Chapter 6 Second Law of Thermodynamics

Transcription

1 Capter 6 Second Law o Termodynamcs Te rst law o termodynamcs s an energy conservaton statement. It determnes weter or not a process can take place energetcally. It does not tell n wc drecton te process can go, even tat tey are energetcally possble. For examples: A rock wll all you lt t up and ten let go, and never come back to your and by tsel. Hot ryng pans loose eat and cool down wen taken o te stove. Ar n a g-pressure tre leaks out rom even a small ole n ts sde to te lower pressure atmospere, and never sees te reverse process. Ice cubes gans eat and melt n a warm room. Second law o termodynamcs s based on uman experence. It doesn t come rom complcated teory and equatons and wll tell you wy some process wll be orbdden altoug tey are energetcally possble. Ts wy we ave to dene te reversblty and rreversblty o a process. A reversble cange (or process) s one n wc te values o P, V, T, and U are well dened durng te cange. I te process s reversed, ten P, V, T, and U wll take on te same values n reverse order. To be reversble, a process must usually be slow (quas-statcally), and te system must be close to equlbrum durng te entre cange. A process s rreversble te system and ts surroundngs can not be returned to ter ntal states, wc s te common case n nature. Pressure Reversble process Pressure Irreversble process Volume Volume Second law o Termodynamcs: Many equvalent statements could be speced or te second law o termodynamcs. One o tem, due to Claussus and states Heat can not be transerred rom a cold reservor to a warm one wtout dong work, 173

2 or n anoter words eat wll not low spontaneously rom a colder to a otter temperature. Entropy (S, [S] = J/K) s a quantty used to measure te degree o dsorder (or randomness) o a system. Irreversble processes suc as spontaneous low o eat rom a otter to a colder body wtout te producton o useul work always ncrease te entropy o an solated system and o te unverse. Hg entropy Hgly dsordered partcles Low entropy Hgly ordered partcles 1. It s a state uncton o te system lke pressure, volume, temperature and energy),.e. t depend only on te ntal and nal state o a system and not on ow t reac tat states Tat s, a system as dente entropy just as t as a dente nternal energy U. 2. Te entropy o a system ncreases wen eat lows nto te system and decreases wen eat lows out. 3. Wen eat dq r enters (or removed rom) a system at absolute temperature T, te cange n entropy, ds, s dened by te rato dq r ds T provded te system canges n a reversble way. Te cange n entropy o a system movng reversble between two equlbrum states and s gven by: dqr S = ds = = S S T arbtrary reversble pat Ts mples tat or an rreversble processes between equlbrum states and, te entropy cange s evaluated usng te last equaton, and any convenent reversble pat wc connects to ; te result s o course pat dependent. 174

3 For termodynamc processes: S + S S system surroundng unverse and te second law o termodynamcs s ormulated as te prncple o ncreasng o entropy. Ts s: Sunverse 0 were te equalty apples to reversble process only. For natural process (rreversble) processes one can say" Natural processes startng n an equlbrum state and endng n anoter proceed n suc a drecton te entropy o te unverse ncreases Cyclc (du = 0) dqr S = = 0 T (Claussus' teorem) Adabatc (dq = 0) dqr dq = 0 S = = 0 T Cangng Pase (dt = 0 but T = constant) Same Pase (e.g. lqud) dq dq = mlv( or ml ) S =. T dq = mc T dt T S = mc = mc ln( ) T T Same Pase (e.g. gases) dq dt nc T S = nc = nc p p p T = dt nc p T S = ncv = ncv T 175 T ln( ) (constant pressure) T T ln( ) (constant volume) T

