Physics 41 Chapter 22 HW Serway 7 th Edition
|
|
- Jeffry Harris
- 6 years ago
- Views:
Transcription
1 yss 41 apter H Serway 7 t Edton oneptual uestons: 1,, 8, 1 roblems: 9, 1, 0,, 7, 9, 48, 54, 55 oneptual uestons: 1,, 8, 1 1 Frst, te effeny of te automoble engne annot exeed te arnot effeny: t s lmted by te temperature of burnng fuel and te temperature of te envronment nto w te exaust s dumped Seond, te engne blok annot be allowed to go over a ertan temperature rd, any pratal engne as frton, nomplete burnng of fuel, and lmts set by tmng and energy transfer by eat ger steam temperature means tat more energy an be extrated from te steam For a onstant temperature eat snk at, and steam at, te effeny of te power plant goes as 1 and s maxmzed for a g 8 (a) en te two sdes of te semondutor are at dfferent temperatures, an eletr potental (voltage) s generated aross te materal, w an drve eletr urrent troug an external rut e two ups at 50 ontan te same amount of nternal energy as te par of ot and old ups ut no energy flows by eat troug te onverter brdgng between tem and no voltage s generated aross te semondutors eat engne must put out exaust energy by eat e old up provdes a snk to absorb output or wasted energy by eat, w as nowere to go between two ups of equally warm water 1 (a) For an expandng deal gas at onstant temperature, te nternal energy stays onstant e gas must absorb by eat te same amount of energy tat t puts out by work en ts entropy ange s ΔS Δ nr ln 1 For a reversble adabat expanson Δ 0, and Δ S 0 n deal gas undergong an rreversble adabat expanson an ave any postve value for ΔS up to te value gven n part (a) roblems: 9, 1, 0,, 4, 7, 9, 48, 54, K (a) 14 K Δ e 14 67% J, eng eng J 588 k Δt 1 s
2 1 n deal gas s taken troug a arnot yle e sotermal expanson ours at 50, and te sotermal ompresson takes plae at 500 e gas takes n 1 00 J of energy from te ot reservor durng te sotermal expanson Fnd (a) te energy expelled to te old reservor n ea yle and te net work done by te gas n ea yle 1 Isotermal expanson at 5 K Isotermal ompresson at K Gas absorbs 1 00 J durng expanson (a) 100 J 74 J 1 5 eng J 459 J O Δ 89 arnot refrg J per 1 J energy removed by eat Seton 5 Gasolne and esel Engnes 7 In a ylnder of an automoble engne, just after ombuston, te gas s onfned to a volume of 500 m and as an ntal pressure of a e pston moves outward to a fnal volume of 00 m and te gas expands wtout energy loss by eat (a) If 140 for te gas, wat s te fnal pressure? How mu work s done by te gas n expandng? (a) f f m f ( a) 44 ka f 00 m d Integratng, m ( 50 )( a)( m ) 1 f 00 m J
3 f 9 Δ S nrln Rln 576 J K ere s no ange n temperature for an deal gas FIG 9 0 (a) Frst, onsder te adabat proess : so lso or L ka 71 ka 150 L nr nr Now, onsder te sotermal proess : K 549 K K ka 100 L 40 L 150 L Next, onsder te adabat proess : 445 ka ut, from above lso onsderng te sotermal proess, 1 Hene, 1 w redues to 100 L Fnally, ka 875 ka 160 L State (ka) (L) (K) For te sotermal proess : Δ Ent n Δ L 40 L 150 L 160 L 160 ln 4 mol 814 J mol K 70 K ln kj 100 so nr For te adabat proess : 0 Δ Ent n ( ) 4 mol ( 814 J mol K ) ( ) K 498 kj and +Δ E nt 0+ ( 498 kj) 498 kj
4 For te sotermal proess : Δ Ent n Δ ln 4 mol 814 J mol K 549 K ln 50 kj 40 and nr Fnally, for te adabat proess : 0 Δ Ent n ( ) 4 mol ( 814 J mol K ) ( ) K kj and +Δ Ent kj kj roess (kj) (kj) Δ Ent (kj) e work done by te engne s te negatve of te work nput e output work gven by te work olumn n te table wt all sgns reversed eng s () eng 156 kj e 07 or 7% 658 kj 549 e or 7% 70 O 0100O 4 arnot yle or arnot effeny arnot yle 9 K K 68 K FIG 4 us, 117 joules of energy enter te room by eat for ea joule of w ork done
