2ndLawHints

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Chapter 20– Second Law of Thermodynamics Hints

1. The 2nd Law of Thermodynamics can be stated in many different ways. The following statements are allversions of the 2nd Law. Show how these are indeed equivalent statements. That is, if one of these is

violated, the others would be violated as well. They have to be either all true or all false!

a) eat tends to flow"# from hot to cold and not the reverse. b) $t is im%ossible to build an engine that is 1&&' efficient.

c) $t is im%ossible to build a %erfect" refrigerator with an infinite coefficient of %erformance (.

The above were discussed in class. It’s pretty obvious that (c), a “perfect refrigerator” that maes heat flow

from cold to hot without effort, directly violates (a). If a “perfect engine” (b) could e!ist it would be used torun a real refrigerator and turn it into a “perfect refrigerator”, so logically, such an engine is alsoimpossible.d) The state of disorder entro%y) of a closed system always tends to increase. "ou will have to wait till theend of the chapter to answer this one.)

#$e use the phrase “heat flows” here for historical reasons going bac to a time when heat was considereda ind of fluid. In reality heat is not a substance so it cannot “flow”, but is rather a process that transfersenergy between ob%ects in thermal contact. $e will continue to say “heat flows” because it is traditionaland is well understood that the phrase is not meant to be literally true.

&ngines, refrigerators, and the 'arnot cycle2. * device called a thermoelectric converter was demonstrated in class. $t has two metal legs and a %ro%eller and it uses a series of semiconductor cells to convert thermal energy to electrical energy. The

device will not wor+ unless one leg is at a higher tem%erature than the other.

a) hy doesn-t this wor+ if the legs are at the same tem%erature even if that tem%erature is very higha) ll engines re*uire a hot heat source and a cold heat source to wor+ the absolute temperatures are notrelevant, it is the temperature difference that counts. b) $n what sense does this intriguing e/%eriment demonstrate the second law of thermodynamicsb) eat engines re*uire a flow of energy to wor+ heat flows naturally from hot to cold temperatures maing it possible to intercept this flow and mae a heat engine.

0. Su%%ose you try to cool the +itchen of your house by leaving the refrigerator door o%en. hat ha%%enshy ould the result be the same if you left o%en a %icnic cooler full of ice /%lain the reason for any

differences. The room will get warmer because the refrigerator e!pels more heat into the outside than istaes out of the inside of the refrigerator. -pening a picnic cooler full of ice is *uite different, since there isno motor doing wor to mae the ice melt. In this situation the melting ice would be removing heat from the surrounding and really cooling the room.

. 3eview the definition of efficiency for an engine and coefficient of %erformance for a refrigerator and a

heat %um%.

e $/0hot 12(0c /0h )+ 3 rf 0cold /$ 0c /(0h 2 0c )+ 3 hp 0hot /$ 0h /(0h 2 0c )

a) hat are the ma/imum and minimum values of these quantities for each devicea) ema!1 and 3 ma!4 (both of which are not possible). The minimum values would be 5ero. b) 4om%are the e and 3 of a reversible engine5refrigerator one that wor+s equally well as one or the other

device between the same hot and cold reservoirs), and show they are related by 3 6 (15e)/e.b) reversible engine2refrigerator would have the same $, 0hot and 0cold re*uirements so you can substituteout these values combining the e*uationse3(0c /0h )(12e)6c) * reversible engine ta+es in 0&& 7 from a hot source and does 1&& 7 of wor+. 8etermine the efficiency of

this engine and the coefficient of %erformance of its refrigerator version.

c) e1/7+ 38

9. 4arnot-s theory shows that in a 4arnot reversible engine5refrigerator :h;:c6 Th;Tc. 8erive the efficiency

(e) formula and the coefficient of %erformance () formula for a 4arnot engine and refrigerator. <or



ma/imum engine efficiency do you want the tem%eratures to be as far a%art as %ossible or as close as

%ossible *nswer the same question for the ( of a 4arnot refrigerator.

9oing the algebra gives e'arnot 12(T c /T h ). :or refrigerator 3 'arnot ;(T h /T c )21<21 , for heat pump 3 'arnot ;12(T c /T h )<21. It should be clear that e increases as the temperatures get farther apart, while 3 increases as thetemperatures get closer together. If you thin about this a bit you can agree that these mae sense6.