4 Isotermal process ( U = T = 0) ence V dv V dq = dw = P V = nrt S = nr nr ln( ) V = V V Heat transer between two masses dened as (m₁,c₁,t₁) and (m₂,c₂,t₂) T T S = mc ln( ) + m c ln( ) T1 T2 were T s te equlbrum temperature. How muc does te entropy ncrease n te ollowng cases? a) 18-gram ce cube at 0 o C s canged to water at 0 o C. [L = ⁵ J/kg] 5 dq ( ) r ml ds1 = = = 22.0 J/K T T b) 18-gram o water at 0 o C s canged to water at 100 o C. [c(water) = 4186 J/(kg.K)] T ds2 = mcln( ) = ln( ) = J/K T c) 18-gram o water at 100 o C s canged to steam at 100 o C. [L v = ⁵ J/kg] 5 dq ( ) r ml v ds3 = = = = J/K T T d) 18-gram ce cube at 0 o C s canged to steam at 100 o C. ds = ds1+ ds2 + ds3 = J/K Two moles o an deal monatomc gas undergo a reversble sotermal expanson rom 0.1 m³ to 0.5 m³ at a temperature o 20 o C. Wat s te cange n entropy o te gas? dv V 0.5 S = nr nrln( ) ln( ) 26.7 J/K V = V = 0.1 = 176

5 One mole o an deal monatomc gas s eated quas-statcally at constant volume rom 100 K to 105 K. Wat s te cange n entropy o te gas? dt T S = ncv ncvln( ) ln( ) 0.61 J/K T = T = Or one can use te ollowng approxmate metod, dq S = = nrcv = 1( 8.314) 0.61 J/K T I 200 g o water at 20 o C s mxed wt 300 g o water at 75 o C, nd: a- te nal equlbrum temperature o te mxture and Frst, calculate te eat lost by 300 g and te eat ganed by te 200 g, Q(lost by 300 g) = Q1 = 300Cwater ( T 75) Q(ganed by 200 g) = Q2 = 200Cwater ( T 20) o Ten, use te act tat Q1+ Q2 = 0, one can solve or T = 53.0 C. b- te cange n entropy o te system S = 300Cwaterln( )+300Cwater ln( ) = ( ) = 7.34 J K J K 177

6 Heat engne: any devce tat works n a cycle, extractng (or absorbng or takes n) eat, Q, rom a ot reservor dong work (W), and exaustng (or rejected or expelled) remanng eat, Q c, nto a cold reservor. Q c makes te pollutons o our envronment. Hot reservor at T Hot reservor at T Q Q Engne W Engne W Q c Q c Cold reservor at T c Cold reservor at T c Heat engne Rergerator Te ecency (e) o te eat engne s gven by: Te power s Energy we get W Q Qc e = = = Energy we pay or Q Q W [Joule] W / t P P [Watt]= e= = t [Seconds] Q / t Q / t 178

7 An automoble engne operates wt an overall ecency o 20%. How many gallons o gasolne are wasted or eac 10 gallons burned? W Q Qc e = = Q Q 10 Qc 0.2 = Qc = 8 gallons 10 An engne absorbs 1600 J rom a ot reservor and expels 1000 J to a cold reservor n eac cycle. a. Wat s te ecency, e, o te engne? W Q Qc e = = = = or 37.5% Q Q 1600 b. How muc work s done, W, n eac cycle? W = Q Qc = = 600 J. c. Wat s te power output o te engne eac cycle lasts or 0.3 s? P W 600 J = = = 2000 Watt. t 0.3 s Carnot cycle: s te most ecent cycle possble or a eat engne. An engne tat operates n accordance to ts cycle between a ot reservor, T, and a cold reservor, T, as te rato Q T = and te maxmum Qc Tc ecency T Tc e = T NOTE THAT: Kelvn Temperature must be used n te above equatons. A Carnot engne as a power output, P, o 200 W and operates between two reservors at 27 o C and 327 o C. 179

8 a- Fnd te maxmum ecency o te engne T Tc ( ) ( ) 300 e = = = = 0.5 T ( ) 600 b- How muc energy s absorbed per our? Use te ecency n te orm: W / t P Q P 200 e = = = = = 400 Watt. Q / t Q / t t e 0.5 For one our, 6 Q = (4.0 10² J/s)(3600 s) = J. c- How muc eat energy s rejected per our? You can use: Qc = Q W = Q Pt J (2 10² W)(3600 s) J. = = Or you can use: T Q = Qc( ) T c 300 = = ( ) J A eat engne operatng between 200 o C and 80 o C aceves 20% o te maxmum possble ecency. a. Wat s te maxmum possble ecency? T Tc emax = ecarnot = = = T 353 b. Wat s te actual ecency? eactual = 0.2 emax = = c. Wat energy nput wll enable te engne to perorm 10⁴ J o work? 4 W W 10 5 e = actual Q J. Q = e = = actual 180

