problemes termodinamica, Ejercicios de Física

¡Descarga problemes termodinamica y más Ejercicios en PDF de Física solo en Docsity! Chapter 19 The Second Law of Thermodynamics Conceptual Problems 5 • An air conditioner’s COP is mathematically identical to that of a refrigerator, that is, WQcrefAC COPCOP == . However a heat pump’s COP is defined differently, as WQhhpCOP = . Explain clearly why the two COPs are defined differently. Hint: Think of the end use of the three different devices. Determine the Concept The COP is defined so as to be a measure of the effectiveness of the device. For a refrigerator or air conditioner, the important quantity is the heat drawn from the already colder interior, Qc. For a heat pump, the ideas is to focus on the heat drawn into the warm interior of the house, Qh. 9 •• Explain why the following statement is true: To increase the efficiency of a Carnot engine, you should make the difference between the two operating temperatures as large as possible; but to increase the efficiency of a Carnot cycle refrigerator, you should make the difference between the two operating temperatures as small as possible. Determine the Concept A Carnot-cycle refrigerator is more efficient when the temperatures are close together because it requires less work to extract heat from an already cold interior if the temperature of the exterior is close to the temperature of the interior of the refrigerator. A Carnot-cycle heat engine is more efficient when the temperature difference is large because then more work is done by the engine for each unit of heat absorbed from the hot reservoir. 17 •• Sketch an SV diagram of the Carnot cycle for an ideal gas. Determine the Concept Referring to Figure 19-8, process 1→2 is an isothermal expansion. In this process heat is added to the system and the entropy and volume increase. Process 2→3 is adiabatic, so S is constant as V increases. Process 3→4 is an isothermal compression in which S decreases and V also decreases. Finally, process 4→1 is adiabatic, that is, isentropic, and S is constant while V decreases. During the isothermal expansion (from point 1 to point 2) the work done by the gas equals the heat added to the gas. The change in entropy of the gas from point 1 (where the temperature is T1) to an arbitrary point on the curve is given by: 1T QS =Δ 385 Chapter 19 386 For an isothermal expansion, the work done by the gas, and thus the heat added to the gas, are given by: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ == 1 1 ln V VnRTWQ Substituting for Q yields: ⎟⎟⎠ ⎞ ⎜⎜ ⎝ ⎛ =Δ 1 ln V VnRS Since , we have: SSS Δ+= 1 ⎟ ⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ += 1 1 ln V VnRSS The graph of S as a function of V for an isothermal expansion shown to the right was plotted using a spreadsheet program. This graph establishes the curvature of the 1→2 and 3→4 paths for the SV graph. V S An SV graph for the Carnot cycle (see Figure 19-8) is shown to the right. V S 1 2 3 4 Estimation and Approximation 23 •• Estimate the maximum efficiency of an automobile engine that has a compression ratio of 8.0:1.0. Assume the engine operates according to the Otto cycle and assume γ = 1.4. (The Otto cycle is discussed in Section 19-1.) Picture the Problem The maximum efficiency of an automobile engine is given by the efficiency of a Carnot engine operating between the same two temperatures. We can use the expression for the Carnot efficiency and the equation relating V and T for a quasi-static adiabatic expansion to express the Carnot efficiency of the engine in terms of its compression ratio. The Second Law of Thermodynamics 389 (a) The cycle is shown to the right: Apply the ideal-gas law to state 1 to find T1: ( )( ) ( ) K300 Kmol atmL108.206mol1.00 L24.6atm1.00 2 11 1 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ⋅ × == −nR VPT The pressure doubles while the volume remains constant between states 1 and 2. Hence: KTT 6002 12 == The volume doubles while the pressure remains constant between states 2 and 3. Hence: KTT 12002 23 == The pressure is halved while the volume remains constant between states 3 and 4. Hence: KTT 6003214 == For path 1→2: 0Δ 1212 == VPW and ( ) kJ74.3K300K600 Kmol J8.314ΔΔ 2 3 122 3 12V12 =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ Chapter 19 390 