=. $n light of the above question discuss the following>

a) * steam5driven turbine is one ma?or com%onent of an electric %ower %lant. hy is it advantageous to

have the tem%erature of the steam as high as %ossiblea) "ou want to ma!imi5e the temperature difference to improve e6 b) * heat %um% is to be installed in a region where the average outdoor tem%erature in the winter months is

5 2&@4. $n view of this, why would it be advisable to %lace the outdoor com%ressor unit dee% in theground hy are heat %um%s not commonly used for heating in cold climates

b) =nderground temperatures vary less than above2ground temperatures, so in winter, undergroundtemperatures would be closer to inside room temperatures (in summer it’s the reverse). aving the hotand cold sources closer in temperature would improve 36

c) hy is your automobile li+ely to burn more gas in winter than in summer

c) -ne would thin the opposite would be true since in the winter there would be a larger differencebetween the woring temperatures of the car engine, but here we need to consider other factors. 'older

temperatures mae lubricants thicer and so there are larger heat loses due to friction between themoving parts6This illustrates that besides temperature differences the reversibility of an engine is animportant factor in determining efficiency.

A. eat flow diagrams" are useful in mastering the conce%ts of engines and the 2nd law.

a) Bse a heat flow diagram to show that if you could build a %erfect engine e61),then you could build a %erfect refrigerator (6C), and vice5versa. This was donein class. >ee if you can reconstruct the logical argument to go with the diagram.

b) Bse a heat flow diagram to show that a reversible engine5refrigerator is the mostefficient device that can be built between two given tem%eratures.

This was done in class. The logical argument is that a reversible engine connected

to its refrigerator version would be able to return all the heat it e!pelled to the cold reservoir bac to the hot reservoir and mae up for the original heat supplied bythe hot reservoir to run the engine. The net 0hot ? and the net 0cold ?, so the heat flow would have been effectively stopped, which is the best one can do in battling thenatural flow of heat. @ow consider an engine that is more efficient than a reversible engine. This enginewould be able to do the wor needed to run the refrigerator above with less heatinput from the hot reservoir. The net effect would be some heat flowing from cold tohot in violation of the 8nd law of thermodynamics. Therefore a reversible engine isthe most efficient engine allowed by nature.

c) <or a numerical e/am%le consider the reversible engine5refrigerator described in c e61;0D (62).

*ssume that it is %ossible to ma+e a more efficient engine that would %roduce 1&& 7 of wor+ by ta+ing inonly 2&& 7 of heat using the same tem%erature sources. hat would be the result if you used the su%er5

engine" to run the original reversible refrigerator (62) ow does that square with the 2nd Law

The result would be a net flow of heat of 1?? A from cold to hot, an direct violation of the 8nd Baw.

E. $magine a s%ecial air filter %laced in a window of a house. *ssume this filter is made such that only airmolecules moving faster than a certain s%eed can e/it the house, and only air molecules moving slower than

that s%eed can enter the house from outside. /%lain why such an air filter would cool the house, and why

the second law of thermodynamics ma+es building such a filter an im%ossible tas+.>uch a filter would cool the house because slower molecules would lower the average inetic energy of theair and therefore lower temperature. >uch a filter couldn’t be built because it would violate the 8nd lawreversing the natural flow of heat without e!ternal wor.

Thot

e61

(FC

Tcold

Thot

ec

( c

Tcold

&ngine2refrigerator reversible pair

Thot

“Cetter” ec

than ( reversible engine T

cold



G. 3eal heat engines, li+e the gasoline engine in a car, always have some friction between their moving

%arts, although lubricants +ee% the friction to a minimum, would a heat engine with com%letely frictionless %arts be 1&&' efficient hy or why not 8oes the answer de%end on whether or not the engine runs on the

4arnot cycle *gain, why or why not @o, even a frictionless engine could not violate the 8nd law because friction has nothing to do with the natural flow of heat. 'arnot engine is not 1??D efficient+ it’s simplythe most efficient engine for a particular temperature difference. Eeal engines have efficiencies lower thanthe 'arnot engine because of non2conservative force (such as friction) and other irreversible features of theengine’s cycle.

1&. *n engine absorbs 1=&& 7 from a hot reservoir and e/%els 1&&& 7 to a cold reservoir in each cycle. a)

hat is the efficiency of the engine e?.7FG b) ow much wor+ is done in each cycle $H?? A c) hat is the average %ower out%ut of the engine if each cycle lasts for &.0& s 8 wattsd) hat fraction of the heat absorbed is e/%elled to the cold reservoir J0c /0hJ?.H8Ge) $f this were a 4arnot reversible engine, what would be the e/treme tem%eratures ratio ould you e/%ect

the actual e/treme tem%eratures ratio Tc;Th) of this engine to be larger or smaller If we assume this engineis as efficient as a 'arnot engine then J0c /0hJ(T c /T h )'arnot ?.H8G. >ince the efficiency of a real engine isalways less than the 'arnot efficiency, H0c /0hJK(T c /T h )real , and the actual (T c /T h ) ratio in the real engine

would be smaller (this means that LT is larger in the real engine).11. * refrigerator has a coefficient of %erformance equal to 0 and consumes && watts of %ower when it-srunning. <or each second that it runs, find

a) the wor+ done by the motor. $LtM?? A b) the thermal energy removed from the inside of the refrigerator, and the thermal energy e/%elled out intothe room. 0c3$18?? A and 0h1H?? A c) if it were a 4arnot refrigerator, and the outside tem%erature was a ty%ical 29 I4, what would be the

tem%erature of the inside of the refrigerator. $s this a realistic tem%erature If this were a 'arnot refrigerator then (0c /0h )(T c /T h )18/1H;T c /(8GN8F7)< T c 88M 32G?O'. This is very unrealistic since typicaltemperatures inside a refrigerator are more lie 1?O', even free5er temperature are only about 2G O'. This

shows how far from the 'arnot ideal real refrigerators actually are.