5 5 n deal refrgerator or deal eat pump s equvalent to a arnot engne runnng n reverse at s, energy s taken n from a old reservor and energy s rejeted to a ot reservor (a) Sow tat te work tat must be suppled to run te refrgerator or eat pump s Sow tat te oeffent of performane of te deal refrgerator s O 5 (a) For a omplete yle, Δ Ent 0 and ( ) 1 e ave already sown tat for a arnot yle (and only for a arnot yle) erefore, e ave te defnton of te oeffent of performane for a refrgerator, O 1000 J 48 Δ Sot 600 K Usng te result from part (a), ts beomes O J Δ Sold 50 K (a) Δ S Δ S +Δ S U ot old 0476 J K e eng e J 417 J () net 417 J 50 J 167 J 1Δ S U 50 K 0476 J K 167 J
6 *54 (a) For te sotermal proess, te work on te gas s ln ( a)( m ) ln J were we ave used and atm a 100 L m FIG 54 Δ a m J 0 and eng J 410 kj Sne s an sotermal proess, Δ Ent, 0 and For an deal monatom gas, R and J 5R nr R R 5 lso, nr R R 5 n Δ 100 R kj R so te total energy absorbed by eat s kj+ 608 kj 14 kj 5 5 n Δ nrδ Δ () J 101 kj (d) eng eng e J J or 89% (e) arnot engne operatng between ot 5060/R and old 1010/R as effeny 1 / 1 1/5 800% e tree-proess engne onsdered n ts problem as mu lower effeny
7 *55 t pont, nr and n 100 mol t pont, nr so t pont, ( )( ) 6 nr and t pont, ( ) nr so e eat for ea step n te yle s R 5R found usng and : FIG 55 n nr n 6 750nR n 6 6nR n 50nR (a) erefore, enterng + 105nR leavng + 850nR () tual effeny, e 0190 (d) arnot effeny, e e arnot effeny s mu ger
11/19/2013. PHY 113 C General Physics I 11 AM 12:15 PM MWF Olin 101
PHY 113 C General Pyss I 11 AM 12:15 PM MWF Oln 101 Plan or Leture 23: Capter 22: Heat engnes 1. ermodynam yles; work and eat eeny 2. Carnot yle 3. Otto yle; desel yle 4. Bre omments on entropy 11/19/2013
More informationPhysics 41 Chapter 22 HW
Pysis 41 apter 22 H 1. eat ine performs 200 J of work in ea yle and as an effiieny of 30.0%. For ea yle, ow mu energy is (a) taken in and (b) expelled as eat? = = 200 J (1) e = 1 0.300 = = (2) From (2),
More informationPHYSICS 212 MIDTERM II 19 February 2003
PHYSICS 1 MIDERM II 19 Feruary 003 Exam s losed ook, losed notes. Use only your formula sheet. Wrte all work and answers n exam ooklets. he aks of pages wll not e graded unless you so request on the front
More informationChapter 18: The Laws of Thermodynamics
Capter 18: e Laws o ermodynams Answers to Even-Numbered Coneptual uestons. (a) Yes. Heat an low nto te system at te same tme te system expands, as n an sotermal expanson o a gas. (b) Yes. Heat an low out
More information#64. ΔS for Isothermal Mixing of Ideal Gases
#64 Carnot Heat Engne ΔS for Isothermal Mxng of Ideal Gases ds = S dt + S T V V S = P V T T V PV = nrt, P T ds = v T = nr V dv V nr V V = nrln V V = - nrln V V ΔS ΔS ΔS for Isothermal Mxng for Ideal Gases
More informationHeat Engines, Entropy, and the Second Law of Thermodynamics
Heat Engnes, Entropy, and te Seond Law o ermodynams HER OULINE.1 Heat Engnes and te Seond Law o ermodynams. Heat umps and Rergerators. Reversble and Irreversble roesses.4 e arnot Engne. Gasolne and esel
More informationPhysics 231 Lecture 35
ysis 1 Leture 5 Main points of last leture: Heat engines and effiieny: eng e 1 Carnot yle and Carnot engine. eng e 1 is in Kelvin. Refrigerators CO eng Ideal refrigerator CO rev reversible Entropy ΔS Computation
More information18. Heat Engine, Entropy and the second law of thermodynamics
8. Heat Engne, Entropy and te seond law o terodynas In nature, ost o proesses are rreversble. due to te seond Law o terodynas Heat alwasys lows ro Hot to old. 8-. Heat Engne and te eond Law o erodynas
More informationEF 152 Exam #3, Fall, 2012 Page 1 of 6
EF 5 Exam #3, Fall, 0 Page of 6 Name: Setion: Guidelines: ssume 3 signifiant figures for all given numbers. Sow all of your work no work, no redit Write your final answer in te box provided - inlude units
More informationFirst Law of Thermodynamics
Frst Law of Thermodynamcs Readng: Chapter 18, Sectons 18-7 to 18-11 Heat and Work When the pston s dsplaced by ds, force exerted by the gas = F = pa, work done by the gas: dw Fds ( pa)( ds) p( Ads) p d.
