hewitt, Problem 2, page 314.

10 Practice The specific heat capacity of ice is about 0.5 cal/g C. Calculate the number of calories it would take to change a 1 g ice cube at absolute zero ( 273 C) to 1 g of boiling water. warming ice: melting: Q 1 = mc ice T = (1 g)(0.5 cal/g C)(273 C) = cal Q 2 = ml f = (1 g) warming water: ( ) ( ) J/kg 1 kg = cal J/cal 1000 g Q 3 = mc water T = (1 g)(1.0 cal/g C)(100 C) = 100 cal Total Q 1 + Q 2 + Q 3 = 320 cal. 1 Hewitt, Problem 2, page 314.

11 Practice The specific heat capacity of ice is about 0.5 cal/g C. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1 g ice cube at absolute zero ( 273 C) to 1 g of boiling water. How does this number of calories required to change the same gram of 100 C boiling water to 100 C steam? 1 Hewitt, Problem 2, page 314.

12 Practice The specific heat capacity of ice is about 0.5 cal/g C. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1 g ice cube at absolute zero ( 273 C) to 1 g of boiling water. How does this number of calories required to change the same gram of 100 C boiling water to 100 C steam? boiling: Q 4 = ml v = (1 g) ( ) ( ) J/kg 1 kg = 540 cal J/cal 1000 g 1 Hewitt, Problem 2, page 314.

13 Practice The specific heat capacity of ice is about 0.5 cal/g C. Supposing that it remains at that value all the way to absolute zero, calculate the number of calories it would take to change a 1 g ice cube at absolute zero ( 273 C) to 1 g of boiling water. How does this number of calories required to change the same gram of 100 C boiling water to 100 C steam? boiling: Q 4 = ml v = (1 g) ( ) ( ) J/kg 1 kg = 540 cal J/cal 1000 g The energy required to transform the water to steam is much bigger than the energy required to heat the ice, convert it to water, and continue heating up to 100 C. 1 Hewitt, Problem 2, page 314.

14 Question Suppose the same process of adding energy to the ice cube is performed as discussed in the last question, but instead we graph the internal energy of the system as a function of energy input. What would this graph look like? 1 Based on Quick Quiz 20.2, Serway & Jewett, page 600.

15 Practice The heat of vaporizations of ethyl alcohol is about 200 cal/g. If 2 kg of this fluid were allowed to vaporize in a refrigerator, show that 5 kg of ice (at 0 C) would be formed from 0 C water. 1 Hewitt, Problem 8, page 314.

16 Practice The heat of vaporizations of ethyl alcohol is about 200 cal/g. If 2 kg of this fluid were allowed to vaporize in a refrigerator, show that 5 kg of ice (at 0 C) would be formed from 0 C water. Hint: in the last problem we melted 1 g of ice and found it required 80 cal. 1 Hewitt, Problem 8, page 314.

17 Practice The heat of vaporizations of ethyl alcohol is about 200 cal/g. If 2 kg of this fluid were allowed to vaporize in a refrigerator, show that 5 kg of ice (at 0 C) would be formed from 0 C water. Hint: in the last problem we melted 1 g of ice and found it required 80 cal. energy needed for vaporization: Q = ml v,ea = (2 kg)(200 cal/g) = cal assuming this same amount of energy was taken from the water: m = Q L f = cal 80 cal/g = 5000 g = 5 kg 1 Hewitt, Problem 8, page 314.

18 Phase Change paths

19 Evaporation evaporation the process by which a liquid changes to a gas at the liquid surface Since changing from a liquid to a gas requires heat, when a liquid evaporates it takes heat from its surroundings. This is why humans sweat in hot weather, pigs wallow puddles, and dogs pant. All are trying to use evaporation of water to reduce body temperature.

20 Evaporation Ben Franklin noticed that a wet shirt kept him feeling cool on a hot day. He decided to experiment to see if the temperature of objects could be lowered by this process. In 1758 he and John Hadley took a mercury thermometer and repeatedly wet the bulb with ether while using bellows to keep air moving over it. Despite it being a warm day, they recorded temperatures as low as 7 F ( 14 C) at the bulb of the thermometer. This is the basic idea behind refrigeration!