12. * 4arnot engine o%erates between two heat reservoirs at tem%eratures T h and T c. *n inventor %ro%osesto increase the efficiency by running one engine between T h and an intermediate tem%erature T and a second

engine between T and T c . The second engine uses the heat e/%elled by the first engine as its own in%ut

energy. 4om%ute the efficiency of this com%osite system, and com%are it to that of the original engine.

=sing the definition of efficiency e1$ 1 /0h and e8$ 8 /0’, where 0’ is the heat e!pelled by the first engineand taen up by the second engine. Cy definition the combined efficiency e18 ($ 1N$ 8 )/0h. >ince $ is e*ual to the difference between heat in and out of each engine, e;(0h 20’)N(0’2 0c )</0h(0h2 0c )/0hThecombined “e” is the same as if a single engine were running between the e!treme temperatures+ there’s no

gain in efficiency from this arrangement.

10. *t %oint * in a 4arnot cycle, 2.0& moles of a monatomic gas have a %ressure of 1&& +Ja, a volume of1&.& liters, and a tem%erature of A2& (. $t e/%ands isothermally to %oint K and then e/%ands adiabatically to

%oint 4, where its volume is 2.& liters. *n isothermal com%ression brings it to %oint 8, where its new

volume is 19.& liters. *n adiabatic %rocess returns the gas to %oint *.

a) 8etermine all the un+nown %ressure, volumes, and tem%eratures by filling in the following table>

a) "ou can determine the unnown values by setting up constant ratios using the isothermal condition(P'onstant) and the various adiabatic conditions (P γ '+ TP γ−1 '+ 1−γ T γ '). The calculations can be

tedious but are straightforward.



P T 1M?? a 1?.? liters F8? 3 C QFG a 1H.? l F8? 3 ' MMG a 8M.? liters GG? 3 9 F11 a 1G.? liters GG? 3

b) 8raw a J vs. gra%h.c) <ind the :, , and MB for each of the ste%s, C, C', '9, and 9.c) =se =7nET/8 to determine internal energies and at all points, use the appropriate e!pressions todetermine the wor done in each process, then find 0’s from the 1 st law

0 $ L= C NH.GQ A 1Mln(1H/1?) A NH.GQ A ? C' ? NG A (1G.F28?.F) A 2G A '9 2G.?7 A 2G.?7 A ? 9 ? 2 G A NG A

d) Show that $net/0in 6 1 R (T '9 /T C ), the 4arnot efficiency.

d) $net/0in 6H.GQ2G.?7); H.GQ 61 R (GG?/F8?)?.87H

1. This is ?ust a variation on the %roblem above. Nne mole of an ideal gas γ 6 1.) is carried through the

4arnot cycle similar to the one described above. *t %oint , the %ressure is 29 atm and the tem%erature is=&& (. *t %oint 4, the %ressure is 1 atm and the tem%erature is && (.

>imilar problem to the one above. "ou must use the gas law as well as the isothermal and adiabaticconditions to find the needed *uantities.a) 8etermine the %ressures and volumes at %oints , C, 4, and 9.a) Sany calculations later P 1.Q l+ P C11. l+ P ' 78. l+ P 9G.MM l+ CM.1M atm+ 9H.?G atm. b) 4alculate the net wor+ done %er cycle.

b) $ net nET Cln(P C /P ) N nET '9ln(P 9 /P ' )(Q.71);H?? ln(11./1.Q) R M?? ln(78./G.MM?<8.H A c) 8etermine the efficiency of an engine o%erating in this cycle. e?.77

19. *n e/%erimental %ower %lant at the Oatural nergy Laboratory of awaii generates electricity from thetem%erature gradient of the ocean. The surface and dee%5water tem%eratures are 2A@4 and =@4, res%ectively.a) hat is the ma/imum theoretical efficiency of this %ower %lant e'arnot ?.?F b) $f the %ower %lant is to %roduce 21& + of %ower, at what rate must heat be e/tracted from the warm

water *t what rate must heat be absorbed by the cold water *ssume the ma/imum theoreticalefficiency. d0h /dt 81?/?.?F 7 S$+ d0c /dt 72?.818.F S$

c) The cold water that enters the %lant leaves it at a tem%erature of 1& I4. hat must be the flow rate of

cold water through the system Pive your answer in +g;h and liters;h.c) d0c /dt 8.F S$ (dm/dt)c LT dm/dt8.F ! 1?H /M1QH(M)?.1HF g/sec H?1 g/hr H?1 l/hr.

