CHAPTER 15: The Laws of Thermodynamics. Answers to Questions

Transcription

1 CAPER 5: he aws f hermdynamics Answers t uestins. If water vapr cndenses n the utside f a cld glass f water, the internal energy f the water vapr has decreased, by an amunt equal t the heat f vaprizatin f the water vapr. eat energy has left the water vapr, causing it t cndense, and heat energy has entered the glass f water, and the air, causing them t get slightly warmer. N wrk is dne, but heat is exchanged.. During cmpressin, wrk is dne n the gas. Assuming that there is n heat flw t r frm the gas (since the prcess is quick), by cnservatin f energy (the first law f thermdynamics) the wrk dne n the gas becmes internal energy f the gas, and s the temperature f the gas is increased. During expansin, wrk is dne by the gas n its surrundings. Again assuming that there is n heat flw t r frm the gas, by cnservatin f energy, the wrk is dne by the gas at the expense f the internal energy f the gas, and s the temperature f the gas is decreased.. Since the prcess is isthermal, there is n change in the internal energy f the gas. hus U 0, and s the heat absrbed by the gas is equal t the wrk dne by the gas. hus 700 J f heat was added t the gas.. It is pssible fr temperature (and thus internal energy) t remain cnstant in a system even thugh there is heat flw int r ut f the system. By the first law f thermdynamics, there must be an equal amunt f wrk dne n r by the system, s that U 0. he isthermal expansin r cmpressin f a gas wuld be an example f this situatin. 5. If the gas is cmpressed adiabatically, n heat enters r leaves frm the gas. he cmpressin means that wrk was dne ON the gas. By the first law f thermdynamics, U, since 0, then U. he change in internal energy is equal t the ppsite f the wrk dne by the gas, r is equal t the wrk dne n the gas. Since psitive wrk was dne n the gas, the internal energy f the gas increased, and that crrespnds t an increase in temperature. his is cnservatin f energy the wrk dne n the gas becmes internal energy f the gas particles, and the temperature increases accrdingly. 6. Mechanical energy can be transfrmed cmpletely int heat. As a mving bject slides acrss a rugh level flr and eventually stps, the mechanical energy f the mving bject has been transfrmed cmpletely int heat. Als, if a mving bject were t be used t cmpress a frictinless pistn cntaining an insulated gas, the kinetic energy f the bject wuld becme internal energy f the gas. A gas that expands adiabatically (withut heat transfer) transfrms internal energy int mechanical energy, by ding wrk n its surrundings at the expense f its internal energy. Of curse, that is an ideal (reversible) prcess. 7. It is pssible t warm the kitchen in the winter by having the ven dr pen. he ven heating elements radiate heat energy int the ven cavity, and if the ven dr is pen, the ven is just heating a bigger vlume than usual. wever, yu cannt cl the kitchen by having the refrigeratr dr pen. he refrigeratr exhausts mre heat than it remves frm the refrigerated vlume, s the rm actually gets warmer with the refrigeratr dr pen. If yu culd have the refrigeratr exhaust int sme ther rm, then the refrigeratr wuld be similar t an air cnditiner, and it culd cl the kitchen, while heating up sme ther space. 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 57