More informationPhysics 207 Lecture 23
ysics 07 Lecture ysics 07, Lecture 8, Dec. Agenda:. Finis, Start. Ideal gas at te molecular level, Internal Energy Molar Specific Heat ( = m c = n ) Ideal Molar Heat apacity (and U int = + W) onstant :
More informationHomework Chapter 21 Solutions!!
Homework Chapter 1 Solutons 1.7 1.13 1.17 1.19 1.6 1.33 1.45 1.51 1.71 page 1 Problem 1.7 A mole sample of oxygen gas s confned to a 5 lter vessel at a pressure of 8 atm. Fnd the average translatonal knetc
More informationP REVIEW NOTES
P34 - REIEW NOTES Capter 1 Energy n Termal Pyss termal equlbrum & relaxaton tme temperature & termometry: fxed ponts, absolute temperature sale P = nrt deal gas law: ( ) ( T ) ( / n) C( T ) ( ) + / n vral
More informationChapter 6 Second Law of Thermodynamics
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
More informationThermodynamics Second Law Entropy
Thermodynamcs Second Law Entropy Lana Sherdan De Anza College May 8, 2018 Last tme the Boltzmann dstrbuton (dstrbuton of energes) the Maxwell-Boltzmann dstrbuton (dstrbuton of speeds) the Second Law of
More informationEF 152 Exam #3, Spring 2016 Page 1 of 6
EF 5 Exam #3, Spring 06 Page of 6 Name: Setion: Instrutions Do not open te exam until instruted to do so. Do not leave if tere is less tan 5 minutes to go in te exam. Wen time is alled, immediately stop
More informationChapters 19 & 20 Heat and the First Law of Thermodynamics
Capters 19 & 20 Heat and te First Law of Termodynamics Te Zerot Law of Termodynamics Te First Law of Termodynamics Termal Processes Te Second Law of Termodynamics Heat Engines and te Carnot Cycle Refrigerators,
More informationUniversity Physics AI No. 10 The First Law of Thermodynamics
Unversty hyscs I No he Frst Law o hermodynamcs lass Number Name Ihoose the orrect nswer Whch o the ollowng processes must volate the rst law o thermodynamcs? (here may be more than one answer!) (,B,D )
More informationLecture 27: Entropy and Information Prof. WAN, Xin
General Pysis I Leture 27: Entropy and Information Prof. WAN, Xin xinwan@zju.edu.n ttp://zimp.zju.edu.n/~xinwan/ 1st & 2nd Laws of ermodynamis e 1st law speifies tat we annot get more energy out of a yli
More informationChapter 5 rd Law of Thermodynamics
Entropy and the nd and 3 rd Chapter 5 rd Law o hermodynamcs homas Engel, hlp Red Objectves Introduce entropy. Derve the condtons or spontanety. Show how S vares wth the macroscopc varables,, and. Chapter
More informationG4023 Mid-Term Exam #1 Solutions
Exam1Solutons.nb 1 G03 Md-Term Exam #1 Solutons 1-Oct-0, 1:10 p.m to :5 p.m n 1 Pupn Ths exam s open-book, open-notes. You may also use prnt-outs of the homework solutons and a calculator. 1 (30 ponts,
More informationChapter 21 - The Kinetic Theory of Gases
hapter 1 - he Knetc heory o Gases 1. Δv 8.sn 4. 8.sn 4. m s F Nm. 1 kg.94 N Δ t. s F A 1.7 N m 1.7 a N mv 1.6 Use the equaton descrbng the knetc-theory account or pressure:. hen mv Kav where N nna NA N
More informationOutline. Unit Eight Calculations with Entropy. The Second Law. Second Law Notes. Uses of Entropy. Entropy is a Property.
Unt Eght Calculatons wth Entropy Mechancal Engneerng 370 Thermodynamcs Larry Caretto October 6, 010 Outlne Quz Seven Solutons Second law revew Goals for unt eght Usng entropy to calculate the maxmum work
More informationGeneral Formulas applicable to ALL processes in an Ideal Gas:
Calormetrc calculatons: dq mcd or dq ncd ( specc heat) Q ml ( latent heat) General Formulas applcable to ALL processes n an Ideal Gas: P nr du dq dw dw Pd du nc d C R ( monoatomc) C C R P Specc Processes:
More informationProblem Set #6 solution, Chem 340, Fall 2013 Due Friday, Oct 11, 2013 Please show all work for credit
Problem Set #6 soluton, Chem 340, Fall 2013 Due Frday, Oct 11, 2013 Please show all work for credt To hand n: Atkns Chap 3 Exercses: 3.3(b), 3.8(b), 3.13(b), 3.15(b) Problems: 3.1, 3.12, 3.36, 3.43 Engel
More informationPhysics 3 (PHYF144) Chap 2: Heat and the First Law of Thermodynamics System. Quantity Positive Negative
Physcs (PHYF hap : Heat and the Frst aw of hermodynamcs -. Work and Heat n hermodynamc Processes A thermodynamc system s a system that may exchange energy wth ts surroundngs by means of heat and work.