d) hat com%ensating factor ma+es this a worthwhile e/%eriment des%ite the low efficiency

d) &ven though the efficiency is low the vast si5e of the oceans could mae this practical. lso there are

no fuel re*uirements, pollution would be low, maybe designs are simpler6there are a lot of other thingsto consider besides efficiency in engine design.

There is an endless number of variations of the “cycle problem”. ere are a few

1=. Nne mole of an ideal diatomic gas is ta+en through the cycle shown.

*ssume that the %rocesses are reversible. 4alculatea) the net wor+ done by the gas. $net $ C N$ C' Q.172M.?MM.? A b) the thermal energy heat) added to the gas.

b) 0in0' N 0 C Q.17 N 1?.11Q.8 A

c) the thermal energy heat) e/%elled by the gas. 0out 0 C' 21M.1 A d) the efficiency of the cycle. e$/0inM.?/1Q.8 ?.88

J atm)

9 *

4 1 K

1& 9& liters)

*

K

8 4



e) hy is this cycle not as efficient as the 4arnot ideal The 'arnot efficiency results from e!tracting theheat at the highest temperature possible and e!pelling the heat at the lowest temperature possible (not %ustbecause it’s reversible). In this cycle heat is added from ' to at less than the highest temperature and heat is e!pelled from C to ' at more that the lowest temp6If all the heat were e!tracted at T C (H?Q 3) and allthe heat e!pelled at T ' (188 3), then the efficiency would be ec12 (1/G)?.Q, but that would re*uire a'arnot cycle.

1A. * gas is ta+en through the cyclic %rocess *K4* shown.a) $f : is negative for the %rocess C' and MB is negative for the %rocess ',

determine the signs of 0, $, and M= associated with each %rocess.

a) ositive 0 C ,$ C , and L= C + all others are negative, $ C' is 5ero. b) <ind the net heat transferred to the system during one com%lete cycle.

b) 0net $ net rea within cycle 1Q A c) $f the cycle is reversed, that is, the %rocess follows the %ath 'C,

what is the net heat transferred %er cycle 51Q A

1E. Nne mole of a monatomic ideal gas is ta+en through the

reversible cycle shown. *t %oint * the %ressure,volume, and tem%erature are Jo, o, and To, res%ectively.

$n terms of 3 and To, find :irst determine T C7T o , T ' HT o , T 98T o.a) the total heat entering the system %er cycle,

a) 01N 087E(8T o )/8 NGE(7T o )/8 (81/8)ET o b) the total heat leaving the system %er cycle,b) 07 N0M7E(2MT o )/8 NGE(2T o )/8 (21F/8)ET oc) the efficiency of an engine o%erating in this reversible cycle, e12(1F/81)?.1

d) 4onsider this same cycle in reverse so it re%resents a refrigerator. hat is its coefficient of %erformance$s this cycle reversible int> 3eview %roblem b. ere 3M.8G. In problem Mb you were ased to provethat 3(12e)/e for a reversible engine. This cycle satisfies this criterion so it is reversible, even though it’s

not a 'arnot cycle. @ote This problem shows that any cycle is reversible if it’s well2defined and one nows the state of the gasat every instant of the cycle. In fact, that is always the assumption when we draw a solid line in a P graph,and all the formulas we have derived for 0, $, and L= tae reversibility for granted. $hat is differentabout the 'arnot cycle is that there are only two e!tremes temperatures involved and the 0h is e!tracted atthe highest temp and all the 0c e!pelled at the lowest temp."ou should reali5e that the efficiencies derived by analy5ing a cycle are always ideal. The efficiency or realengines are determined e!perimentally, by actually testing the engine, not by analy5ing a cycle.

1G. Qour te/tboo+ has a nice illustration of a basic refrigerator. The mechanism consists of two sets of coils

a condenser outside the refrigerator and an evaporator inside) containing a refrigerant fluid that is %artially

liquefied and va%oriRed as it flows through the coils. * compressor com%resses the fluid adiabatically whenit is mostly gaseous, raising its tem%erature and %ressure and delivers it to the condenser at a tem%erature

higher than the surrounding room tem%erature. *s the fluid %asses through the condenser it gives u% heat to

the outside and the fluid liquefies. Then an e!pansion valve allows the fluid toe/%and adiabatically, lowering its %ressure and cooling it below the tem%erature

inside the refrigerator. *s the cooled fluid flows through the eva%orator it

absorbs heat from the inside of the refrigerator and va%oriRes ?ust before itenters the com%ressor where the cycle starts all over again.

a) $n the J diagram shown, what sections corres%ond to each of the

%rocesses described above C compression+ C' condensation+ '9 e!pansion+ 9 vapori5ation.