2 Chapter 5 he aws f hermdynamics 8. his definitin f efficiency is nt useful, because with this definitin, if the exhaust heat is less than the wrk dne (which is pssible), the efficiency wuld exceed unity. Efficiency shuld be cmparing t the heat input, nt the heat utput. 9. (a) In an internal cmbustin engine, the high temperature reservir is the ignited gas-air mixture. he lw temperature reservir is the gases exhausted frm the cylinder int the atmsphere. (b) In a steam engine, the high temperature reservir is the heated, high-pressure steam frm the biler. he lw temperature reservir is the lw-pressure steam frm the exhaust. 0. he efficiency f a Carnt engine is given by Eq. 5-5, e. Bth a decrease in and an increase in wuld cause the value f t decrease, increasing the efficiency. Since, the 0C change is a larger percentage f change fr, and s will change the fractin mre than 0 0 the same numeric increase in the denminatr. Nte e and e denminatr, and s e e.. Bth efficiencies have the same numeratr, but e has a larger. utilize the thermal energy in the cean waters, a heat engine wuld need t be develped that perated between tw different temperatures. If surface temperature water was t be bth the surce and the exhaust, then n wrk culd be extracted. If the temperature difference between surface and deep cean waters were t be used, there wuld be cnsiderable engineering bstacles, high expense, and ptential envirnmental difficulties invlved in having a heat engine that cnnected surface water and deep cean water. ikewise, if the difference in temperature between trpical water and arctic r Antarctic water were t be used, the same type f majr difficulties wuld be invlved because f the large distances invlved.. (a) If a gas expands adiabatically, there is n heat transfer, and therefre S 0 by Eq. 5-8, S. (b) If a gas expands isthermally, there is n change in its internal energy, and the gas des wrk n its surrundings. hus by the first law f thermdynamics, there must be heat flw int the gas, and s S 0 the entrpy f the gas increases.. he adiabatic expansin results in n change in entrpy, since there is n heat transfer. he isthermal expansin requires heat flw int the gas t cmpensate fr the wrk that the gas des, and s the entrpy f the gas S increases mre fr the isthermal expansin.. (a) he ersin f sil due t water flw ver the grund. (b) he xidatin f varius metals (cpper, zinc, irn, etc.) when left expsed t the air. (c) he cnversin f mechanical energy t heat energy by frictin; i.e., a sliding bject decreasing in speed and eventually stpping, and the surfaces f cntact getting warmer. (d) A pile f cmpst decmpsing. he reverse f these prcesses is nt bserved. 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 58

3 Giancli Physics: Principles with Applicatins, 6 th Editin 5. kg f liquid irn has mre entrpy, because the atms in liquid irn are less rdered than thse in slid irn. Als, heat had t be added t slid irn in rder t melt it, and S. 6. (a) If the lid is remved frm a bttle f chlrine gas, the gas mlecules will diffuse ut f the muth f the bttle, and eventually spread ut unifrmly in whatever vlume t which they are cnfined. (b) he reverse prcess, that f individual chlrine gas mlecules in a clsed vlume spntaneusly entering a small vlume, never happens. he prbability f the gas mlecules all entering the bttle is infinitesimal cmpared t the prbability f the gas mlecules being unifrmly spread thrughut the rm. he reverse prcess wuld require a spntaneus decrease in entrpy. (c) Sme ther examples f irreversibility: the shuffling f an rdered deck f cards; the diffusin f dye in a liquid; the tppling f buildings during an earthquake. 7. Any air cnditiner-type heat engine will remve heat frm the rm ( the lw temperature input). rk ( ) is input t the device t enable it t remve heat frm the lw temperature regin. By the nd law f thermdynamics (cnservatin f energy), there must be a high-temperature exhaust heat which is larger than. Perhaps the inventr has sme clever methd f having that exhaust heat mve int a well-insulated heat sink, like a cntainer f water. But eventually the additin f that heat int the device will cause the device t heat up warmer than the rm, and then heat will be transferred t the rm. One very simple device that culd d what is described in the prblem wuld be a fan blwing ver a large blck f ice. eat frm the rm will enter the ice; cl air frm near the surface f the ice can be blwn ut f the bx by a fan. But after the ice melts, the nly end result is that the fan mtr wuld heat the air. 8. (a) An empty perfume bttle is placed in a rm cntaining perfume mlecules, and all f the perfume mlecules mve int the bttle frm varius directins at the same time. (b) ater n the sidewalk calesces int drplets, are prpelled upward, and rise int the air. (c) Ppcrn is placed in the refrigeratr, and it unpps, changing backed int uncked kernels. (d) A huse gt warmer in the winter while the utdrs gt clder, due t heat mving frm the utdrs t inside the huse. 9. hile the state f the papers has gne frm disrder t rder, they did nt d s spntaneusly. An utside surce (yu) caused the increase in rder. Yu had t prvide energy t d this (thrugh yur metablic prcesses), and in ding s, yur entrpy increased mre than the entrpy f the papers decreased. he verall effect is that the entrpy f the universe increased, satisfying the secnd law f thermdynamics. 0. he first statement, Yu can t get smething fr nthing, is a whimsical way f saying that energy is cnserved. Fr instance, ne way t write the st law is U. his says that wrk dne by a system must have a surce either heat is input t the system r the internal energy f the system is lwered. It csts energy either heat energy r internal energy t get wrk dne. Anther way t say this is that n heat engine can be built which puts ut mre energy in the frm f wrk than it extracts in the frm f heat r internal energy. he secnd statement, Yu can t even break even, reflects the fact that a cnsequence f the nd law is that there is n heat engine that is 00% efficient. Even thugh the st law is satisfied by an engine that takes in 00 J f heat and utputs 00 J f wrk, the nd law says that is impssible. If 00 J f heat were taken in, less than 00 J f wrk can be utput frm the heat engine, even if it is an ideal heat engine. Sme energy will be lst as exhaust energy. 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 59

