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Chapter 16 Thermodynamics 第十六章 熱學 What is heat? Heat can flow from a hotter body to a colder body. In the eighteenth century, heat was believed to be an invisible, massless fluid called “caloric”. The caloric theory assumes that caloric is a conserved quantity and that the particles of caloric repel each other but are attracted to the particles of ordinary matter. Heat flow Heat is the energy transferred between a system and its environment because of a temperature difference that exists between them. Raising temperature Heat capacity In 1760 Joseph Black was the first person to realize that the rise in temperature of a body could be used to determine the quantity of heat absorbed by it. Q Heat capacity T A commonly used (non-SI) unit of heat is the calorie, which is defined as the quantity of heat required to raise the temperature of 1 g of water from 14.5 °C to 15.5 °C. 1 calorie 4.186 J A British unit for heat is Btu, that is the amount of heat required to raise the temperature of 1 lb of water from 63 °F to 64 °F. 1 BTU 252.5 calorie 1057 J Specific heat The amount of heat required to raise a temperature change of ΔT of a material is proportional to its mass. Q mcT The proportional constant c is defined to be the specific heat of the material. 1 Q c m T The molar specific heat is defined as C = Mc, where M is the molar mass. Q nCT where n is the number of moles in the material. Specific heat of some materials Specific heat of water Example 1 During a bout with the flu an 80-kg man ran a fever of 39.0°C instead of the normal body temperature of 37.0°C. Assuming that the human body is mostly water, how much heat is required to raise his temperature by that amount? Q mcT 80 4180 2 J 6.7 105 J 160 Cal Determining specific heat The method mixtures Q1 Q2 0 m1c1T1 m2c2T2 0 T1 T f T1 T2 T f T2 Latent heat Joseph Black also recognize that adding heat to a system does not always result in a change of its temperature. This happens when the material in the system undergoes a first-order phase change. The latent heat L is defined to be the heat needed to convert the material of one unit mass from one phase to another at its phase transition temperature. Q mL is the heat needed for a sample of mass m. Latent heats and phase transitions of water The surrounding air is at room temperature, but this ice–water (watervapor) mixture remains at 0°C (100°C) until all of the ice (water) has melted (evaporated) and the phase change is complete. Latent heats of some materials Generation of heat by friction In 1798 Benjamin Thompson noticed that a large amount of heat was generated during the process of boring of cannons. It seemed to him that heat could be continually produced by mechanical work. The mechanical equivalent of heat James Joule first performed the mechanical work-heat conversion experiment in 1845, and he repeated the experiment several times. His final result after 40 years of work is 778 ft·lb of work is equivalent to 1 Btu. The mechanical equivalent of heat 1Btu 778 ft lb 778 (12 2.54 /100) (9.8 / 2.2)m N 1056 J 1Btu 252.5calorie 1calorie 4.183J Modern day value: 1calorie 4.186J Heat is energy transferred between two bodies as a consequence of a temperature difference between them. Signs for heat and work in thermodynamics Work done during volume change dW Fdx pAdx pdV Work done during volume change V2 W pdV V1 Example 2 An ideal gas undergoes an isothermal (constant-temperature) expansion at temperature T, during which its volume changes from V1 to V2. How much work does the gas do? V2 W pdV V1 V2 W V1 pV nRT nRT V2 dV nRT n( ) V V1 V2 p1 W nRT n( ) nRT n( ) V1 p2 Path between thermodynamic states Heat added in a thermodynamic process Internal energy and the first law of thermodynamics First law of thermodynamics: U Q W Internal energy is a function of thermodynamic states U Q W While Q and W depend on the path, U = Q W is independent of path. The change in internal energy of a system during any thermodynamic process depends only on the initial and final states, not on the path leading from one to the other. The internal energy of a cup of coffee depends on just its thermodynamic state— how much water and ground coffee it contains, and what its temperature is. It does not depend on the history of how the coffee was prepared—that is, the thermodynamic path that led to its current state. Cyclic processes and isolated systems A cyclic process is a process that eventually returns a system to its initial state. U Q W 0 Q W For an isolated system there are no work done and no heat exchange with its surroundings. Q W 0 U 0 Infinitesimal change of