More information3B SCIENTIFIC PHYSICS
3B SCIENTIFIC PHYSICS Peltier Heat Pump 0076 Instrution manual 05/7 TL/JS Transport ase Semati view 3 Stirrer unit 4 Connetor for stirrer unit 5 Connetor for power supply 6 Stirring rod old side 7 Peltier
More informationMaximum work for Carnot-like heat engines with infinite heat source
Maximum work for arnot-like eat engines wit infinite eat soure Rui Long and Wei Liu* Sool of Energy and Power Engineering, Huazong University of Siene and enology, Wuan 4374, ina orresponding autor: Wei
More informationPhysics 4B. A positive value is obtained, so the current is counterclockwise around the circuit.
Physcs 4B Solutons to Chapter 7 HW Chapter 7: Questons:, 8, 0 Problems:,,, 45, 48,,, 7, 9 Queston 7- (a) no (b) yes (c) all te Queston 7-8 0 μc Queston 7-0, c;, a;, d; 4, b Problem 7- (a) Let be the current
More informationThermodynamics and Gases
hermodynamcs and Gases Last tme Knetc heory o Gases or smple (monatomc) gases Atomc nature o matter Demonstrate deal gas law Atomc knetc energy nternal energy Mean ree path and velocty dstrbutons From
More informationESCI 341 Atmospheric Thermodynamics Lesson 11 The Second Law of Thermodynamics
ESCI 341 Atmosperi ermodynamis Lesson 11 e Seond Law of ermodynamis Referenes: Pysial Cemistry (4 t edition), Levine ermodynamis and an Introdution to ermostatistis, Callen HE SECOND LAW OF HERMODYNAMICS
More informationTHE SECOND LAW OF THERMODYNAMICS
HE SECOND LAW OF HERMODYNAMICS 9 EXERCISES Setions 9. and 9.3 e Seond Law of ermodynamis and Its Appliations 3. INERPRE is problem requires us to alulate te effiieny of reversible eat engines tat operate
More informationChemical Engineering Department University of Washington
Chemcal Engneerng Department Unversty of Washngton ChemE 60 - Exam I July 4, 003 - Mass Flow Rate of Steam Through a Turbne (5 onts) Steam enters a turbne at 70 o C and.8 Ma and leaves at 00 ka wth a qualty
More informationNonequilibrium Thermodynamics of open driven systems
1 Boynam Otal Imagng Center (BIOPIC) 2 Beng Internatonal Center for Matematal Resear (BICMR) Peng Unversty, Cna Nonequlbrum ermoynams of oen rven systems Hao Ge A sngle boemal reaton yle B + AP 1 C + ADP
More informationPhysics 240: Worksheet 30 Name:
(1) One mole of an deal monatomc gas doubles ts temperature and doubles ts volume. What s the change n entropy of the gas? () 1 kg of ce at 0 0 C melts to become water at 0 0 C. What s the change n entropy
More informationExample
Chapter Example.- ------------------------------------------------------------------------------ sold slab of 5.5 wt% agar gel at 78 o K s.6 mm thk and ontans a unform onentraton of urea of. kmol/m 3.