:2

0Jo K 4

2Jo

:1

:0

Jo

* :

8

o

2o

J +Ja)

K=

2 * 4

= 19 m0

J

4 K

8 *



b) /%lain why the flat sections in the gra%h are isothermal as well as isobaric. hy doesn-t this violate the

gas law J6n3T The fluid is undergoing a phase change during those processes and is not behaving liean ideal gas6c) ow would you determine the ( of this refrigerator "ou would have to determine 0hot and 0cold . >incethe heat is transferred during the phase changes you would have to now the latent heat of vapori5ation ofthis fluid “B” and the mass of the fluid undergoing the phase changes, then 0mB.d) hat would the area under the gra%h mean in this diagram >ame as any 2P graph, the wor done bythe fluid.e) $s the 1st Law :56MB) valid in this %roblem "es, the 1 st Baw is a version of the wor2energy principle

that is a fundamental principle in all phenomena. owever L= would not e*ual n' v LT because the fluid isnot an ideal gas, and 0 would depend on the latent heat.

The following problems feature some well2nown engines

2&. 3eview the gasoline engine-s idealiRed Ntto cycle discussed in the te/t and in class. 8raw a gra%h of the

cycle and ma+e sure you understand the derivation of the efficiency formula and the meaning of thecom%ression ratio". ere-s a ty%ical gasoline engine %roblem.

a) * gasoline engine has a com%ression ratio of = and uses a gas for which γ 6 1.. hat is the

efficiency of the engine The compression ratio is the larger initial volume P 1 divided by the smaller,compressed volume P 8.

a) e-tto12 (P 1 / P 8 )12γ

12 (H)2?.M

?.G1 b) $f the actual efficiency is 19', what fraction of the fuel is wasted as a result of friction and

unavoidable heat losses *ssume com%lete combustion of the air5fuel mi/ture.) 7HDc) ach cycle the engine ta+es in 1.=1 1& 7 of heat obtained by burning gasoline with a heat of

combustion of .=& 1& 7;g. ow much heat is discarded in each cycle

c) t 1G D efficiency, $?.1G(1H.1 A)8.M8 A and 0c 0h 2$1H.128.M817.F A d) hat mass of fuel is burned in each cycle 1.H1/M.H??.7G g e) $f the engine goes through =&.& cycles %er second, what is its %ower out%ut in +ilowatts and in

horse%ower $/Lt8.M8 !H?1MG $1M hp.

21. *n idealiRed 8iesel engine o%erates in a cycle +nown as theair2standard 9iesel cycle. <uel is s%rayed into the cylinder at the %oint of ma/imum com%ression, C. 4ombustion occurs during the

e/%ansion C 4, which is a%%ro/imated as an isobaric %rocess.

The rest of the cycle is the same as in the gasoline engine. Showthat the efficiency of an engine o%erating in this idealiRed

8iesel cycle is> e =1− 1

γ

T D−T

A

T C −T

B

ere 0h 0 C' n' (T ' RT C ) and 0c 0 9 n' P (T RT 9 ). Aust plug into e 12 J0c /0h J and γ ' /' P 6

@ote It is possible to derive an e!pression for the efficiency of the 9iesel engine in terms of thecompression ratios (P /P C ) and (P /P ' ) but the proof is algebraically messy. It involves eliminating thetemperatures from the formula by relating them to the volumes using adiabatic and isobaric relationships. It is not something I would as you to do on a test, but if you’re interested chec out 'hallenge roblem 78.

22. The gra%h illustrates the Stirling engine cycle that o%erates between two isotherms T 1 and T 8 T 8 KT 1 ).*ssuming that the o%erating gas is an ideal monatomic gas, calculate the efficiency of this engine whoseconstant5volume %rocesses occur at the volumes P o and P o /r, where “r” is the com%ression ratio.

a) ow does this cycle differ from the Ntto cyclea) The adiabatics are replaced with isothermal processes.

J K 4 *diabatic

8 *diabatic

*



b) *ssuming an ideal monatomic gas, determine 0, $, and L=for each of the %rocesses> ab, bc, cd, and da.