states U Q W dW pdV dU dQ dW dU dQ pdV Example 3 You propose to eat a 900-calorie hot fudge sundae (with whipped cream) and then run up several flights of stairs to work off the energy you have taken in. How high do you have to climb? Assume that your mass is 60.0 kg. One food calorie equals 1000 cal, or 1 kcal. 1cal 4.190J Q W mgh Q h 6.41km mg Example 4 The figure below shows a pV-diagram for a cyclic process, one in which the initial and final states are the same. It starts at point a and proceeds counterclockwise in the pV-diagram to point b, then back to a, and the total work is W = 500J. (a) Why is the work negative? (b) Find the change in internal energy and the heat added during this process. The work done equals the area under the curve, with the area taken as positive for increasing volume and negative for decreasing volume. The positive work done by the system from a to b is less than the absolute value of the negative work by the system from b to a. U 0 Q W 500 J 500 J of heat coming out of the system. Example 5 A series of thermodynamic processes is shown in the pV-diagram of the attached figure. In process ab, 150 J of heat is added to the system, and in process bd, 600 J of heat is added. Find (a) the internal energy change in process ab; (b) the internal energy change in process abd (shown in light blue); and (c) the total heat added in process acd (shown in dark blue). U Q W In process ab: In process bd: is valid for each process. W 0 U ab Qab Wbd pb (Vd Vd ) U bd Qbd Wbd U ad U ab U bd U bd Qacd U ad Wac Example 6 One gram of water (1 cm3) becomes 1671 cm3 of steam when boiled at a constant pressure of 1 atm (1.013 105 Pa). The heat of vaporization at this pressure is LV = 2.256 106 J/kg. Compute (a) the work done by the water when it vaporizes and (b) its increase in internal energy. W p(V2 V1 ) 1.013 105 1670 106 J 169 J Q mLV 2256 J U Q W 2087 J Types of thermodynamic processes (a) adiabatic process: no heat transferred into or from the system. Q0 U W (b) isochoric process: a constant-volume process. W 0 U Q (c) isobaric process: a constant pressure process. W p(V2 V1 ) U Q p(V2 V1 ) (d) Isothermal process: a constant temperature process. T 0 (e) Adiabatic free expansion: a free expansion without exchanging heat or doing work to the system’s surrounding. Q W 0 while V increases. What kind of process is it? When the cork is popped on a bottle of champagne, When the partition is broken, When the soup is heated in a pot, Ideal gases The internal energy of an ideal gas depends only on its temperature, not on its volume or pressure. Molar specific heat at constant volume: CV Since the work done by the gas is zero, we have U Q nCV T Molar specific heat at constant pressure: Cp For n moles of a gas: Q nC p T Since the work done by the gas is pV, we have U Q W nC p T pV nCV T nC p T pV nCV T nC p T nRT C p CV R Molar specific heats of some gases The ratio of heat capacities, Cp/CV, is denoted by . Cp CV Example 7 A typical dorm room or bedroom contains about 2500 moles of air. Find the change in the internal energy of this much air when it is cooled from 23.9C to 11.6C at a constant pressure of 1.00 atm. Treat the air as an ideal gas with = 1.400. U nCV T C R V CV CV Cp R CV 1 Adiabatic processes for an ideal gas In an adiabatic process: dQ 0 dU pdV From PV = nRT, we have pdV Vdp nRdT CV ( pdV Vdp) RnCV dT RpdV (CV R) pdV CV Vdp 0 dV CV R CV CV Cp pV const. dp 0 V p or TV 1 const. nCV dT pdV Isothermal work and free expansion Isothermal process: the system is kept in contact with a single heat bath at temperature T when it expands or contracts. Isothermal expansion Free expansion Isothermal work by ideal gas dW pdV V2 W pdV V1 nRT p V dV W nRT V1 V V2 W nRT n( ) V1 V2 Example 8 The compression ratio of a diesel engine is 15 to 1; this means that air in the cylinders is compressed to 1/15 of its initial volume (Fig. 19.21). If the initial pressure is 1.01 105 Pa and the initial temperature is 27C, find the final pressure and the temperature after compression. Air is mostly a mixture of diatomic oxygen and nitrogen; treat it as an ideal gas with = 1.40. The piston compresses quickly, so it is an adiabatic process. p1V1 p2V2 1 T1V1 1 T2V2 V1 p2 p1 ( ) V2 V1 1 T2 T1 ( ) V2 Example 9 In Example 8, how much work does the gas do during the compression if the initial volume of the cylinder is 1.00 L. Assume that CV for air is 20.8 J/mole·K and = 1.40. V1 p p1 ( ) V V2 W pdV V1 V2 W V1 V1 V2 p1 ( ) dV p1V1 V dV V1 V W p1V1 1 2 V 1 1 V 1 p1V1 p2V2 1 Speed of sound v dP B V dV B Newton assumed that the compressions and rarefactions occur isothermally: dP nRT B V V ( 2 ) P dV V v B P nM / V