4 Chapter 5 he aws f hermdynamics. In an actin mvie, seeing a building r car g frm an explded state t an un-explded state. In a mvie with vehicle crashes, seeing tw cllided vehicles separate frm each ther, becming unwrecked as they separate. atching smene un-write smething n a piece f paper mving a pen ver paper, taking away written marks as the pen mves.. he synthesis f cmplex mlecules frm simple mlecules des invlve a decrease in entrpy f the cnstituent mlecules, since they have becme mre structured r rdered. wever, the mlecules are nt a clsed system. his prcess des nt ccur spntaneusly r in islatin. he living rganism in which the synthesis prcess ccurs is part f the envirnment that must be cnsidered fr the verall change in entrpy. he living rganism will have an increase in entrpy that is larger than the decrease in entrpy f the mlecules, and s verall, the secnd law is still satisfied, and the entrpy f the entire system will increase. Slutins t Prblems. Use the first law f thermdynamics, Eq. 5-, and the definitin f internal energy, Eq. -. Since the wrk is dne by the gas, it is psitive. (a) Since the temperature des nt change, U 0 (b) U U J.0 0 J. (a) he wrk dne by a gas at cnstant pressure is fund frm Eq Pa 5 5 atm 8. m.0 m J 6. 0 J P V atm (b) he change in internal energy is calculated frm the first law f thermdynamics 86 J 5 6 U 00 kcal J 5. 0 J kcal. Fr the drawing f the graph, the pressure is given relative t the starting pressure, which is taken t be P. 0 A Segment A is the cling at cnstant pressure. P P B Segment B is the isthermal expansin. V Segment A is the cmpressin at cnstant pressure..5 P atm Segment B is the isthermal expansin. A Segment C is the pressure increase at cnstant vlume. 0.5 B C 0 V Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 60

5 Giancli Physics: Principles with Applicatins, 6 th Editin 5. Segment A is the isthermal expansin. Since the temperature and the amunt f gas are cnstant, the quantity PV nr is cnstant. Since the pressure is reduced by a factr f.5, the vlume will increase by a factr f.5, t a final vlume f.5. Segment B is the cmpressin at cnstant pressure. Segment C is the pressure increase at cnstant vlume P atm A.0 C.0.0 B 0.0 V (a) Since the cntainer has rigid walls, there is n change in vlume. P V 0 J (b) Use the first law f thermdynamics t find the change in internal energy. U 65 kj 0 65 kj 7. (a) Since the prcess is adiabatic, 0 J (b) Use the first law f thermdynamics t find the change in internal energy. U 0 850J 850 J (c) Since the internal energy is prprtinal t the temperature, a rise in internal energy means a rise in temperature. 8. (a) rk is nly dne in the expansin at cnstant pressure, since there must be a vlume change t have wrk dne Pa 0 m P V.0 atm J atm (b) Use the first law f thermdynamics t find the heat flw. Ntice that the temperature change ver the entire prcess is 0, s there is n change in internal energy. U 0 79 J 9. Since the expansin is adiabatic, there is n heat flw int r ut f the gas. Use the first law f thermdynamics t calculate the temperature change. U nr 0 nr 7500 J.5 ml 8.5J ml K 0K.0 0 K 0. (a) N wrk is dne during the first step, since the vlume is cnstant. he wrk in the secnd step is given by P V. P V.0 0 Pa 0 m 5. atm J atm (b) Since there is n verall change in temperature, U 0 J (c) he heat flw can be fund frm the first law f thermdynamics. U U J=.5 0 J int the gas 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 6