9 Heat pumps and Rergerator: A eat pump, or rergerator, s a devce tat moves eat rom a reservor at a low temperature, Q, to one at ger temperature, Q. Mecancal work, W, must be suppled to te eat pump n order to accompls ts. Te coecent o perormance (COP) o a rergerator s dened by: c COP wat we want Qc Qc Tc = = = = wat we pay or W Q Q T T c c Te coecent o perormance (COP) o a eat pump s dened by: 1 Q Q T COP(eat pump) = = = = e W Q Q T T c c Wat s te coecent o perormance o a eat pump tat brngs eat rom te outdoors at o C nto + 22 o C ouse? (Hnt: te eat pump does work, W, wc s also avalable to eat te ouse.) T COP (eat pump) = T T c ( ) 295 = = = 11.8 ( ) (273 3)

10 True-False Questons 1- Lqud water as less entropy tan ce. 2- Te entropy o a system can never decrease. 3- Crystals are less dsordered tan gases 4- Te most ecent cyclc process s te Carnot cycle. T 5- A rergerator (eat pump) s a eat engne workng n reverse. T 6- It s mpossble to construct a eat engne wc does work wtout rejectng some eat to a cold reservor. T 7- Entropy s a quantty used to measure te degree o dsorder n a system. T 8- Te total entropy decreases or any system tat undergoes an rreversble process. F 9- Te total entropy o a system ncreases only t absorbs eat. F 10- No eat engne as ger ecency tan Carnot ecency. T 11- Te ecency o te deal engne sould be greater tan one. F 12- No cange n entropy or a system goes n reversble cyclc process. T 13- I te steam s condensed, ts entropy wll decrease. T 14- Te coecent o perormance o rergerator sould be less tan one. F 15- Isolated systems tend toward dsorder and entropy s a measure o ts dsorder. T 16- Te entropy o te unverse decreases n any process. F 17- Te ecency o eat engnes can be 100%. F 18- Heat engnes can ave ecency ger tan Carnot engne workng between te same two temperatures. F 19- To calculate te ecency o deal engne te temperature sould be n Celsus. F 20- Te termal ecency o an deal engne can be = 1.0. F 21- For any process te cange n entropy o a closed system < 0. F 182

11 Supplementary Problems One mole o an deal monatomc gas s eated quas-statcally at constant volume rom 100 K to 105 K. Wat s te cange n entropy o te gas? (a) 0.18 J/K. (b) 0.26 J/K. (c)@ 0.61 J/K. (d) 1.03 J/K. (e) 1.39 J/K. Suppose tat 10 kg o water at 50 o C s mxed wt an equal amount o water at 10 o C. Wen termal equlbrum s reaced, wat s te cange n entropy o te mxture? (a) 250 J/K (b) 130 J/K (c) 246 J/K (d) 551 J/K (e)@ 183 J/K Fve moles o an deal datomc gas (Cp = 7R/2) s taken troug an sovolumetrc process. I te nal pressure s ve tmes te ntal pressure, wat s te cange n entropy o te gas? (a) (b) (c) (d)@ (e) 234 J/K -234 J/K -167 J/K 167 J/K -151 J/K Fnd te cange n entropy wen 100 g o ce at 0 o C s eated slowly to 80 o C. (a) 85 cal/k (b) 25 cal/k (c) 62 cal/k (d) 12 cal/k (e)@ 55 cal/k 183

12 Te let-and sde o te contaner sown n Fgure 2 contans 5 moles o ntrogen gas, n termal equlbrum wt te rgt and sde, wc contans 3 moles o ydrogen gas. Te two sdes are separated by a partton, and te contaner s nsulated. Ater te partton s broken, wat s te cange n entropy o te system? (a) 34 J/K (b) 58 J/K (c) zero (d) 12 J/K (e)@ 49 J/K A contaner olds 240 g o water at 8 o C. Te contaner s placed n a rergerator mantaned at - 5 o C. Calculate te cange n entropy o te water ater t s n termal equlbrum wt te rergerator. (a) 331 J/K (b)@ -331 J/K (c) -254 J/K (d) -172 J/K (e) 254 J/K A 10 kg pece o ce at 0 o C s canged slowly and reversbly to water at 70 o C. Wat s te cange n entropy o te Ice? (a) J/K. (b) J/K. (c)@ J/K. (d) J/K. (e) J/K. 184