More informationReview of Classical Thermodynamics
Revew of Classcal hermodynamcs Physcs 4362, Lecture #1, 2 Syllabus What s hermodynamcs? 1 [A law] s more mpressve the greater the smplcty of ts premses, the more dfferent are the knds of thngs t relates,
More informationMain components of the above cycle are: 1) Boiler (steam generator) heat exchanger 2) Turbine generates work 3) Condenser heat exchanger 4) Pump
Introducton to Terodynacs, Lecture -5 Pro. G. Cccarell (0 Applcaton o Control olue Energy Analyss Most terodynac devces consst o a seres o coponents operatng n a cycle, e.g., stea power plant Man coponents
More informationLecture 27: Entropy and Information Prof. WAN, Xin
General Pysis I Leture 27: Entropy and Information Prof. WAN, Xin xinwan@zju.edu.n ttp://zimp.zju.edu.n/~xinwan/ Outline Introduing entropy e meaning of entropy Reversibility Disorder Information Seleted
More informationCHAPTER 7 ENERGY BALANCES SYSTEM SYSTEM. * What is energy? * Forms of Energy. - Kinetic energy (KE) - Potential energy (PE) PE = mgz
SYSTM CHAPTR 7 NRGY BALANCS 1 7.1-7. SYSTM nergy & 1st Law of Thermodynamcs * What s energy? * Forms of nergy - Knetc energy (K) K 1 mv - Potental energy (P) P mgz - Internal energy (U) * Total nergy,
More informationLecture 26 Finite Differences and Boundary Value Problems
4//3 Leture 6 Fnte erenes and Boundar Value Problems Numeral derentaton A nte derene s an appromaton o a dervatve - eample erved rom Talor seres 3 O! Negletng all terms ger tan rst order O O Tat s te orward
More informationOn Adaptive Control of Simulated Moving Bed Plants. Plants Using Comsol s Simulink Interface. Speaker: Marco Fütterer
daptve Smulated Movng ed Plants Usng Comsol s Smulnk Interfae Speaker: Maro Fütterer Insttut für utomatserungstehnk Otto-von-Guerke Unverstät Unverstätsplatz, D-39106 Magdeburg Germany e-mal: maro.fuetterer@ovgu.de
More informationPhysics 207 Lecture 27
hyscs 07 Lecture 7 hyscs 07, Lecture 7, Dec. 6 Agenda: h. 0, st Law o Thermodynamcs, h. st Law o thermodynamcs ( U Q + W du dq + dw ) Work done by an deal gas n a ston Introducton to thermodynamc cycles
More information= T. (kj/k) (kj/k) 0 (kj/k) int rev. Chapter 6 SUMMARY
Capter 6 SUMMARY e second la of termodynamics leads to te definition of a ne property called entropy ic is a quantitative measure of microscopic disorder for a system. e definition of entropy is based
More informationApplied Physics Research; Vol. 7, No. 5; 2015 ISSN E-ISSN Published by Canadian Center of Science and Education
Appled Pyss Resear; Vol. 7, No. 5; 2015 ISSN 1916-9639 E-ISSN 1916-9647 Publsed by Canadan Center of Sene and Eduaton oward te rue Seond Law Part II: e Untenable Assumpton of Current Seond Law ermodynams
More informationForce = F Piston area = A
CHAPTER III Ths chapter s an mportant transton between the propertes o pure substances and the most mportant chapter whch s: the rst law o thermodynamcs In ths chapter, we wll ntroduce the notons o heat,
More informationIntroduction to Statistical Methods
Introducton to Statstcal Methods Physcs 4362, Lecture #3 hermodynamcs Classcal Statstcal Knetc heory Classcal hermodynamcs Macroscopc approach General propertes of the system Macroscopc varables 1 hermodynamc
More informationLecture 3 Examples and Problems
Lecture 3 Examles and Problems Mechancs & thermodynamcs Equartton Frst Law of Thermodynamcs Ideal gases Isothermal and adabatc rocesses Readng: Elements Ch. 1-3 Lecture 3, 1 Wllam Thomson (1824 1907) a.k.a.
More informationPhysics 202 Homework 5
Physics 202 Homework 5 Apr 29, 2013 1. A nuclear-fueled electric power plant utilizes a so-called boiling water reac- 5.8 C tor. In this type of reactor, nuclear energy causes water under pressure to boil
More informationThe Second Law of Thermodynamics
Capter 6 Te Seond Law of Termodynamis In te last two apters of tis book we applied te first law of termodynamis to losed and open systems onsidering bot quasistati and non-quasi-stati proesses. A question
More informationPhysics 2B Chapter 17 Notes - Calorimetry Spring 2018
Physs 2B Chapter 17 Notes - Calormetry Sprng 2018 hermal Energy and Heat Heat Capaty and Spe Heat Capaty Phase Change and Latent Heat Rules or Calormetry Problems hermal Energy and Heat Calormetry lterally
More informationStructure and Drive Paul A. Jensen Copyright July 20, 2003
Structure and Drve Paul A. Jensen Copyrght July 20, 2003 A system s made up of several operatons wth flow passng between them. The structure of the system descrbes the flow paths from nputs to outputs.
More informationNAME and Section No.