The change in the internal energy of the system as it goes from state 1 to state 2 is given by the 1st law of thermodynamics: oninintΔ WQE += Because : 012 =W kJ 74.3Δ 1212int, == QE For path 2→3: ( )( ) kJ99.4 atmL J101.325L24.6L49.2atm2.00Δ 2323on −=⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −−=−=−= VPWW ( ) kJ5.12K600K1200 Kmol J8.314ΔΔ 2 5 232 5 23P23 =−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ Apply to obtain: oninintΔ WQE += kJ 5.7kJ 99.4kJ 5.12Δ 23 int, =−=E For path 3→4: 03434 =Δ= VPW and ( ) kJ48.7K0021K600 Kmol J8.314ΔΔΔ 2 3 342 3 34V34int,34 −=−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ ==== TRTCEQ Apply to obtain: oninintΔ WQE += kJ 48.70kJ 48.7Δ 34 int, −=+−=E For path 4→1: ( )( ) kJ49.2 atmL J101.325L2.94L24.6atm1.00Δ 4141on =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ −−=−=−= VPWW and ( ) kJ24.6K600K003 Kmol J8.314ΔΔ 2 5 412 5 41P41 −=−⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ === TRTCQ Apply to obtain: oninintΔ WQE += kJ 75.3 kJ 49.2kJ 24.6Δ 41 int, −= +−=E The Second Law of Thermodynamics 391 onW inQ For easy reference, the results of the preceding calculations are summarized in the following table: Process , kJ , kJ ( )oninintΔ WQE += , kJ 1→2 0 3.74 3.74 2→3 −4.99 12.5 7.5 3→4 0 −7.48 −7.48 4→1 2.49 −6.24 −3.75 (b) The efficiency of the cycle is given by: ( ) 2312 4123 in by QQ WW Q W + −+− ==ε Substitute numerical values and evaluate ε: %15 kJ5.12kJ3.74 kJ2.49kJ4.99 ≈ + − =ε Remarks: Note that the work done per cycle is the area bounded by the rectangular path. Note also that, as expected because the system returns to its initial state, the sum of the changes in the internal energy for the cycle is zero. Second Law of Thermodynamics 39 •• A refrigerator absorbs 500 J of heat from a cold reservoir and releases 800 J to a hot reservoir. Assume that the heat-engine statement of the second law of thermodynamics is false, and show how a perfect engine working with this refrigerator can violate the refrigerator statement of the second law of thermodynamics. Determine the Concept The following diagram shows an ordinary refrigerator that uses 300 J of work to remove 500 J of heat from a cold reservoir and releases 800 J of heat to a hot reservoir (see (a) in the diagram). Suppose the heat-engine statement of the second law is false. Then a ″perfect″ heat engine could remove energy from the hot reservoir and convert it completely into work with 100 percent efficiency. We could use this perfect heat engine to remove 300 J of energy from the hot reservoir and do 300 J of work on the ordinary refrigerator (see (b) in the diagram). Then, the combination of the perfect heat engine and the ordinary refrigerator would be a perfect refrigerator; transferring 500 J of heat from the cold reservoir to the hot reservoir without requiring any work (see (c) in the diagram).This violates the refrigerator statement of the second law. Chapter 19 394 Noting that V1 = 22.4 L, evaluate V3: ( ) L30.6 atm1 atm1.55L22.4 1.4 1 3 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ =V Substitute numerical values in equation (1) and evaluate T3 and t3: ( ) K373 L22.4 L30.6K2733 ==T and C10027333 °=−= Tt (b) Process 1→2 takes place at constant volume (note that γ = 1.4 corresponds to a diatomic gas and that CP – CV = R): ( ) kJ3.12 K273K423 Kmol J8.314 ΔΔC 2 5 122 5 12V12 = −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = == TRTQ Process 2→3 takes place adiabatically: 023 =Q Process 3→1 is isobaric (note that CP = CV + R): ( ) kJ2.91 K373K732 Kmol J8.314 ΔΔC 2 7 122 7 31P31 −= −⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⋅ = == TRTQ (c) The efficiency of the cycle is given by: inQ W =ε (2) Apply the first law of thermodynamics to the cycle: oninintΔ WQE += or, because (the system begins and ends in the same state) and 0cycle int, =ΔE ingas by theon QWW =−= . Evaluating W yields: kJ0.21kJ2.910kJ3.12 312312 =−+= ++==∑ QQQQW Substitute numerical values in equation (2) and evaluate ε : %7.6 kJ3.12 kJ0.21 ==ε The Second Law of Thermodynamics 395 (d) Express and evaluate the efficiency of a Carnot cycle operating between 423 K and 273 K: 35.5% K234 K73211 h c C =−=−= T Tε *Heat Pumps 49 • As an engineer, you are designing a heat pump that is capable of delivering heat at the rate