0 $ L= ab nET 1ln(1/r) nET 1ln(1/r) ?bc n' P (T 82T 1 ) ? n' P (T 82T 1 )cd nET 8ln(r) nET 8ln(r) ?da n' P (T 12T 8 ) ? n' P (T 12T 8 )

c) 4om%are the heat transfers during the constant volume %rocesses to show that they cancel out and do notrequire an e/ternal source of energy. This is called regeneration.)

c) It’s obvious from the chart that the isochoric processes are e*ual in value and opposite in sign. Thismeans that these energies can be e!changed within the gas itself and re*uire no e!ternal heat source or heat sin. This allows us to leave them out of the efficiency calculation altogether.d) 8etermine the efficiency of this engine and com%are to a 4arnot engine o%erating between the

same two e/treme tem%eratures.

d) Beaving out the heat e!changes in the isochoric processes, e 12;nET 1Jln(1/r)J/nET 8 )Jln(r)J<12(T 1 /T 8 ), %ust lie the 'arnot cycle, but this assumes the isochoric energy e!changes are truly internal and do notaffect the outside environment.e) $f Sterling engines are so efficient, why are they not used more often, to %ower cars, for e/am%le

Isotheral processes are necessarily slow and that maes the >terling engine less powerful. $e want cars to go fast so we sacrifice efficiency to power. The same is true about the 'arnot cycle which is why we don’t see any real engines using it. @ote of interest The >tirling cycle is used in e!ternal combustion engines where heat is supplied by burning fuel outside the cylinder (instead of e!plosively inside the cylinder as in the -tto cycle). :or this reason>tirling2cycle engines are *uieter than -tto2cycle engines, since there are no intae and e!haust valves (ama%or source of engine noise). $hile small >tirling engines are used for a variety of purposes, >tirlingengines for automobiles have not been successful because they are larger, heavier, more e!pensive, and less powerful than conventional automobile engines given their si5e.

:inally6&ntropyU

20. * cou%le of conce%tual questions about entro%y.

a) $f you run a movie film bac+wards, it is as if the direction of time were reversed. hat scenes would

%rovide clues that the movie was running in reverse and what scenes would not hat laws) are clearlyviolated when the movie is run in reverse -nly the 8nd lawU... b) Some critics of biological evolution claim that it violates the 2nd Law of Thermodynamics, since

evolution involves sim%le life forms develo%ing into more com%le/ and more highly ordered organisms./%lain why this is not a valid argument against evolution. The 8nd law applies to the entire =niverse andother “closed systems”. single living organism or species is not a closed system+ it is constantlye!changing energy with its environment and the entropy decreases due to the evolution of comple! life formsis more than off2set by the increases in entropy of the rest of the environment.

2. $n theory any %rocess can be %erformed reversibly or very close to reversibly not ?ust isothermal andadiabatic) if the %ressure, volume, and tem%erature can be infinitesimally changed at every state of the

%rocess.

a) 8escribe how one could conduct a%%ro/imate) reversible isochoric, isobaric, isothermal, and adiabatic

%rocesses.a) Eeversible processes re*uire complete control, so the state of the system (,P, VT) is well2nown at alltimes and change only infinitesimally. very important point is that in reversible processes heat istransferred between ob%ects at the same temperature (or varying in temperature by an infinitesimal amount). In an engine the heat is e!changed between the gas going through the cycle and the e!ternal heat sources,T hot and T cold . In the reversible isothermal and adiabatic processes, heat flows in or out of the gas as the pressure is changed infinitesimally by adding or subtracting weight from the piston. In the reversible

isochoric and isobaric processes the temperature of the gas and the heat sources must not vary more than

J c

T2

d

b

T1

a

o;r

o



an infinitesimal amount even as the temperature changes+ this can be done by having a fine thermostatcontrol that changes the temperature of the heat source slowly as the heat flows in or out of the container6 b) 8etermine e/%ressions for calculating the entro%y changes in each of those reversible %rocesses and %utthem in the chart below> =se the definition of L>0/T or Wd0/T6:or e!ample L> P Wn' P dT/Tn' P ln(T/T o )6

$sochoric $sobaric $sothermal *diabatic

4onstant

3atio

LP? /T constant

L?P/T constant

LT? P constant

0? P γ constant

J vs. Pra%h

MS n' P ln(T/T o )and T/T o/ o

n' ln(T/T o )and T/T oP/P o

nEln(P/P o )and P/P o o /

&

c) Show that for any %rocess, the change entro%y is due to either a volume change and;or a tem%erature

change.c) s you can see from the e!pressions in the chart, the terms all have either a volume ratio or a temp. ratio.The temperature and volume ratios can also be related to pressure ratios of course. It’s also interesting to note that, since ' ' P NE L> L> P NL> T .d) Show that in an adiabatic %rocess, the entro%y change due to the volume change is equal and o%%osite to

the entro%y change due to the tem%erature change hence MS6&).d) >ince entropy is a state function, any process can be replaced with an e*uivalent series of processes. Theadiabatic process is e*uivalent to an isochoric process from T o to T and an isothermal process form P o to P, so L> adiaL> P NL> T n' P ln(T/T o ) N nEln(P/P o ) >ince (T/T o ) (P o /P) γ 21 for adiabatic processes, L> adin' P ln(P o /P) γ 21 NnEln(P/P o ) L> adin;' P (12γ )NE<ln(P o /P)n;' P 2' NE<ln(P o /P)?