13 Consder 5 moles o an solated deal gas ntally at a pressure P1 and volume V1. Te gas expands reely to a nal pressure P2 and volume V2. I te entropy cange durng ts process s 16.6 J/K, ten te rato o te nal pressure to te ntal pressure s: (a)@ 0.67 (b) 3.0 (c) 0.33 (d) 0.50 (e) 1.5 Wen an deal gas s subjected to a reversble adabatc compresson process, wc one o te ollowng statements s TRUE: (a) Te gas rejects eat. (b) No work s done on te gas. (c)@ Te entropy o te gas does not cange. (d) Te gas absorbs eat. (e) Te nternal energy o te gas does not cange. Wat s te cange n entropy o 200-g o water as ts temperature ncreases rom 0 o C to 50 o C. (a) J/K. (b)@ J/K. (c) J/K. (d) J/K. (e) J/K g o water at 15.0 o C are converted slowly nto ce at o C. Wat s te cange o entropy o water? (a) J/K (b) J/K (c)@ J/K (d) J/K (e) J/K 185

14 An deal monatomc gas s conned to a cylnder by a pston. Te pston o s slowly pused n so tat te gas temperature remans at 27 C. Durng te compresson, 750 J o work s done on te gas. Te cange n te entropy o te gas s: (a) 3.0 J/K. (b)@ J/K. (c) 2.5 J/K. (d) Zero. (e) J/K. One mole o an deal monatomc gas expands at constant pressure to tree tmes ts ntal volume. Wat s te cange o entropy o te gas n ts process? (a)@ J/K (b) J/K (c) J/K (d) J/K (e) J/K Te cange n entropy s zero or (a) reversble processes durng wc no work s done. (b) reversble sotermal processes. (c) reversble sobarc processes. (d) all adabatc processes. (e)@ reversble adabatc processes. A 4.0-kg pece o ron at 800 K s dropped nto a lake wose temperature s 280 K. Assume tat te lake s so large tat ts temperature rse s neglgble. Fnd te cange n te entropy o te lake. [specc eat o ron = 0.11 kcal/kg.k]. (a) kcal/k (b) kcal/k (c)@ kcal/k (d) zero (e) kcal/k 186

15 A sample o an deal monatomc gas undergoes te reversble process A to B dsplayed n te T-S dagram sown n gure 6. Te process s : (a) a cange o pase. (b) an sotermal compresson. (c) a constant-volume process. (d)@ an sotermal expanson. (e) a ree expanson. Fve moles o an deal gas undergo a reversble sotermal compresson rom volume V to volume V/2 at temperature 30 o C. Wat s te cange n te entropy o te gas? (a) 29 J/K. (b) -81 J/K. (c)@ -29 J/K. (d) -18 J/K. (e) 18 J/K. Wc o te ollowng statements s correct? (a) For an adabatc process te cange n entropy s negatve t s done rreversbly. (b) Te ecency o a Carnot engne s 100%. (c) For an sotermal expanson te cange n entropy o an deal gas s zero. (d) A Carnot engne does not reject any eat as waste. (e)@ For an adabatc process te cange n entropy s zero t s done reversbly. 187

16 One mole o a monatomc deal gas s taken rom an ntal state () to a nal state () as sown n gure 1. Calculate te cange n entropy o te gas or ts process. (a) 1.25 J/K. (b)@ 36.5 J/K. (c) 11.2 J/K. (d) 22.5 J/K. (e) 25.0 J/K. You mx two samples o water, A and B. Sample A s 100 g at 20 o C and sample B s also 100 g but at 80 o C. Calculate te cange n te entropy o sample B. (a) 8.9 cal/k. (b) 9.7 cal/k (c) cal/k. (d) zero. (e)@ cal/k. Fve moles o an deal monatomc gas are allowed to expand sobarcally. Te ntal volume s 20.0 cm 3 and te nal volume s 100 cm 3. Fnd te cange n entropy o te gas. (a) 67.0 J/K (b) 100 J/K (c) 152 J/K (d) 52.0 J/K (e)@ 167 J/K 188