21 Phase Diagrams 1 A typical phase diagram. The dashed green line shows the unusual behavior of water. Diagram by Matthieumarechal, Wikipedia.

22 Liquids below their melting point & above their boiling point Phase changes require the bond structure of the substance to change.

23 Liquids below their melting point & above their boiling point Phase changes require the bond structure of the substance to change. As a liquid is cooled, it will generally start to freeze at one point before another. (Symmetry breaking.) For example, ice cubes in a freezer freeze from the top down. As water freezes solid crystals form and spread throughout. If the water is very pure and cooled without shaking, it can be cooled below 0 C without freezing. This is called supercooling and can happen in some other liquids also under the right conditions.

24 Liquids below their melting point & above their boiling point The same thing can happen when water (or other liquids) are heated just above their boiling points. As the liquid starts to boil, bubbles of vapor form inside it. This happens most easily at defects in the edges of the container. If the container is very smooth and not shaken, the only way the liquid can vaporize is from the middle. Surface tension opposes this. The liquid can be heated a bit above its boiling point. This is called superheating. It is used in bubble chambers to observe charged particles.

25 Heat and Work We now take a closer look at the first law of thermodynamics. We will study thermodynamic systems. These systems are in thermodynamic equilibrium internally. This means that every part of the system will be at the same temperature, and if the system is a gas, it will all be at the same pressure.

26 Heat and Work We now take a closer look at the first law of thermodynamics. We will study thermodynamic systems. These systems are in thermodynamic equilibrium internally. This means that every part of the system will be at the same temperature, and if the system is a gas, it will all be at the same pressure. In particular, for our present discussion we will be considering an ideal gas. (Can use PV = nrt.)

27 Heat and Work Ideal gas clarification. We make the following assumptions in the ideal gas model: the volume of the gas particles is negligible compared to the total gas volume molecules are identical hard spheres (will relax this later) collisions between molecules are elastic there are no intermolecular forces (aside from hard-sphere collisions) there are no long-range forces from the environment (can be relaxed) A real gas is behaves as an idea gas when it is at high temperature and low density (far from condensation).

28 Variables The variables we will use can be broken into types: state variables describe system s state / properties T, P, V, and E int. transfer variables describing energy transferred into our out of the system Q, W

29 Variables The variables we will use can be broken into types: state variables describe system s state / properties T, P, V, and E int. transfer variables describing energy transferred into our out of the system Q, W intensive variables variables that don t change value when the system is doubled in size P, T, ρ, c extensive variables variables that double their value when the system is doubled in size V, E int, m, C

30 ergy is entering or leaving ndary Work represents done a change on a gas given state of the system, Imagine compressing a gas in a piston quasi-statically (meaning er variable. slowly In enough this section, so the gas remains in thermal equilibrium). namic systems, work. Work r 7, and here we investigate gas contained in a cylinder e gas occupies a volume V d on the piston. If the pise exerted by the gas on the he force exerted by the pisiston inward and compress e system to remain essenoint of application of the A iston is pushed downward dy d S r 5 dy j^ (Fig. 20.4b), the rk in Chapter 7, 2PA dy is discussion. Because A dy a b work done on the gas as How much work is done Figure on20.4 the gas? Work is done on 1 (20.8) a gas contained in a cylinder at a Figure form Serway & pressure Jewett. P as the piston is P V

31 rvation process ransfer leaving change system, Work done on a gas ection,. Work stigate ylinder lume V the pison the the pismpress essenof the nward b), the P A V dy Definition of work: W = F dr For this system: W = ( F j) dy j = F dy = P A dy = P dv se A dy gas as (20.8) ositive.. If the a Figure 20.4 Work is done on a gas contained in a cylinder at a pressure P as the piston is pushed downward so that the gas is compressed. b Vf W = P dv V i