Chemstry 391 Fall 2007 Exam I KEY (Monday September 17) 1. (25 Ponts) ***Do 5 out of 6***(If 6 are done only the frst 5 wll be graded)*** a). Defne the terms: open system, closed system and solated system
More informationAnnouncements. Exam 4 - Review of important concepts
Announcements 1. Exam 4 starts Friday! a. Available in esting Center from Friday, Dec 7 (opening time), up to Monday, Dec 10 at 4:00 pm. i. Late fee if you start your exam after 4 pm b. Covers C. 9-1 (up
More informationDissipated energy and Entropy Production for an Unconventional Heat Engine: The Stepwise Circular Cycle
sspated energy and Entropy roducton for an nconventonal Heat Engne: he Stepwse rcular ycle Francesco d Lberto, Raffaele astore and Fulvo erugg partmento Scenze Fsche - nverstà d apol Federco II - SI-R,
More informationTEST 5 (phy 240) 2. Show that the volume coefficient of thermal expansion for an ideal gas at constant pressure is temperature dependent and given by
ES 5 (phy 40). a) Wrte the zeroth law o thermodynamcs. b) What s thermal conductvty? c) Identyng all es, draw schematcally a P dagram o the arnot cycle. d) What s the ecency o an engne and what s the coecent
More informationTemperature. Chapter Heat Engine
Chapter 3 Temperature In prevous chapters of these notes we ntroduced the Prncple of Maxmum ntropy as a technque for estmatng probablty dstrbutons consstent wth constrants. In Chapter 9 we dscussed the
More informationfind (x): given element x, return the canonical element of the set containing x;
COS 43 Sprng, 009 Dsjont Set Unon Problem: Mantan a collecton of dsjont sets. Two operatons: fnd the set contanng a gven element; unte two sets nto one (destructvely). Approach: Canoncal element method:
More informationVoltammetry. Bulk electrolysis: relatively large electrodes (on the order of cm 2 ) Voltammetry:
Voltammetry varety of eletroanalytal methods rely on the applaton of a potental funton to an eletrode wth the measurement of the resultng urrent n the ell. In ontrast wth bul eletrolyss methods, the objetve
More informationPhysical Chemistry I for Biochemists. Chem340. Lecture 16 (2/18/11)
hyscal Chemstry I or Bochemsts Chem34 Lecture 16 (/18/11) Yoshtaka Ish Ch4.6, Ch5.1-5.5 & HW5 4.6 Derental Scannng Calormetry (Derental hermal Analyss) sample = C p, s d s + dh uson = ( s )Kdt, [1] where
More informationImage classification. Given the bag-of-features representations of images from different classes, how do we learn a model for distinguishing i them?
Image classfcaton Gven te bag-of-features representatons of mages from dfferent classes ow do we learn a model for dstngusng tem? Classfers Learn a decson rule assgnng bag-offeatures representatons of
More informationthe first derivative with respect to time is obtained by carefully applying the chain rule ( surf init ) T Tinit
.005 ermal Fluids Engineering I Fall`08 roblem Set 8 Solutions roblem ( ( a e -D eat equation is α t x d erfc( u du π x, 4αt te first derivative wit respect to time is obtained by carefully applying te
More informationPhysical Chemistry I for Biochemists. Lecture 18 (2/23/11) Announcement
Physcal Chestry I or Bochests Che34 Lecture 18 (2/23/11) Yoshtaka Ish Ch5.8-5.11 & HW6 Revew o Ch. 5 or Quz 2 Announceent Quz 2 has a slar orat wth Quz1. e s the sae. ~2 ns. Answer or HW5 wll be uploaded
More information3-1 Introduction: 3-2 Spontaneous (Natural) Process:
- Introducton: * Reversble & Irreversble processes * Degree of rreversblty * Entropy S a state functon * Reversble heat engne Carnot cycle (Engne) * Crteron for Eulbrum SU,=Smax - Spontaneous (Natural)
More informationBidimensional Analysis of a Thermoelectric Module using Finite Element Techniques
Bdmensonal Analyss of a Termoeletr Module usng Fnte Element Tenques *Antono Arenas, Jorge Vázquez, Rafael Palaos Unversdad Pontfa Comllas Esuela Téna Superor de ngenería *Departamento de Fludos y Calor,
More informationExercise 10: Theory of mass transfer coefficient at boundary
Partle Tehnology Laboratory Prof. Sotrs E. Pratsns Sonneggstrasse, ML F, ETH Zentrum Tel.: +--6 5 http://www.ptl.ethz.h 5-97- U Stoffaustaush HS 7 Exerse : Theory of mass transfer oeffent at boundary Chapter,
More informationS = = = nrln = 10.0 mol ln = 35.9
hy 212: General hysics II 1 hapter 20 orksheet (2 nd Law of hermodynamics & eat Engines) Entropy: 1. A sample of 10.0 moles of a monatomic ideal gas, held at constant temperature (1000), is expanded from
More informationGravity Drainage Prior to Cake Filtration
1 Gravty Dranage Pror to ake Fltraton Sott A. Wells and Gregory K. Savage Department of vl Engneerng Portland State Unversty Portland, Oregon 97207-0751 Voe (503) 725-4276 Fax (503) 725-4298 ttp://www.e.pdx.edu/~wellss
More informationA quote of the week (or camel of the week): There is no expedience to which a man will not go to avoid the labor of thinking. Thomas A.