29. 4onsider 1 gram of su%er cold ice at 51& I4 in a insulated container at 1 atm. eat from an e/ternal

source is added to the container and the ice goes through all the %hase changes until you end u% with 1 gram

of water va%or at 11&I4. In this problem I will use cal/3 as entropy units for convenience.a) 8etermine the entro%y changes throughout the entire %rocess. 3ecall> d: 6 mcdT and L6:;m.

L> icemc ln(T/T o )(1)(?.G) ln(8F7/8H7)?.?1QF cal/3?.?FQ A/3+

L> melt mB/TQ?/8F7?.87 cal/3+ L> water mc ln(T/T o )(1)(1) ln(7F7/8F7)?.718 cal/3+ L> vapmB/TGM?/7F71.MG cal/3+ L> steammc ln(T/T o )(1)(?.G) ln(7Q7/7F7)?.?178 cal/3+ or L> steamn' ln(T/T o )(1/1Q)(E/8)

ln(7Q7/7F7) ?.?GG? A/3?.?178 cal/3 b) $f the heat is being transferred from a heat source that has a constant tem%erature of 11&I4, what is the

entro%y change of the heat source 8oes the combined entro%y change satisfy the 2nd law ofthermodynamics

L> source 20total /T 2F7?/7Q7 21.1 cal/3 and L> ice/water/steamN8.? cal/3, so the overall entropy change is Nwhich is consistent with the 8nd law6

2=. Nne mole of 2 gas is contained in the left5hand side of the containershown, which has equal volumes left and right. The right5hand side is evacuated.

hen the valve is o%ened, the gas streams into the right side, without o%%osition.a) 8oes the tem%erature of the gas change /%lain.

a) The gas e!pands but it does @- wor, so no heat is enters the system and its temperature doesn’t change. It’s pressure does decease to X the original pressure. b) This %rocess is called an irreversible free e/%ansion". 4om%are this %rocess to reversible isothermal

e/%ansion and the reversible adiabatic e/%ansion for the same volume change. hat are the similarities and

differences

b) This process resembles the isothermal reversible process more than any other in that the final state of the system has the same temperature and pressure. It has only a superficial resemblance to the reversible

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adiabatic process in that no heat enters the system, but in the reversible adiabatic e!pansion the finaltemperature and pressure are different.c) ow would you find the entro%y change of this irreversible free e/%ansion"c) To find the entropy change it’s necessary to find a reversible process with the same initial and final states. In this case, that’s an isothermal process. >o, L> free e!pansion L> T nEln(P/P o )Eln8G.FH A/

2A. * 2.&5liter container has a center %artition that divides it into two equal %arts,

as shown. The left side contains 2 gas and the right side contains N2 gas. Koth

gases are at room tem%erature and at atmos%heric %ressure. The %artition is

removed and the gases are allowed to mi/ freely.a) hat is the entro%y increase

a) There is no temperature change but each gas now has a larger volume to move in. The entropy changecan be determined from the volume change of each gas, independent of the other gas. That is, L> T8nEln(P/P o ) 8(?.G)(Q.71)ln8 G.FH A/ b) hat is the entro%y increase if the tem%eratures were different Let-s say hydrogen had twice the

tem%erature of the o/ygen.b) The entropy change above would still apply, but in addition there would be entropy changes due to thetemperature changes of the gases. >o we need the e*uilibrium temperature, which is 1.GT o!ygen , you can thin of this as a two2step process, first let the gases come to thermal e*uilibrium, then remove the partition andlet them mi!. The entropy changes due to the temperature changes will be L> -8n' P ln(T/T o )(?.G)

(GE/8)ln(1.G) M.81 A/3, plus L> 8n' P ln(T/T o )(?.G)(GE/8)ln(?.FG) 27.?? A/3

The net L>G.FH NM.8127.??H.F A/3.c) hat is the entro%y increase if the gases were the same but the tem%eratures of the sam%les were

different

c) In this case the volume increase would not change the entropy so (a) would not apply. -nly thetemperature changes would change the entropy net L>M.8127.??1.81 A/3.

2E. *n ob?ect of mass m1, s%ecific heat c1, and tem%erature T 1 is %laced in contact with a second ob?ect of

mass m8, s%ecific heat c8, and tem%erature T 8 T 1.a) 8etermine an e/%ression for the equilibrium tem%erature T.a) n old calorimetry problem, T(m1c1T 1 N m8c8T 8 )/(m1c1N m8c8 ).

b) Show that the entro%y change MS of the system is> L> m1c1 ln(T/T 1 ) N m8c8 ln(T/T 8 ).b) This follows straight from the entropy e!pression of L> WmcdT/T6c) 4an the e/%ression above ever give a value less that & /%lain.

c) This is an irreversible process so entropy must increase+ that means the net entropy change described bythe e!pression in (b) must always give a N answer. This is hard to prove algebraically, but it’s easy to see innumerical e!amples lie the ones in the ne!t problem.