32 e gas be Work done on a gas fer variservation a process transfer r leaving Here, the volume decreases, so the work done on the gas is positive. a change e system, s section, rk. Work vestigate cylinder volume V f the pisas on the y the piscompress in essenn of the ownward 0.4b), the ause A dy e gas as (20.8) P A a V dy Figure 20.4 Work is done on a gas contained in a cylinder at a pressure P as the piston is b Work done on a gas Vf W = P dv The work done Von a gas i equals the negative of the area under the PV curve. The area The work is negative donehere is the because area under the the P-V curve volume (with is decreasing, the appropriate resulting sign). in positive work. P f P i P V f f V i i V To e ing I dep at e imp of d cur N PV d T i d F par illu

33 same initial volume, temperature, and pressure, and is assumed to be ideal. In Figure 20.7a, the gas is thermally insulated from its surroundings except at the bottom of Work done depends on the process the gas-filled region, where it is in thermal contact with an energy reservoir. An energy reservoir Different is a source paths of energy or processes that is considered to go from to (V be i, so P great i ) to (V that f, P a finite f ) require transfer of energy to or from the reservoir does not change its temperature. The piston is held different amounts of work. A constant-pressure compression followed by a constant-volume process A constant-volume process followed by a constantpressure compression An arbitrary compression P P P P f f P f f P f f P i V f V i i V P i a b c V f In all of the processes shown above, there are temperature changes during the process. 1 Figure form Serway & Jewett. V i i V P i V f V i i V

34 Heat transfer also depends on the process This process happens at constant temperature: 20.5 The First Law of The gas is initially at temperature T i. The hand reduces its downward force, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature T i. The g initial tempe T i an conta by a th memb with v above Energy reservoir at T i Energy reservoir at T i a b c Heat Q is transferred to the gas, and the gas does positive work on its surroundings. (Equivalently, negative work is done on the gas.) Figure 20.7 Gas in a cylinder. (a) The gas is in contact with an energy reservoir. The walls of the cylin base in contact with the reservoir is conducting. (b) The gas expands slowly to a larger volume. (c) The half of a volume, with vacuum in the other half. The entire cylinder is perfectly insulating. (d) The gas

35 Heat transfer also depends on the process This process also happens at constant temperature, and has the same start and end points, (V i, P i ) to (V f, P f ): 20.5 The First Law of Thermodynamics 603 The hand reduces its downward force, allowing the piston to move up slowly. The energy reservoir keeps the gas at temperature T i. The gas is initially at temperature T i and contained by a thin membrane, with vacuum above. The membrane is broken, and the gas expands freely into the evacuated region. ir at T i No heat is transferred to the gas, and the gas does no work. c d

36 First Law of Thermodynamics Reminder: Internal energy, E int or U The energy that a system has as a result of its temperature and all other molecular motions, effects, and configurations, when viewed from a reference frame at rest with respect to the center of mass of the system. 1st Law The change in the internal energy of a system is equal to the sum of the heat added to the system and the work done on the system. E int = W + Q This is just the conservation of energy assuming only the internal energy changes.

37 Applying the 1st Law Some special cases of interest: for an isolated system, W = Q = 0 E int = 0 for a process (V i, P i ) to (V i, P i ), a cycle, E int = 0 Q = W

38 Applying the 1st Law: new Vocabulary There are infinitely many paths (V i, P i ) to (V f, P f ) that we might consider. Ones that are of particular interest for modeling systems in engines, etc. are processes that keep a variable constant. There is a technical name for each kind of process that keeps a variable constant.