6 Chapter 5 he aws f hermdynamics. (a,c) See diagram. (b) he wrk dne is fund frm Eq. 5-. P V 55N m 8.00m.00m P N m A.7 0 J he change in internal energy depends n the temperature change, which can be related t the ideal gas law, PV nr. U nr nr nr PV PV P V.7 0 J.0 0 J (d) he change in internal energy nly depends n the initial and final temperatures. Since thse temperatures are the same fr prcess (B) as they are fr prcess (A), the internal energy change is the same fr prcess (B) as fr prcess (A),.0 0 J B 00 V m Fr the path ac, use the first law f thermdynamics t find the change in internal energy. U 6 J 5 J 8 J ac ac ac Since internal energy nly depends n the initial and final temperatures, this U applies t any path that starts at a and ends at c. And fr any path that starts at c and ends at a, U U 8 J ca ac (a) Use the first law f thermdynamics t find. abc U U 8J 8 J 76 J abc abc abc abc abc abc (b) Since the wrk alng path bc is 0, P V P V V. Als nte that the wrk abc ab b ab b b a alng path da is 0. P V P V V P V V cda cd c cd c d c b a b abc 8 J J (c) Use the first law f thermdynamics t find. abc U U 8J J 5 J cda cda cda cda cda cda (d) As fund abve, U U U U 8 J c a ca ac (e) Since U U 5 J U U 5 J U U U U U 5 J U 5 J J. d c d c da a d a c ca Use the first law f thermdynamics t find 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 6. da U U J 0 J da da da da da da. e are given that 80 J and 55 J. ac ac (a) Use the first law f thermdynamics t find U U U a c ca U U 80J 55J 5 J ca ac ac ac (b) Use the first law f thermdynamics t find. cda U U U 5 J 8 J 6 J cda cda cda cda cda cda ca cda (c) Since the wrk alng path bc is 0, P V P V V. abc ab a ab a b a P V P V V.5P V V J 95 J abc ab a ab a b a d c d cda (d) Use the first law f thermdynamics t find abc U U 5 J 95J 0 J abc abc abc abc abc abc

7 Giancli Physics: Principles with Applicatins, 6 th Editin (e) Since U U 0 J U U 0 J, we have the fllwing. a b b a U U U U U 0 J U 0 J 5 J 0 J 5 J. bc c b c a ac Use the first law f thermdynamics t find. bc U U 5 J 0 5 J bc bc bc bc bc bc 7. In Example 5-8, the ttal energy transfrmed was.5 0 J. e will subtract away the energy fr hur f desk wrk and add in the energy fr hur f running. 7 7 Energy.5 0 J 5 J s 50 J s 600 s h.5 0 J 600 Cal 5. Fllw the pattern set in Example 5-8. Find the average rate by dividing the ttal energy fr the day by hurs. 8.0 h 70 J s 8.0 h 5J s.0 h 0 J s Avg. Energy h h 5J s.5 h 60J s 0.5 h 50J s 6. Frm able 5-, the change in metablic rate if ne hur f sleeping is exchanged fr light activity is an additin f 0 watts 70 watts = 60 watts. Nte that this increased rate is nly applicable fr ne hur per day. J 600 s h 65 day kg fat.0 lb kg 5. kg lbs 7 s h day y 0 J kg 7. he efficiency f a heat engine is given by Eq J e 0.8 8% 00 J 800 J 8. he efficiency f a heat engine is given by Eq J e 0.0 0%.0 kcal 86 J kcal 9. he maximum (r Carnt) efficiency is given by Eq. 5-5, with temperatures in Kelvins K e 0. % K 0. he Carnt efficiency is given by Eq. 5-5, with temperatures in Kelvins. 0 7 K e 699K 6 C 0 C e 0.8. he maximum (r Carnt) efficiency is given by Eq. 5-5, with temperatures in Kelvins K e K hus the ttal pwer generated can be fund as fllws. Actual Pwer tal Pwer max. eff. perating eff. 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 6

8 Chapter 5 he aws f hermdynamics Actual Pwer.G tal Pwer 5.66 G max. eff. perating eff Exhaust Pwer tal Pwer Actual Pwer 5.66 G. G.6 G J s 600s h.6 0 J h. Calculate the Carnt efficiency fr the given temperatures. 77 K e % ideal 9 K. his is a perfect Carnt engine, and s its efficiency is given by Eqs. 5- and 5-5. Equate these tw expressins fr the efficiency. e t K t 7 K 0 C 5. 0 J s 680kcal s 86J kcal. his is a perfect Carnt engine, and s its efficiency is given by Eqs. 5- and 5-5. Use these tw expressins t slve fr the rate f heat utput. 5 7 K e 0.6 e e 0 7 K t t e he efficiency f a heat engine is given by Eq. 5-. e e t t e 550 M M 6. Find the exhaust temperature frm the riginal Carnt efficiency, and then recalculate the intake temperature fr the new Carnt efficiency, using the same exhaust temperature. e e K K e 9 K 69 C 60 C 59.6 K e Find the intake temperature frm the riginal Carnt efficiency, and then recalculate the exhaust temperature fr the new Carnt efficiency, using the same intake temperature K e 0 K e Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 6