17 10.0 kg o water at zero o C are mxed wt 10.0 kg o water at 100 o C. Te specc eat o water s 4.19 kj/kg.k. Te total cange n entropy o te system s (a)@ 1.02 kj/k (b) 6.03 kj/k (c) 7.05 kj/k (d) 13.1 kj/k (e) zero One mole o an deal gas undergoes an sotermal expanson n wc ts volume ncreases to ve tmes ts ntal value. Wat s te cange o entropy o te gas n ts process? (a) 12.5 J/K (b) 4010 J/K (c) 20.1 J/K (d)@ 13.4 J/K (e) zero Two moles o an deal gas undergo an adabatc ree expanson rom an ntal volume o 0.6 L to 1.3 L. Calculate te cange n entropy o gas. V 1.3 S = nr ln( ) = ln( ) (a) J/K. V 0.6 (b) -5.3 J/K. J (c) 16.6 J/K. = (d)@ 12.9 J/K. K (e) zero. System A (one klogram o ce at 0 o C ) s added to system B (one klogram o water at 100 o C ) n an nsulator contaner. Calculate te total cange n entropy o system A. (a) (b) (c) (d)@ (e) Innte kj/k kj/k kj/k kj/k. 189

18 A 5.00-kg block o copper s at 296 K. I t s eated suc tat ts entropy ncreases by 1.07 kj/k, wat s te nal temperature? [Te specc eat o copper s 386 J/(kg*K)] (a) 760 K. (b) 273 K. (c) 310 K. (d)@ 515 K. (e) 100 K. T 3 T S = mcln = 5 386ln T 296 T = e = K. Wc o te ollowng statements are CORRECT: 1. Two objects are n termal equlbrum tey ave te same temperature. 2. In an sotermal process, te work done by an deal gas s equal to te eat energy 3. In an adabatc process, no eat enters or leaves te system. 4. Te termal ecency o an deal engne can be = For any process te cange n entropy o a closed system < 0. (a)@ 1, 2, and 3. (b) 3 and 5. (c) 4 and 5. (d) 1, 2 and 5. (e) 1 and 4. Wc o te ollowng statements are WRONG: 1. Te ecency o te deal engne s greater tan one. 2. Te cange n entropy s zero or reversble sotermal processes. 3. In cyclc processes, te cange n entropy s zero. 4. I steam s condensed, ts entropy wll decrease. 5. I ce s melted, ts entropy wll decrease. (a) 1, 2 and 4. (b)@ 1, 2 and 5. (c) 2, 3 and 4. (d) 1, 3 and 5. (e) 1, 2 and

19 Wc o te ollowng statements are true? o (I) Temperatures tat der by 10 C must der by 18 (II) 0 o C s te lowest temperature tat one can reac. (III) Heat conducton reers to te transer o termal energy between objects n contact. (IV) Te entropy o a system never decreases. (V) Heat s a orm o energy. (a)@ I, III, and V. (b) II, III, and V. (c) I, III, and IV. (d) I, II, and IV. (e) II, III, and IV. o F. An deal eat pump s used to absorb eat rom te outsde ar at -10 o C and transers t nto a ouse at a temperature o 30 o C. Wat s te eat energy transerred nto te ouse 5.0 kj o work s done on te eat pump? (a) 20 kj (b) 76 kj (c) 18 kj (d) 12 kj (e)@ 38 kj Wc one o te ollowng statements s WRONG? (a)@ Te entropy o te unverse remans constant n all processes. (b) Perect engnes do not exst because tey volate te second law o termodynamcs. (c) No real engne s more ecent tan Carnot engne. (d) In an solated system, te entropy ncreases or a rreversble process and remans constant or a reversble process. (e) Te cange n entropy o a system depends only on te ntal and nal states. 191

20 One mole o an deal monatomc gas (C v = 3R/2)s taken troug te cycle sown n Fgure 1. I T a = 590 K, T b = 450 K and T c = 300 K, calculate te ecency o an engne operatng n ts cycle. (a) 0.45 (b) 0.28 (c) 0.08 (d) 0.55 (e)@ 0.14 A eat engne as a monatomc gas as te workng substance and ts operatng cycle s sown by te P-V dagram n Fgure 1. In one cycle, 18.2 kj o eat energy s absorbed by te engne. Fnd te ecency o te eat engne. (a) 0.31 (b) 0.25 (c) 0.55 (d) 0.22 (e)@