of 20 kW to a house. The house is located where, in January, the average outside temperature is –10ºC. The temperature of the air in the air handler inside the house is to be 40ºC. (a) What is maximum possible COP for a heat pump operating between these temperatures? (b) What must the minimum power of the electric motor driving the heat pump be? (c) In reality, the COP of the heat pump will be only 60 percent of the ideal value. What is the minimum power of the electric motor when the COP is 60 percent of the ideal value? Picture the Problem We can use the definition of the COPHP and the Carnot efficiency of an engine to express the maximum efficiency of the refrigerator in terms of the reservoir temperatures. We can apply the definition of power to find the minimum power needed to run the heat pump. (a) Express the COPHP in terms of Th and Tc: ch h h c h c h hh HP 1 1 1 1 COP TT T T T Q Q QQ Q W Q c − = − = − = − == Substitute numerical values and evaluate COPHP: 3.6 26.6 K263K313 K133COPHP = = − = (b) The COPHP is also given by: motor out HPCOP P P = ⇒ HP out motor COP P P = Substitute numerical values and evaluate Pmotor: kW2.3 6.26 kW20 motor ==P Chapter 19 396 (c) The minimum power of the electric motor is given by: ( )maxHP, c HP c min COPεε dt dQ dt dQ P == where HPε is the efficiency of the heat pump. Substitute numerical values and evaluate Pmin: ( )( ) kW3.5 6.2660.0 kW20 min ==P Entropy Changes 53 • You inadvertently leave a pan of water boiling away on the hot stove. You return just in time to see the last drop converted into steam. The pan originally held 1.00 L of boiling water. What is the change in entropy of the water associated with its change of state from liquid to gas? Picture the Problem Because the water absorbed heat in the vaporization process its change in entropy is positive and given by T Q S OHby absorbed OH 2 2 Δ = . See Table 18-2 for the latent heat of vaporization of water. The change in entropy of the water is given by: T Q S OHby absorbed OH 2 2 Δ = The heat absorbed by the water as it vaporizes is the product of its mass and latent heat of vaporization: vv OHby absorbed 2 VLmLQ ρ== Substituting for yields: OHby absorbed 2 Q T VLS vOH2Δ ρ = Substitute numerical values and evaluate : OH2ΔS ( ) K kJ05.6 K 373 kg kJ2257L 00.1 L kg 00.1 Δ OH 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ =S 57 •• A system completes a cycle consisting of six quasi-static steps, during which the total work done by the system is 100 J. During step 1 the system absorbs 300 J of heat from a reservoir at 300 K, during step 3 the system absorbs 200 J of heat from a reservoir at 400 K, and during step 5 it absorbs heat from a The Second Law of Thermodynamics 399 Entropy and ″Lost″ Work 63 •• A a reservoir at 300 K absorbs 500 J of heat from a second reservoir at 400 K. (a) What is the change in entropy of the universe, and (b) how much work is lost during the process? Picture the Problem We can find the entropy change of the universe from the entropy changes of the high- and low-temperature reservoirs. The maximum amount of the 500 J of heat that could be converted into work can be found from the maximum efficiency of an engine operating between the two reservoirs. (a) The entropy change of the universe is the sum of the entropy changes of the two reservoirs: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−= +−=Δ+Δ=Δ ch ch chu 11 TT Q T Q T QSSS Substitute numerical values and evaluate ΔSu: ( ) J/K0.42 K300 1 K400 1J500Δ u = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −−=S (b) Relate the heat that could have been converted into work to the maximum efficiency of an engine operating between the two reservoirs: hmaxQW ε= The maximum efficiency of an engine operating between the two reservoir temperatures is the efficiency of a Carnot device operating between the reservoir temperatures: h c Cmax 1 T T −== εε Substitute for εmax to obtain: h h c1 Q T TW ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= Substitute numerical values and evaluate W: ( ) J125J500K400 K3001 =⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=W Chapter 19 400 General Problems 67 • An engine absorbs 200 kJ of heat per cycle from a reservoir at 500 K and releases heat to a reservoir at 200 K. Its efficiency is 85 percent of that