9 Giancli Physics: Principles with Applicatins, 6 th Editin e e 0 K K 8 C 50 C 8. Fr each engine, the efficiency is given by e 0.60e. hus Carnt e 0.6e C e 0.6e C 0 7 K K 90 7 K 0 7 K Fr the first engine, the input heat is frm the cal. e e and e. cal cal Fr the secnd energy, the input heat is the utput heat frm the first engine. e e e e cal Add the tw wrk expressins tgether, and slve fr e e e e e e e cal cal cal t cal cal t e e e e e e e e. cal Calculate the rate f cal use frm the required rate f input energy, cal t kg 9. 0 J s 9. 0 J s 58.6 kg s.6 0 kg s J t. cal 9. he cefficient f perfrmance fr a refrigeratr is given by Eq. 5-6c, with temperatures in Kelvins. 5 7 K COP K 5 7 K 0. he cefficient f perfrmance fr a refrigeratr is given by Eq. 5-6c, with temperatures in Kelvins. COP 7.0 COP 7 K 59.9K C COP 8.0. he cefficient f perfrmance fr a refrigeratr is given by Eq. 5-6c, with temperatures in Kelvins. Use that expressin t find the temperature inside the refrigeratr. COP 5.0 COP 9 7 K 5 K C COP 6.0. he efficiency f a perfect Carnt engine is given by Eqs. 5- and 5-5. Equate these tw expressins t slve fr the wrk required. e ; e 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 65

10 Chapter 5 he aws f hermdynamics (a) (b) 800 J 0 J J 50 J he heat t be remved,, is the latent heat f fusin fr the ice, s m V. fusin O O fusin he wrk dne in ne hur is.0 kilwatt-hur. he COP fr a refrigeratr is COP. Equate the tw expressins fr m COP V COP O fusin O O fusin watt hur kg m. 0 J kg and slve fr the vlume. 600 s hur.0 0 m 76 COP, and s. he COP fr a heat pump is COP and the efficiency is e. hus they are reciprcals f each ther. S if the efficiency is 0.5, the COP is eat energy is taken away frm the water, s the change in entrpy will be negative. he heat transfer is the mass f the steam times the latent heat f vaprizatin. S m kg.6 0 J kg vap 7 00 K.5 0 J K 6. he heat added t the water is fund frm mc. Use the average temperature f 50 C in the apprximate entrpy calculatin. S mc.00 kg 86J kg C 00C 7 50 K.0 0 J K 7. eat energy is taken away frm the water, s the change in entrpy will be negative. he heat taken away frm the water is fund frm m. Nte that.00 m f water has a mass f.00 0 kg. S m fusin fusin kg. 0 J kg 7K 6. 0 J K 8. here are three terms f entrpy t cnsider. First, there is a lss f entrpy frm the water fr the freezing prcess, S. Secnd, there is a lss f entrpy frm that newly-frmed ice as it cls t 0 C, S. hat prcess has an average temperature f 5 C. Finally, there is a gain f entrpy by the great deal f ice, S, as the heat lst frm the riginal mass f water in steps and ges 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 66

11 Giancli Physics: Principles with Applicatins, 6 th Editin int that great deal f ice. Since it is a large quantity f ice, we assume that its temperature des nt change during the prcesses. S S S m mc fusin ice kg. 0 J kg fusin ice 7K m mc.00 0 kg 00J kg C 0C 5 7 K kg. 0 J kg 00J k S S S S.78 0 J K 5 0 J K 0 7 K g C J K 0C J K J K J K J K.60 0 J K 9. Energy has been made unavailable in the frictinal stpping f the sliding bx. e take that lst kinetic energy as the heat term f the entrpy calculatin. S mv i 0.0 kg.0m s 9 K 0.5J K Since this is a decrease in availability, the entrpy f the universe has increased. 0. ake the energy transfer t use as the initial kinetic energy f the rck, because this energy becmes unusable after the cllisin it is transferred t the envirnment. e assume that the rck and the envirnment are bth at temperature. 0 S S KE 0. he same amunt f heat that leaves the high temperature heat surce enters the lw temperature bdy f water. S S S lw high high lw lw high S.86 J 7.50cal s t t cal 7 7 K 0 7 K.5 0 J K s. he same amunt f heat that leaves the high temperature water will enter the lw temperature water. Since the tw masses f water are the same, the equilibrium temperature will be the midpint between the tw initial temperatures, 5 C. he cl water average temperature is 0 C 5 C 7.5 C, and the warm water average temperature is S S S mc high lw lw high 60 C 5 C 5.5 C. 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 67