21 A Carnot engne wose low temperature reservor s at 7 o C as an ecency o 50%. It s desred to ncrease te ecency to 70%. By ow many degrees sould te temperature o te g temperature reservor be ncreased wle te cold reservor remans at te same temperature? (a) 742 K (b) 560 K (c)@ 373 K (d) 280 K (e) 434 K A eat engne absorbs J per cycle rom a ot reservor wt an ecency o 25% and executes 3.15 cycles per second. Wat s te power output o te eat engne? (a) W. (b) W. (c) W. (d) W. (e)@ W. One mole o an deal gas s taken troug te cycle sown n te T-S dagram o gure (2). Calculate te ecency o te cycle. (a) (b) (c) (d) (e)@

22 An deal engne absorbs eat at 527 o C and rejects eat at 127 o C. I t as to produce useul mecancal work at te rate o 750 Watts, t must absorb eat at te rate o: (a) 2250 Watts. (b) 527 Watts. (c) 750 Watts. (d) 375 Watts. (e)@ 1500 Watts. Specy te CORRECT statement: (a) (b) (c) Te entropy o te unverse decreases n any process. To calculate te ecency o deal engne te temperature sould be n Celsus. Heat engnes can ave ecency ger tan Carnot engne workng between te same two temperatures. (d) Te ecency o eat engnes can be 100%. (e)@ Isolated systems tend toward dsorder and entropy s a measure o ts dsorder. An deal engne, wose low-temperature reservor s at 27 o C, as an ecency o 20%. By ow muc sould te temperature o te gtemperature reservor be ncreased to ncrease te ecency to 50%? (a) 20 K. (b)@ 225 K. (c) 975 K. (d) 88 K. (e) 300 K. An deal eat engne as a power output o 200 W. Te engne operates between two reservors at 300 K and 600 K. How muc energy s absorbed per our? (a) J. (b)@ J. (c) J. (d) J. (e) J. 194

23 A Carnot eat engne operates between two reservors wose temperatures are 27 o C and 127 o C. I we want to double te ecency o te eat engne, wat sould be te temperature o te ot reservor? Assume te temperature o te cold reservor s kept constant. (a)@ 600 K (b) 1200 K (c) 800 K (d) 1500 K (e) 900 K One mole o an deal gas s taken troug te reversble cycle sown n gure 1, wt Q 1 = 6.0 kj, Q 2 = 30 kj, Q 3 = 18 kj and Q 4 = 10 kj. Wat s te ecency o ts cycle? (a) 0.54 (b) 0.44 (c)@ 0.22 (d) 0.33 (e) 0.29 A Carnot eat engne absorbs 70.0 kj as eat and expels 55.0 kj as eat n eac cycle. I te low-temperature reservor s at 120 o C, nd te temperature o te g-temperature reservor. (a) 35.8 o C (b) 393 o C (c) 153 o C (d)@ 227 o C (e) 450 o C 195

24 A eat engne operates between 600 K and 300 K. In eac cycle t takes 100 J rom te ot reservor, loses 25 J to te cold reservor, and does 75 J o work. Ts eat engne volates: (a) Te rst law but not te second law o te termodynamcs. (b)@ Te second law but not te rst law o termodynamcs. (c) Neter te rst law nor te second law. (d) Conservaton o energy. (e) Bot, te rst law and te second law o termodynamcs. One mole o a datomc deal gas s taken troug te cycle sown n Fgure 2. Process b->c s adabatc, P a = 0.3 atm, P b = 3.0 atm, V b = m 3, and Vc = 4.0 V b. Wat s te ecency o te cycle? (a) 74%. (b)@ 53%. (c) 34%. (d) 28%. (e) 12%. A Carnot engne as an ecency o 20%. It operates between two constant-temperature reservors derng n temperature by 70.0 K. Wat s te temperature o te HOT reservor? (a)@ 350 K. (b) 280 K. (c) 400 K. (d) 300 K. (e) 70 K. 196