of a Carnot engine working between the same reservoirs. (a) What is the efficiency of this engine? (b) How much work is done in each cycle? (c) How much heat is released to the low-temperature reservoir during each cycle? Picture the Problem We can use the definition of efficiency to find the work done by the engine during each cycle and the first law of thermodynamics to find the heat released to the low-temperature reservoir during each cycle. (a) Express the efficiency of the engine in terms of the efficiency of a Carnot engine working between the same reservoirs: ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −== h c C 185.085.0 T T εε Substitute numerical values and evaluate ε : %51510.0K500 K200185.0 ==⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=ε (b) Use the definition of efficiency to find the work done in each cycle: ( )( ) MJ0.10 kJ 102kJ200.5100h = === QW ε (c) Apply the first law of thermodynamics to the cycle to obtain: kJ89 kJ021kJ002cycleh,cyclec, = −=−= WQQ 73 •• (a) Which of these two processes is more wasteful? (1) A block moving with 500 J of kinetic energy being slowed to rest by sliding (kinetic) friction when the temperature of the environment is 300 K, or (2) A reservoir at 400 K releasing 1.00 kJ of heat to a reservoir at 300 K? Explain your choice. Hint: How much of the 1.00 kJ of heat could be converted into work by an ideal cyclic process? (b) What is the change in entropy of the universe for each process? Picture the Problem All 500 J of mechanical energy are lost, i.e., transformed into heat in process (1). For process (2), we can find the heat that would be converted to work by a Carnot engine operating between the given temperatures and subtract that amount of work from 1.00 kJ to find the energy that is lost. In Part (b) we can use its definition to find the change in entropy for each process. (a) For process (2): inCrecoveredmax,2 QWW ε== The Second Law of Thermodynamics 401 The efficiency of a Carnot engine operating between temperatures Th and Tc is given by: h c C 1 T T −=ε and hence in h c recovered 1 QT TW ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −= Substitute for Cε to obtain: ( ) J250kJ1.00 K 400 K 3001recovered =⎟ ⎠ ⎞ ⎜ ⎝ ⎛ −=W or 750 J are lost. Process (1) produces more waste heat. Process (2) is more wasteful of available work. (b) Find the change in entropy of the universe for process (1): J/K1.67 K300 J500ΔΔ 1 === T QS Express the change in entropy of the universe for process (2): ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −Δ= Δ + Δ −=Δ+Δ=Δ hc ch ch2 11 TT Q T Q T QSSS Substitute numerical values and evaluate ΔS2: ( ) J/K833.0 K400 1 K300 1kJ1.00Δ 2 = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=S 75 •• A heat engine that does the work of blowing up a balloon at a pressure of 1.00 atm absorbs 4.00 kJ from a reservoir at 120ºC. The volume of the balloon increases by 4.00 L, and heat is released to a reservoir at a temperature Tc, where Tc < 120ºC. If the efficiency of the heat engine is 50% of the efficiency of a Carnot engine working between the same two reservoirs, find the temperature Tc. Picture the Problem We can express the temperature of the cold reservoir as a function of the Carnot efficiency of an ideal engine and, given that the efficiency of the heat engine is half that of a Carnot engine, relate Tc to the work done by and the heat input to the real heat engine. Using its definition, relate the efficiency of a Carnot engine working between the same reservoirs to the temperature of the cold reservoir: h c C 1 T T −=ε ⇒ ( )Chc 1 ε−= TT Chapter 19 404 The work done during the isobaric compression CD is: ( ) ( )( ) kJ5.680 Latm J101.325Latm56.09L75.78L19.7atm1.00CDCCD −= ⋅ ×⋅−=−=−= VVPW Express and evaluate the work done during the constant-volume process DA: 0DA =W Substitute numerical values and evaluate W: kJ97.6kJ972.6 0kJ5.680kJ5.737kJ915.6 == +−+=W 83 ••• A common practical cycle, often used in refrigeration, is the Brayton cycle, which involves (1) an adiabatic compression, (2) an isobaric (constant pressure) expansion,(3) an adiabatic expansion, and (4) an isobaric compression back to the original state. Assume the system begins the adiabatic compression at temperature T1, and transitions to temperatures T2, T3 and T4 after each