12 Chapter 5 he aws f hermdynamics.0 kg 86 J kg C 5C 9.J K K K. he equilibrium temperature is fund using calrimetry, frm chapter. he heat lst by the aluminum is equal t the heat gained by the water. m c m c Al Al ial f O O f io f m c m c Al Al ial O O io m c m c Al Al O O.8 kg 900J kg C 0.0 C.0 kg 86J kg C 0 C.8 kg 900J kg C.0 kg 86J kg C he amunt f heat lst by the aluminum, and gained by the water, is m c.5 C.0 kg 86J kg C.5 C 0 C.9 0 J O O f io In calculating the entrpy change, we will need t use estimates fr the temperatures f the water and the aluminum since their temperatures are nt cnstant. e will use their average temperatures. O avg 0 C.5 C.5 C 0 C.5 C 7.5 C S S S Al Al avg O Al avg O avg.9 0 J.J K.5 7 K K. (a) e 550 J 00 J 0.50 e 650 K 970 K 0.0 actual ideal hus e e % f ideal actual ideal (b) he heat reservirs d nt change temperature during the peratin f the engine. here is an entrpy lss frm the input reservir, because it lses heat, and an entrpy gain fr the utput reservir, because it gains heat. Nte that 00 J 550 J 650 J. S S S input utput 00 J 650 J 970 K 650 K 0.7 J K (c) Fr the Carnt engine, the exhaust energy will be e. Carnt S S S 0 input utput A numeric calculatin might give a very small number due t nt keeping all digits in the calculatin. 5. hen thrwing tw dice, there are 6 pssible micrstates. (a) he pssible micrstates that give a ttal f 5 are: ()(), ()(), ()(), and ()(). hus the prbability f getting a 5 is 6 9. (b) he pssible micrstates that give a ttal f are: (5)(6) and (6)(5). hus the prbability f getting an is Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 68

13 Giancli Physics: Principles with Applicatins, 6 th Editin 6. A macrstate is a set f 5 cards frm the deck, as given in the prblem. Fr example, fur aces and a king is a macrstate. w jacks, tw queens, and an ace is a macrstate. A micrstate is a specific set f cards that meets the criterin f a certain macrstate. Fr example, the set (ace f spades, ace f clubs, ace f hearts, ace f diamnds, king f spades) is a micrstate f the macrstate f aces and a king. he prblem then is asking fr the relative number f micrstates fr the given macrstates. (a) here are nly micrstates fr this macrstate, crrespnding t the particular suit t which the king belngs. (b) Since every card is specified, there is nly micrstate fr this macrstate. (c) here are 6 pssible jack pairs, (spade/club, spade/heart, spade/diamnd, club/heart, club/diamnd, and heart/diamnd), 6 pssible queen pairs, and pssible aces, s there are 6 x 6 x = card cmbinatins r micrstates fr this macrstate. (d) here are 5 pssibilities fr the first card, 8 pssibilities fr the secnd card, and s n. It is apparent that there are many mre micrstates fr this macrstate than fr any f the ther listed macrstates. hus in rder f increasing prbability, we have (b), (a), (c), (d). 7. Frm the table belw, we see that there are a ttal f Macrstate 6 6 micrstates. Pssible Micrstates ( = heads, = tails) Number f micrstates 6 heads, 0 tails 5 heads, tails 6 heads, tails 5 heads, tails 0 heads, tails 5 heads, 5 tails 6 0 heads, 6 tails (a) he prbability f btaining three heads and three tails is 0 6. (b) he prbability f btaining six heads is he required area is 0 h day m 6 m day 9 h Sun 0. A small huse with 000 ft f flr space, and a rf tilted at 0 m, wuld have a rf area f 000ft 0 m, cs 0.8 ft which is abut twice the area needed, and s the cells wuld fit n the huse. But nt all parts f the rf wuld have 9 hurs f sunlight, s mre than the minimum number f cells wuld be needed. 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 69