RT M The correct process is adiabatic: B P v RT M The second law of thermodynamics tells us that heat naturally flows from a hot body (such as a freshly cooked ear of corn) to a cold one (such as a pat of butter). Is it ever possible for heat to flow from a cold body to a hot one? Irreversible processes An egg is dropped onto a floor, a pizza is baked, a car is driven into a lamppost, large waves erode a sandy beach, —these one-way processes are irreversible, meaning that they cannot be reversed by means of only small changes in their environment. Reversible processes Reversible processes are thus equilibrium processes, with the system always in thermodynamic equilibrium. The limit of a heat engine Industrial revolution of the West Steam engine and Carnot’s analysis Heat engines Any device that transforms heat partly into work or mechanical energy is called a heat engine. Usually, a quantity of matter inside the engine undergoes inflow and outflow of heat, expansion and compression, and sometimes change of phase. We call this matter the working substance of the engine. A cyclic process is a sequence of processes that eventually leaves the working substance in the same state in which it started. All heat engines absorb heat from a source at a relatively high temperature, perform some mechanical work, and discard or reject some heat at a lower temperature. A heat reservoir is a large thermodynamic system that can exchange heat with the working substance without changing the reservoir’s temperature. Heat engines and thermal efficiency The thermal efficiency of a heat engine is defined as the work output divided by the heat input: W QH W QH QC QH QC 1 QC QH Example 10 A gasoline engine in a large truck takes in 10,000 J of heat and delivers 2000 J of mechanical work per cycle. The heat is obtained by burning gasoline with heat of combustion LC = 5.0 104 J/g. (a) What is the thermal efficiency of this engine? (b) How much heat is discarded in each cycle? (c) How much gasoline is burned in each cycle? (d) If the engine goes through 25 cycles per second, what is its power output in watts? In horsepower? (e) How much gasoline is burned per second? Per hour? W 2000 J 0.2 QH 10000 J QH mgasoline 0.20 g LC QC QH W 8000 J P (25cycles / s) (2000 J / cycle ) 50kW gas burned per second : (0.20 g / cycle ) (25cycle / s) 5.0 g / s The gasoline engine (Otto Cycle) The gasoline engine (Otto Cycle) The six steps in the idealized Otto cycle: 1. Intake stroke (x to a): VxrV 2. Compression stroke (a to b): rVV 3. Ignition (b to c): V 4. Power stroke (c to d): V rV x 5. Exhaust (d to a): rV 6. Exhaust stroke (a to x): rVVx Efficiency of Otto cycle QH nCV (Tc Tb ) 0 QC nCV (Ta Td ) 0 1 QC QH Td Ta 1 Tc Tb For the two adiabatic processes ab and cd: Ta (rV ) 1 TbV 1 1 1 r 1 Td (rV ) 1 TcV 1 for an ideal Otto cycle. 0.56 for r 8 and 1.4, while 0.35 for real engines. The Diesel engine The four steps in the Diesel cycle: 1. Compression stroke (a to b): rVV 2. Fuel injection and ignition (b to c): V Vc at constant Pb 3. Power stroke (c to d): VcrV 4. Cooling (d to a): rV Typical theoretical efficiency of an Diesel engine is 0.65 to 0.70. Real efficiencies of heat engines Gasoline engines: about 56%. Diesel engines: about 65% to 70%. Refrigerators A refrigerator as a heat engine operating in reverse, i.e. it takes heat from a cold place (the inside of the refrigerator) and gives it off to a warmer place (usually the air in the room where the refrigerator is located). The coefficient of performance K is defined as QC K W QH W QC QC K QC QH Coefficient of performance (COP) of a heat pump is defined as: K QH W QH QC QH Principle of the mechanical refrigeration cycle. An air conditioner QC H K W P where H is the heat removal rate and P is the power needed, and K is typically 3. EER is the coefficient of performance with H given in BTU/H and P in watt (W). 1W 3.413BTU / H The second law of thermodynamics It is impossible to build a heat engine that converts heat completely to work—that is, an engine with 100% thermal efficiency. The “engine” statement or the Kelvin-Planck statement of the second law: It is impossible for any system to undergo a process in which it absorbs heat from a reservoir at a single temperature and converts the heat completely into mechanical work, with the system ending in the same state in which it began. The “refrigerator” statement or the Clausius statement of the second law: It is impossible for any process to have as its sole result the transfer of heat from a cooler to a hotter body. The Kelvin-Planck statement of the second law It is impossible for a heat engine that operates in a cycle to convert