A quote of the week (or camel of the week): here s no expedence to whch a man wll not go to avod the labor of thnkng. homas A. Edson Hess law. Algorthm S Select a reacton, possbly contanng specfc compounds
More informationA Theorem of Mass Being Derived From Electrical Standing Waves (As Applied to Jean Louis Naudin's Test)
A Theorem of Mass Beng Derved From Eletral Standng Waves (As Appled to Jean Lous Naudn's Test) - by - Jerry E Bayles Aprl 4, 000 Ths paper formalzes a onept presented n my book, "Eletrogravtaton As A Unfed
More informationSection A 01. (12 M) (s 2 s 3 ) = 313 s 2 = s 1, h 3 = h 4 (s 1 s 3 ) = kj/kgk. = kj/kgk. 313 (s 3 s 4f ) = ln
0. (a) Sol: Section A A refrigerator macine uses R- as te working fluid. Te temperature of R- in te evaporator coil is 5C, and te gas leaves te compressor as dry saturated at a temperature of 40C. Te mean
More informationEarlier Lecture. This gas tube is called as Pulse Tube and this phenomenon is called as Pulse Tube action.
31 1 Earlier Leture In te earlier leture, we ave seen a Pulse Tube (PT) ryoooler in wi te meanial displaer is removed and an osillating gas flow in te tin walled tube produes ooling. Tis gas tube is alled
More informationIsothermal vs. adiabatic compression
Isothermal vs. adabatc comresson 1. One and a half moles of a datomc gas at temerature 5 C are comressed sothermally from a volume of 0.015 m to a volume of 0.0015 m. a. Sketch the rocess on a dagram and
More informationME 300 Exam 2 November 18, :30 p.m. to 7:30 p.m.
CICLE YOU LECTUE BELOW: Frst Name Last Name 1:3 a.m. 1:3 p.m. Nak Gore ME 3 Exam November 18, 14 6:3 p.m. to 7:3 p.m. INSTUCTIONS 1. Ths s a closed book and closed notes examnaton. You are provded wth
More informationMath 324 Advanced Financial Mathematics Spring 2008 Final Exam Solutions May 2, 2008
Mat 324 Advanced Fnancal Matematcs Sprng 28 Fnal Exam Solutons May 2, 28 Ts s an open book take-ome exam. You may work wt textbooks and notes but do not consult any oter person. Sow all of your work and
More information8-4 P 2. = 12 kw. AIR T = const. Therefore, Q &
8-4 8-4 Air i compreed teadily by a compreor. e air temperature i mataed contant by eat rejection to te urroundg. e rate o entropy cange o air i to be determed. Aumption i i a teady-low proce ce tere i
More informationHandout 12: Thermodynamics. Zeroth law of thermodynamics
1 Handout 12: Thermodynamics Zeroth law of thermodynamics When two objects with different temperature are brought into contact, heat flows from the hotter body to a cooler one Heat flows until the temperatures
More informationThermodynamics. AP Physics B
Thermodynamics AP Physics B ork done by a gas Suppose you had a piston filled with a specific amount of gas. As you add heat, the temperature rises and thus the volume of the gas expands. The gas then
More informationCHEMISTRY Midterm #2 answer key October 25, 2005
CHEMISTRY 123-01 Mdterm #2 answer key October 25, 2005 Statstcs: Average: 70 pts (70%); Hghest: 97 pts (97%); Lowest: 33 pts (33%) Number of students performng at or above average: 62 (63%) Number of students
More informationFeedforward neural network. IFT Réseaux neuronaux
Feedforard neural netork IFT 725 - Réseau neuronau Septemer Astrat6, 22 ugolaroelle@userrookeasepte verste de Serrooke Sep Mat for my sldes Feedforard neural netork ARTIFICIAL NEURON aroelle@userrookea
More informationCarnot Cycle - Chemistry LibreTexts
CARNOT CYCLE The Carnot cycle has the greatest efficiency possible of an engine (although other cycles have the same efficiency) based on the assumption of the absence of incidental wasteful processes
More informationMAE140 - Linear Circuits - Fall 13 Midterm, October 31
Instructons ME140 - Lnear Crcuts - Fall 13 Mdterm, October 31 () Ths exam s open book. You may use whatever wrtten materals you choose, ncludng your class notes and textbook. You may use a hand calculator
More informationtechnische universiteit eindhoven Analysis of one product /one location inventory control models prof.dr. A.G. de Kok 1
TU/e tehnshe unverstet endhoven Analyss of one produt /one loaton nventory ontrol models prof.dr. A.G. de Kok Aknowledgements: I would lke to thank Leonard Fortun for translatng ths ourse materal nto Englsh
More informationChemistry 163B Free Energy and Equilibrium E&R ( ch 6)
Chemstry 163B Free Energy and Equlbrum E&R ( ch 6) 1 ΔG reacton and equlbrum (frst pass) 1. ΔG < spontaneous ( natural, rreversble) ΔG = equlbrum (reversble) ΔG > spontaneous n reverse drecton. ΔG = ΔHΔS
More informationScientific Research of the Institute of Mathematics and Computer Science 1(11) 2012, 23-30
Please te ts artle as: Grażyna Kałuża, Te numeral soluton of te transent eat onduton roblem usng te latte Boltzmann metod, Sentf Resear of te Insttute of Matemats and Comuter Sene,, Volume, Issue, ages
More informationPulse Coded Modulation
Pulse Coded Modulaton PCM (Pulse Coded Modulaton) s a voce codng technque defned by the ITU-T G.711 standard and t s used n dgtal telephony to encode the voce sgnal. The frst step n the analog to dgtal
More informationAdvanced Circuits Topics - Part 1 by Dr. Colton (Fall 2017)
Advanced rcuts Topcs - Part by Dr. olton (Fall 07) Part : Some thngs you should already know from Physcs 0 and 45 These are all thngs that you should have learned n Physcs 0 and/or 45. Ths secton s organzed
More informationChapter 8 Thermodynamic Relations
Chapter 8 Thermodynami Relations 8.1 Types of Thermodynami roperties The thermodynami state of a system an be haraterized by its properties that an be lassified as measured, fundamental, or deried properties.