14 Chapter 5 he aws f hermdynamics 9. (a) Assume that there are n dissipative frces present, and s the energy required t pump the water t the lake is just the gravitatinal ptential energy f the water. U grav mgh kg s 0.0 h 9.80m s 5 m. 0 h (b) 6. 0 kh 6. 0 k h 0.75 h 7. 0 k 50. e assume that the electrical energy cmes frm the 00% effective cnversin f the gravitatinal ptential energy f the water. mgh m V P gh gh t t t kg m 5m s 9.8m s 5 m Accrding t the heat figures prvided by the inventr, the engine is 50% efficient: t.50 M e t.00 M he ideal engine efficiency at the perating temperatures is given by Eq K e 0.9 ideal 5 K hus his engine is nt pssible, even if it were ideal. S yes, there is smething fishy abut his claim. is engine is better than ideal. 5. (a) he wrk dne at cnstant pressure is P V. P V.00 atm.0 0 Pa atm. m.9 m 5 P atm J. 0 J (b) Use the first law f thermdynamics. U (c) See the adjacent graph J. 0 J. 0 J V (a) (b) (c) 0 J 5 cycles rk s cylinders.96 0 J s.0 0 J s cycle cylinder t e t e e J s J s.96 0 J s s.7 min J s.6 0 J s 5. (a) he heat that must be remved frm the water is fund in three parts cling the liquid water t the freezing pint, freezing the liquid water, and then cling the ice t the final temperatures. 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 70

15 Giancli Physics: Principles with Applicatins, 6 th Editin m c c liquid liquid fusin ice ice 5 86J kg C 5C. 0 J kg kg.67 0 J 00J kg C 7C he Carnt efficiency can be used t find the wrk dne by the refrigeratr. e 5 7 K J.88 0 J.9 0 J 7 7 K (b) Use the cmpressr wattage t calculate the time. P t t P.88 0 J 0 8.9s. min 55. (a) Calculate the Carnt efficiency fr an engine perated between the given temperatures. 7 K e % ideal 7+7 K (b) Such an engine might be feasible in spite f the lw efficiency because f the large vlume f fuel (cean water) available. Ocean water wuld appear t be an inexhaustible surce f heat energy. (c) he pumping f water between radically different depths wuld prbably mve smaller seadwelling creatures frm their natural lcatin, perhaps killing them in the transprt prcess. Mixing the water at different temperatures will als disturb the envirnment f sea-dwelling creatures. here is a significant dynamic f energy exchange between the cean and the atmsphere, and s any changing f surface temperature water might affect at least the lcal climate, and perhaps als cause larger-scale climate changes. 56. ake the energy transfer t use as the initial kinetic energy f the cars, because this energy becmes unusable after the cllisin it is transferred t the envirnment. S m s. 0 kg 95km h mv i.6km h 0 7 K.6 0 J K 57. (a) he equilibrium temperature is fund using calrimetry, frm chapter. he heat lst by the water is equal t the heat gained by the aluminum. m c m c O O io f Al Al f ial f m c m c Al Al ial O O io m c m c Al Al O O 0. kg 900J kg C 5 C 0. kg 86J kg C 50 C 0. kg 900J kg C 0. kg 86J kg C (b) he amunt f heat lst by the aluminum, and gained by the water, is m c.55 C 5 C 0. kg 86J kg C 50 C.55 C.9 0 J O O io f 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 7

16 Chapter 5 he aws f hermdynamics In calculating the entrpy change, we need t use estimates fr the temperatures f the water and the aluminum since their temperatures are nt cnstant. e will use their average temperatures. O avg 50 C.55 C 7.6 C 5 C.55 C 9.78 C S S S Al Al avg O O avg J K 0.58J K Al avg.9 0 J 58. he COP fr an ideal heat pump is given by Eq K (a) COP K (b) K K 7 7 COP t t COP s J 7. 0 J 59. he efficiency is given by e needed. t e t t 5 hp 76 hp.9 0 J s t.0 0 kcal gal 90 km 86 J h.9 0 J s J s t, and s the input pwer and the useful pwer are t gal km h kcal 600 s 0. % J s 60. Find the riginal intake temperature frm the riginal Carnt efficiency, and then recalculate the intake temperature fr the new Carnt efficiency,, using the same exhaust temperature. e e e e 7 K 0 K 70K e e Nte that there is NO wrk dne as the gas ges frm state A t state B r state D t state C, because there is n vlume change. In general, the wrk dne can be fund frm the area under the PV curve representing the prcess under cnsideratin. (a) P V V ADC A C A (b) P V V ABC C C A (c) P P V V AC C A C A 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 7