its input heat completely into work. The Clausius statement of the second law It is impossible for a cyclic device to transfer heat continuously from a cold body to a hot body without the input of work or other effect on the environment. Equivalence of the Kelvin-Planck and Clausius statements Equivalence of the Kelvin-Planck and Clausius statements Reversible and irreversible processes For a process to be reversible three conditions must be satisfied: (a) It must be quasistatic. (b) There must be no friction. (c) Any transfer of heat must occur at constant temperature, or be associated with infinitesimal temperature difference. Irreversible and reversible engines The temperature of the firebox of a steam engine is much higher than the temperature of water in the boiler, so heat flows irreversibly from firebox to water. Carnot’s quest to understand the efficiency of steam engines led him to the idea that an ideal engine with only reversible processes. The Carnot Cycle In 1824, Sadi Carnot devised a reversible cycle of operations, called Carnot cycle. Efficiency of the Carnot cycle a to b: c to d: Vb QH nRTH n( ) Va V QC nRTC n( c ) Vd For an adiabatic process, we can show: TV 1 const. TH Vb 1 TCVc 1 Vb Vc Va Vd TH Va 1 TCVd 1 QC QH TC TH Carnot efficiency: C 1 TC TH Example 11 A Carnot engine takes 2000 J of heat from a reservoir at 500 K, does some work, and discards some heat to a reservoir at 350 K. How much work does it do, how much heat is discarded, and what is the efficiency? QC QH TC TH TC QC QH 1400 J TH W QH QC 600 J C 1 TC 0.30 TH Example 12 Suppose 0.200 mol of an ideal diatomic gas ( = 1.40) undergoes a Carnot cycle with temperatures 227°C and 27°C. The initial pressure is 10.0 105 Pa, and during the isothermal expansion at the higher temperature the volume doubles. (a) Find the pressure and volume at each of points a, b, c, and d in the pV-diagram. (b) Find Q, W, and U for each step and for the entire cycle. (c) Determine the efficiency directly from the results of part (b), and compare it with the result from C 1 TC / TH . pV nRT Va nRTa / pa 8.3110 4 m3 Vb 2Va 1.66 10 3 m3 pb paVa / Vb 5.00 105 Pa TV 1 const. QH nRTH n(Vb / Va ) 0.200 8.314 500 n(2) J Carnot refrigerator K QC W K Carnot QC QH QC TC TH TC Example 13 If the cycle described in Example 3 is run backward as a refrigerator, what is its coefficient of performance? K Carnot TC 300 1.5 TH TC 500 300 Carnot’s Theorem All reversible engines operating between two given reservoirs have the same efficiency. No cyclical heat engine has a greater efficiency than a reversible engine operating between the same two temperatures. The Kelvin temperature scale QC QH TC TH T Q QR TR If we choose the reference temperature to be 273.16 K at the triple point of water, then it follows that the Kelvin and ideal-gas scales are identical. Entropy The second law of thermodynamics, as we have stated it, is rather different in form from many familiar physical laws. It is not an equation or a quantitative relationship but rather a statement of impossibility. However, the second law can be stated as a quantitative relationship with the concept of entropy. We have talked about several processes that proceed naturally in the direction of increasing disorder. Entropy provides a quantitative measure of disorder. Reversible cycles Any reversible cycle can be approximated by a series of Carnot cycles as shown in the left figure. For a Carnot cycle, we have: QC QH TC TH or QH QC 0 TH TC Thus for a reversible cycle, we have: dQ Q T T 0 Thermodynamic definition of entropy b dQ b dQ dQR R R T a T a T 0 path I path II Thus we can define a state function S as the potential energy from a conservative force: S S f Si i or dS dQR T f dQR T Example 14 One kilogram of ice at 0C is melted and converted to water at 0C. Compute its change in entropy, assuming that the melting is done reversibly. The heat of fusion of water is Lf = 3.34 105 J/kg. S i f dQR Q 3.34 105 J 1.22 103 J / K T T 273K Example 15 One kilogram of water at 0C is heated to 100C. Compute its change in entropy. S i f T f mcdT Tf dQR mcn( ) Ti T T Ti 373 S 1.00kg (4.19kJ / kg K ) n( ) 1.31103 J / K 273 Reversible process for an ideal gas The first law: dQ dU dW nRT dQ nCV dT pdV nCV dT dV V dQ dT dV nCV nR T T V f dQ Tf Vf dT dV S nCV nR i T Vi i T T V S nCV n( Tf Ti ) nRn( Vf Vi ) Example 16 A gas expands adiabatically and reversibly. What is its change in entropy? In an adiabatic process, no heat enters or leaves the system. Hence dQ = 0 and there is no change in entropy in this reversible process. Every reversible adiabatic process is a constant entropy process. For this reason, reversible