25 A eat engne operates n a Carnot cycle between reservors o temperatures 127 o C and 727 o C. It s ound tat 20 J o eat s expelled to te cold reservor n every cycle. Wat s te work done per cycle? (a) 40 J (b) 20 J (c) 50 J (d)@ 30 J (e) 10 J An 8.0-MW electrc power plant as an ecency o 30 %. It loses ts waste eat to te envronment. How muc eat s lost to te envronment per second? (a) 23 MJ (b) 5.6 MJ (c) 2.4 MJ (d)@ 19 MJ (e) 8.0 MJ A car engne delvers 8.6 kj o work per cycle. I ts ecency s 30%, nd te energy lost by te engne per cycle. (a) 24 kj. (b)@ 20 kj. (c) 8.6 kj. (d) 26 kj. (e) 14 kj. A eat engne as a termal ecency o 20%. It runs 2 revolutons per second and delvers 80 W. For eac cycle nd te eat dscarged to te cold reservor. (a) 40 W. (b) 61 W. (c) 200 W. (d) 121 W. (e)@ 160 W. W 80 W 40 (per cycle) = = 40 J, QH = = = 200 J 2 ε 0.2 Q = Q W = = 160 J C H 197

26 Wc one o te ollowng statements s WRONG? (a) A rergerator s a eat engne workng n reverse. (b) Te most ecent cyclc process s te Carnot cycle. (c)@ Te total entropy decreases or any system tat undergoes an rreversble process. (d) Entropy s a quantty used to measure te degree o dsorder n a system. (e) It s mpossble to construct a eat engne wc does work wtout rejectng some eat to a cold reservor. Wc one o te ollowng statements s WRONG? (a) No eat engne as ger ecency tan Carnot ecency. (b) Termal energy cannot be transerred spontaneously rom a cold object to a ot object. (c) Ater a system as gone troug a reversble cyclc process, ts total entropy does not cange. (d) A eat pump works lke a eat engne n reverse. (e)@ Te total entropy o a system ncreases only t absorbs eat. Wat s te coecent o perormance o a rergerator tat absorbs 40 cal/cycle at low temperature and expels 51 cal/cycle at g temperature? (a) 4.6. (b) (c) 2.3. (d)@ 3.6. (e) An deal reregerator as a coecent o perormance (COP) o 5. I te temperature nsde te reezer s - 20 o C, wat s te temperature at wc eat s rejected? (a) (b)@ (c) (d) (e) -20 o C 31 o C -45 o C 20 o C 36 o C 198

27 Wc one o te ollowng statements s WRONG? (a) No eat engne as ger ecency tan Carnot ecency. (b) A rergerator works lke a eat engne n reverse. (c) Termal energy cannot be transerred spontaneously rom a cold object to a ot object. (d)@ Te total entropy o a system ncreases only t absorbs eat. (e) Ater a system as gone troug a reversble cyclc process, ts total entropy does not cange. Wat s te coecent o perormance o an deal rergerator te temperatures o te two reservors are 10 o C and 27 o C. (a)@ 7.1 (b) 1.5 (c) 0.5 (d) 8.0 (e) 6.5 A Carnot rergerator as a coecent o perormance equal to 6. I te rergerator expels 80 J o eat to a ot reservor n eac cycle, nd te eat absorbed rom te cold reservor. (a) 21 J. (b) 15 J. (c) 30 J. (d) 5 J. (e)@ 69 J. A Carnot rergerator as a coecent o perormance equal to 5. Te rergerator absorbs 120 J o eat rom a cold reservor n eac cycle. How muc eat s expelled to te ot reservor? (a) 720 J (b) 480 J (c) 600 J (d)@ 144 J (e) 125 J 199

28 Wat mass o water at 0 o C can a reezer make nto ce cubes n one our, te coecent o perormance o te rergerator s 3.0 and te power nput s 0.2 Klowatt? (a) 0.4 kg. (b) 1.9 kg. (c) 3.0 kg. (d) 9.2 kg. (e)@ 6.5 kg. Durng one cycle, a Carnot rergerator does 200 J o work to remove 600 J rom ts cold compartment. How muc energy per cycle s exausted to te ktcen as eat? (a)@ 800 J. (b) 200 J. (c) 225 J. (d) 600 J. (e) 450 J. 200