17 Giancli Physics: Principles with Applicatins, 6 th Editin 6. (a) he exhaust heating rate is fund frm the delivered pwer and the efficiency. Use the utput energy with the relatinship mc Vc t calculate the vlume f air that is heated. e e t t e mc Vc t mc t V t t t c he change in air temperature is V t t t c 7.0C. he heated air is at a cnstant pressure f atm s day.kg m.0 0 J kg C 7.0C km m day 8 km day m (b) If the air is 00 m thick, find the area by dividing the vlume by the thickness. Vlume 8km A 90 km thickness 0. km his wuld be a square f apprximately 6 miles t a side. hus the lcal climate fr a few miles arund the pwer plant might be heated significantly. 6. (a) he exhaust heating rate can be fund frm the delivered pwer P and the Carnt efficiency. hen use the relatinship between energy and temperature change, mc, t calculate the temperature change f the cling water. e t t P m V mc t c c t t Equate the tw expressins fr t, and slve fr. V P t V c t P c K.0 0 kg m 7m s 86J kg C 65 K 85 K (b) he additin f heat per kilgram fr the dwnstream water is 5.0K 5.C t c. e use the average temperature f the river water fr the calculatin:. Nw the entrpy 0 increase can be calculated. S c 0 86J kg C 5.0K K 77 J kg K 6. (a) Calculate the Carnt efficiency by e and cmpare it t the 5% actual efficiency. e 85 7 K 95 7 K % Carnt 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 7

18 Chapter 5 he aws f hermdynamics hus the engine s relative efficiency is e e % actual Carnt (b) ake the stated 00 hp as the useful pwer btained frm the engine. Use the efficiency t calculate the exhaust heat. P e t hp hp 600 s Pt e e h J s h 9 9 kcal J.5 0 J.6 0 kcal 86 J 65. he net frce n the pistn must be 0, and s the weight f the pistn must be equal t the net frce exerted by the gas pressures n bth sides f the pistn. See the free-bdy diagram. F F F mg 0 P A P A mg 0 P inside utside inside utside air air mg 5 P.0 atm.0 0 inside utside atm m A Pa atm Pa 0.0 kg 9.8m s e see that the weight f the pistn is negligible cmpared t the pressure frces. hen the gas is heated, we assume that the inside pressure des nt change. Since the weight f the pistn des nt change, and the utside air pressure des nt change, the inside air pressure cannt change. hus the expansin is at a cnstant pressure, and s the wrk dne can be calculated. Use this with the first law f thermdynamics t find the heat required fr the prcess. U nr PV U P V U P V P V P V PA y 00 J Pa m.0 0 m 66. (a) Multiply the pwer times the time times the mass per Jule relatinship fr the fat. 7 95J s 600s h h d.0 kg fat.7 0 J 0.8kg d 0.kg d Futside air Finside air mg (b).0 kg d 0.8 kg.5 d 67. he radiant energy is the heat t be remved at the lw temperature. It can be related t the wrk necessary thrugh the efficiency. e t t t 500 t Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 7

19 Giancli Physics: Principles with Applicatins, 6 th Editin t t t savings K K 68. find the mass f water remved, find the energy that is remved frm the lw temperature reservir frm the wrk input and the Carnt efficiency. hen use the latent heat f vaprizatin t determine the mass f water frm the energy required fr the cndensatin. Nte that the heat f vaprizatin used is that given in chapter fr evapratin at 0 C. e m vapr m vapr s 7 8 K.6 kg 7 K J kg 005 Pearsn Educatin, Inc., Upper Saddle River, NJ. All rights reserved. his material is prtected under all cpyright laws as they currently exist. N prtin f this material may be reprduced, in any frm r by any means, withut permissin in writing frm the 75