2G. The following are a cou%le of numerical versions of the %roblem above.a) 4alculate the increase in entro%y of the Bniverse when you add 2& g of 9@4 cream to 2&& g of =&@4

coffee. The s%ecific heat of cream and coffee is .2 A/gY'O.a) T eGG O' net L> 8? ln;(GGN8F7)/8FQ< N 8?? ln(78Q/777)17.218.F1.1 cal/3

b) * 1.&5+g iron horseshoe is ta+en from a furnace at G&&@4 and dro%%ed into .& +g of water at 1&@4. $f noheat is lost to the surroundings, determine the total entro%y change.

b) "ou need cironMF? A/g and cwater M1? A/g to be able to get T e7M O' net L> MF? ln;(7MN8F7)/(??N8F7)< N M(M1?) ln(7?F/8Q7)2H7?N17H7 F77 A/3

0&. *nother useful gra%h is a TS5diagram. This problem shows how a different graph from a P2graph can be useful6a) Pra%h a 4arnot cycle, %lotting (elvin tem%erature vertically and entro%y

horiRontally. This is called a tem%erature5entro%y diagram, or TS5diagram.

b) Show that the area under any curve re%resenting a reversible %ath in atem%erature5entro%y diagram re%resents the heat transferred during the %rocess.

b) The graph’s area WTd> WT(d0/T) Wd0 0c) 8erive from your diagram the e/%ression for the thermal efficiency of

&.9 mole &.9 mole

2

N2

T * K

4 8

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a 4arnot cycle.

c) 0hot T C L> and 0cold 2T '9 L> e12(T '9 L> /T C L>)12(T '9 /T C )12(T c /T h )d) 8raw a tem%erature5entro%y diagram for the Stirling cycle, described

in Jroblem 22. Bse this diagram to relate the efficiency of the 4arnot and

Stirling cycles. The two cycles have the same “e”6

'hallenge problems

01. The ma/imum %ower that can be e/tracted by a wind turbine from an air stream is a%%ro/imately 6d 8

v7

, where d is the blade diameter, v is the wind s%eed, and the constant 6 &.9 Us0

;m9

.a) /%lain the de%endence of on d and on v by considering a cylinder of air that %asses over the

turbine blades in time t . This cylinder has diameter d, length B 6 v t, and density ρ . b) The Vod59K wind turbine at (aha+u on the awaiian island of Nahu has a blade diameter of GA m

slightly longer than a football field) and sits ato% a 9E5m tower. $t can %roduce 0.2 V of electric

%ower. *ssuming 29' efficiency, what wind s%eed is required to %roduce this amount of %ower

Pive your answer in m;s and in +m;h.c) 4ommercial wind turbines are commonly located in or downwind of mountain %asses. hy

a) The inetic energy/time of the volume of wind passing through the bladesis (L3/Lt) X(Lm/Lt)v8 X( ρ Zd 8 L!/MLt)v8 X( ρ Z/M) d 8v7.b) 78!1?H ?.8G ma! ?.8G( d 8v7 ) ?.8G( &.9) F 8 )v7

v7? m/s1?Q m/hr.

c) >trong winds develop through mountain passes6

02. $t is %ossible to derive an e/%ression for efficiency for a 8iesel cycle in terms of the com%ression ratio.

4onsider that the cycle starts at %oint ( see Jroblem 21) with air at tem%erature T .a) $f the tem%erature at %oint 4 is T ' , derive an e/%ression for the efficiency of the cycle in terms of the

com%ression ratios, r 1 (P /P C ) and r 8 (P /P ' ). *ssume an ideal gas. This is a messy and long algebraic %roblem. Qou have to eliminate the tem%eratures in the e/%ression derived in %roblem 21 using adiabatic

and isobaric relationshi%s.

a) The answer I got is e 12 γ 221;(r 82γ 2 r 12γ )/(r 8212 r 121 )<, where r 1P /P C and r 8P /P ' , are the compressionand e!pansion ratios for the adiabatic portions of the 9iesel cycle illustrated in problem 81. In addition the

compression ratios can be related as follows r 821 r 1 γ−2 (T /T ' )<. [oing on the web you can find a detail solution. It’s mostly messy algebra. b) hat is the efficiency if T W 0&& (, Tc 6 G9& (, γ 6 1.&, and r 6 21 :ormula plug6

T

* K

4 8S

d v

L!

Bvt

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