leg of the cycle. (a) Sketch this cycle on a PV diagram. (b) Show that the efficiency of the overall cycle is given by ε = 1− T4 − T1( ) T3 − T2( ) . (c) Show that this efficiency, can be written as ε = 1− r 1−γ( ) γ , where r is the pressure ratio Phigh/Plow of the maximum and minimum pressures in the cycle. Picture the Problem The efficiency of the cycle is the ratio of the work done to the heat that flows into the engine. Because the adiabatic transitions in the cycle do not have heat flow associated with them, all we must do is consider the heat flow in and out of the engine during the isobaric transitions. (a) The Brayton heat engine cycle is shown to the right. The paths 1→2 and 3→4 are adiabatic. Heat Qh enters the gas during the isobaric transition from state 2 to state 3 and heat Qc leaves the gas during the isobaric transition from state 4 to state 1. 1 2 3 4 P V ⇓ ⇓ hQ cQ The Second Law of Thermodynamics 405 (b) The efficiency of a heat engine is given by: in ch in Q QQ Q W − ==ε (1) During the constant-pressure expansion from state 1 to state 2 heat enters the system: ( )23PPh23 Δ TTnCTnCQQ −=== During the constant-pressure compression from state 3 to state 4 heat enters the system: ( )41PPc41 Δ TTnCTnCQQ −−=−=−= Substituting in equation (1) and simplifying yields: ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) ( )23 14 23 4123 23P 41P23P 1 TT TT TT TTTT TTnC TTnCTTnC − − −= − −+− = − −−−− =ε (c) Given that, for an adiabatic transition, , use the ideal-gas law to eliminate V and obtain: constant 1 =−γTV constant 1 =−γ γ P T Let the pressure for the transition from state 1 to state 2 be Plow and the pressure for the transition from state 3 to state 4 be Phigh. Then for the adiabatic transition from state 1 to state 2: 1 high 2 1 low 1 −− = γ γ γ γ P T P T ⇒ 2 1 high low 1 TP PT γ γ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Similarly, for the adiabatic transition from state 3 to state 4: 3 1 high low 4 TP PT γ γ − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = Subtract T1 from T4 and simplify to obtain: ( )23 1 high low 2 1 high low 3 1 high low 14 TT P P T P PT P PTT −⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ =− − −− γ γ γ γ γ γ Chapter 19 406 Dividing both sides of the equation by T3 − T2 yields: γ γ 1 high low 23 14 − ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ = − − P P TT TT Substitute in the result of Part (b) and simplify to obtain: ( ) γ γ γ γ γ γ ε − −− −= ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ −=⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ −= 1 1 low high 1 high low 1 11 r P P P P where low high P P r = 89 ••• The English mathematician and philosopher Bertrand Russell (1872- 1970) once said that if a million monkeys were given a million typewriters and typed away at random for a million years, they would produce all of Shakespeare’s works. Let us limit ourselves to the following fragment of Shakespeare (Julius Caesar III:ii): Friends, Romans, countrymen! Lend me your ears. I come to bury Caesar, not to praise him. The evil that men do lives on after them, The good is oft interred with the bones. So let it be with Caesar. The noble Brutus hath told you that Caesar was ambitious, And, if so, it were a grievous fault, And grievously hath Caesar answered it . . . Even with this small fragment, it will take a lot longer than a million years! By what factor (roughly speaking) was Russell in error? Make any reasonable assumptions you want. (You can even assume that the monkeys are immortal.) Picture the Problem There are 26 letters and four punctuation marks (space, comma, period, and exclamation point) used in the English language, disregarding capitalization, so we have a grand total of 30 characters to choose from. This fragment is 330 characters (including spaces) long; there are then 30330 different possible arrangements of the character set to form a fragment this long. We can use this number of possible arrangements to express the probability that one monkey will write out this passage and then an estimate of a monkey’s typing speed to approximate the time required for one million monkeys to type the passage from Shakespeare. Assuming the monkeys type at random, express the probability P that one monkey will write out this passage: 33030 1 =P