adiabatic processes are also called isentropic processes. The increase in disorder resulting from the gas occupying a greater volume is exactly balanced by the decrease in disorder associated with the lowered temperature and reduced molecular speeds. Example 17 A thermally insulated box is divided by a partition into two compartments, each having volume V. Initially, one compartment contains n moles of an ideal gas at temperature T, and the other compartment is evacuated. We then break the partition, and the gas expands to fill both compartments. What is the entropy change in this free-expansion process? Since W = 0, we have ΔU = 0. For an ideal gas, it means that Tf = Ti = T. To figure out ΔS we need to find a reversible process that links the final state to the initial state. It is a quasistatic isothermal expansion. S g nRn( Vf Vi ) Note that there is no change in its surrounding, so ΔSe = 0. Su S g S e 0 Example 18 For the Carnot engine in Example 11, find the total entropy change in the engine during one cycle. QC 1400J QH S H 4J / K TH QC S C 4 J / K TC STotal S H SC 0 Entropy in irreversible processes In an idealized, reversible process involving only equilibrium states, the total entropy change of the system and its surroundings is zero. But all irreversible processes involve an increase in entropy. Unlike energy, entropy is not a conserved quantity. The entropy of an isolated system can change, it can never decrease. Example 19 Suppose 1.00 kg of water at 100C is placed in thermal contact with 1.00 kg of water at 0C. What is the total change in entropy? Assume that the specific heat of water is constant at 4190 J/kg·K over this temperature range. S H Tf Ti SC Tf Ti STotal 323mcdT dQR 373 mcn( ) 373 T T 323 323 mcdT dQR 323 mcn( ) 273 T T 273 323 323 S H SC mcn( ) 102 J / K 273 373 Entropy and the second law In a reversible process the entropy of an isolated system stays constant; in an irreversible process the entropy increases. S 0 Clausius statement leads to ΔS 0 and vice versa. The availability of energy One can convert work into heat completely, but cannot convert heat into work completely. As matter of fact, convert some of the heat at one temperature into work requires the existence of another reservoir at a lower temperature. We note that: 1. The degradation of energy is associated with the transition of the system from an ordered to a disordered state. 2. In any natural (irreversible) process, some energy becomes unavailable to do useful work. Entropy and disorder The second law requires that an isolated system tends to evolve towards states of higher entropy. A local decrease in entropy of a system can occur at the expense of a greater increase in entropy in its environment. The second law is often called “time’s arrow” because it tells us that in (irreversible) natural processes an isolated system always evolves toward states of higher entropy. One prediction of the second law about the evolution of the universe is that it will evolve toward a state of thermodynamic equilibrium characterized by uniform density, temperature and pressure. This gloomy state of maximum entropy is usually referred to as the “heat death”. Summary_1 Heat: a transfer of energy Specific heat: Q mcT nC T Latent heat: Q mL Work done in a quasistatic process: W Vf Vi PdV First law of thermodynamics: U Q W Internal energy of an ideal gas: U nCv T Relationship between molar specific heats: C p Cv R Quasistatic adiabatic: PV const. Summary_2 The thermal efficiency of a heat engine is defined as the work output divided by the heat input: QL W 1 QH QH The second law of thermodynamics: Kelvin-Plank: It is impossible for a heat engine that operates in a cycle to convert its heat completely into work. Clausius: It is impossible for a cyclic device to transfer heat continuously from a cold body to a hot body without the input of work or other effect on the environment. In a reversible process the entropy of an isolated system stays constant; in an irreversible process the entropy increases. Summary_3 Entropy is a state function. The change in entropy S between a initial state to a final state is defined by any reversible process that links the two equilibrium states. S S f Si i f dQR T Entropy is a measure of the disorder in a system. The second law states that the natural (irreversible) processes tend to evolve to states of greater disorder, or from states of low probability to states of high probability. Further reading & homework • The Feynman lectures on Physics, Vol. 1, Chapter 44, 45, 46 • Questions: 16.1, 16.5, 16.8, 16.10, 16.16 • Problems: 16.13, 16.17, 16.18, 16.20, 16.25, 16.30, 16.31, 16.35, 16.36