39 Applying the 1st Law: new Vocabulary Adiabatic process is a process where no heat is transferred into or out of the system. Q = 0 This type of transformation occurs when the gas is in a thermally insulated container, or the process is very rapid, so there is no time for heat transfer. Since Q = 0: E int = W

40 Applying the 1st Law: new Vocabulary Isobaric process is a process that occurs at constant pressure. P i = P = P f This type of transformation occurs when the gas is free to expand or contract by coming into force equilibrium with a constant external environmental pressure. In this case, the expression for work simplifies: W = Vf V i P dv = P(V f V i )

41 Applying the 1st Law: new Vocabulary Isovolumetric process is a process that occurs at constant volume. V i = V = V f In an isovolumetric process the work done is zero, since the volume never changes. W = 0 E int = Q

42 Applying the 1st Law: new Vocabulary Isothermal process is a process that occurs at constant temperature. T i = T = T f This kind of transformation is achieved by putting the system in thermal contact with a large constant-temperature reservoir. In an isothermal process, assuming no change of phase (staying an ideal gas!): E int = 0

43 Isothermal Expansion of an Ideal Gas Since T = 0, PV = nrt reduces to: PV = a where a is a constant. We could also write this as: 606 P = a VChapter 20 The First Law of Thermody Plotting this function: P i P f P i Isotherm PV = constant The curve is a hyperbola. f Isothermal Exp Suppose an ideal g This process is des hyperbola (see App stant indicates that Let s calculate th The work done on t process is quasi-stat V i V f V

44 Isothermal Expansion of an Ideal Gas Since nrt is constant, the work done on the gas in an isothermal expansion is: Vf W = V i Vf P dv nrt = V i V dv ( ) Vf = nrt ln V i ( ) Vi W = nrt ln V f

45 Questions Which path is isobaric? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

46 Questions Which path is isobaric? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D Path D. A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

47 Questions Which path is isothermal? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

48 Questions Which path is isothermal? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D Path C. A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

49 Questions Which path is isovolumetric? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

50 Questions Which path is isovolumetric? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D Path A. A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

51 Questions Which path is adiabatic? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

52 Questions Which path is adiabatic? final state. Figure 20.9 The PV diagram for an isothermal expansion of an ideal gas from an initial state to a P Because T is c n and R: To evaluate th result at the in D Path B. A B C T 1 T 2 T 3 T 4 Figure (Quick Quiz 20.4) Identify the nature of paths A, B, 1 Based on Quick C, Quiz and 20.4, D. Serway & Jewett, page 606. V Numerically, t shown in Figu done on the g the work done Q uick Quiz 2 ric, isotherm

53 Example 20.6: Boiling water This example illustrates how we can apply these ideas to liquids and solids also, and even around phase changes, as long as we are careful. Suppose 1.00 g of water vaporizes isobarically at atmospheric pressure ( Pa). Its volume in the liquid state is V i = V liq = 1.00 cm 3, and its volume in the vapor state is V f = V vap = 1671 cm 3. Find the work done in the expansion and the change in internal energy of the system. Ignore any mixing of the steam and the surrounding air; imagine that the steam simply pushes the surrounding air out of the way.

54 Example 20.6: Boiling water V i = V liq = 1.00 cm 3 V f = V vap = 1671 cm 3 L v = J/kg Work done,

55 Example 20.6: Boiling water V i = V liq = 1.00 cm 3 V f = V vap = 1671 cm 3 L v = J/kg Work done, W = P(V f V i ) = ( Pa)( ) 10 6 m 3 = 169 J Internal energy, E int?

56 Example 20.6: Boiling water V i = V liq = 1.00 cm 3 V f = V vap = 1671 cm 3 L v = J/kg Work done, W = P(V f V i ) = ( Pa)( ) 10 6 m 3 = 169 J Internal energy, E int? Know W, must find Q: Q = L v m = ( J/kg)(1 3 kg) = 2260 J So, E int = W + Q = 2.09 kj

57 Summary latent heat more about phase changes work, heat, and the first law of thermodynamics applying the first law in various cases Homework Serway & Jewett: Read chapter 20 an look at the examples. prev: Ch 20, onward from page 615. OQs: 3, 5, 7; CQs: 3, 11; Probs: 1, 3, 9, 13, 19 new: Ch 20, OQs: 1, 13; Probs: 25, 27, 29, 31, 35, 39, 41, 59, 71, 77

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