More informationA Theorem of Mass Being Derived From Electrical Standing Waves (As Applied to Jean Louis Naudin's Test)
A Theorem of Mass Beng Derved From Eletral Standng Waves (As Appled to Jean Lous Naudn's Test) - by - Jerry E Bayles Aprl 5, 000 Ths Analyss Proposes The Neessary Changes Requred For A Workng Test Ths
More informationLecture #4 Capacitors and Inductors Energy Stored in C and L Equivalent Circuits Thevenin Norton
EES ntro. electroncs for S Sprng 003 Lecture : 0/03/03 A.R. Neureuther Verson Date 0/0/03 EES ntroducton to Electroncs for omputer Scence Andrew R. Neureuther Lecture # apactors and nductors Energy Stored
More informationClass: Life-Science Subject: Physics
Class: Lfe-Scence Subject: Physcs Frst year (6 pts): Graphc desgn of an energy exchange A partcle (B) of ass =g oves on an nclned plane of an nclned angle α = 3 relatve to the horzontal. We want to study
More informationCharged Particle in a Magnetic Field
Charged Partle n a Magnet Feld Mhael Fowler 1/16/08 Introduton Classall, the fore on a harged partle n eletr and magnet felds s gven b the Lorentz fore law: v B F = q E+ Ths velot-dependent fore s qute
More informationElectrochemistry Thermodynamics
CHEM 51 Analytcal Electrochemstry Chapter Oct 5, 016 Electrochemstry Thermodynamcs Bo Zhang Department of Chemstry Unversty of Washngton Seattle, WA 98195 Former SEAC presdent Andy Ewng sellng T-shrts
More informationTP A SOLUTION. For an ideal monatomic gas U=3/2nRT, Since the process is at constant pressure Q = C. giving ) =1000/(5/2*8.31*10)
T A SOLUTION For an deal monatomc gas U/nRT, Snce the process s at constant pressure Q C pn T gvng a: n Q /( 5 / R T ) /(5/*8.*) C V / R and C / R + R 5 / R. U U / nr T (/ ) R T ( Q / 5 / R T ) Q / 5 Q
More informationChapter 20 The First Law of Thermodynamics
Chapter he Frst aw o hermodynamcs. developng the concept o heat. etendng our concept o work to thermal processes 3. ntroducng the rst law o thermodynamcs. Heat and Internal Energy Internal energy: s the
More informationand Statistical Mechanics Material Properties
Statstcal Mechancs and Materal Propertes By Kuno TAKAHASHI Tokyo Insttute of Technology, Tokyo 15-855, JAPA Phone/Fax +81-3-5734-3915 takahak@de.ttech.ac.jp http://www.de.ttech.ac.jp/~kt-lab/ Only for
More information4.2 Chemical Driving Force
4.2. CHEMICL DRIVING FORCE 103 4.2 Chemcal Drvng Force second effect of a chemcal concentraton gradent on dffuson s to change the nature of the drvng force. Ths s because dffuson changes the bondng n a
More informationORDINARY DIFFERENTIAL EQUATIONS EULER S METHOD
Numercal Analss or Engneers German Jordanan Unverst ORDINARY DIFFERENTIAL EQUATIONS We wll eplore several metods o solvng rst order ordnar derental equatons (ODEs and we wll sow ow tese metods can be appled
More informationCHEM 305 Solutions for assignment #4
CEM 05 Solutions for assignment #4 5. A heat engine based on a Carnot cycle does.50 kj of work per cycle and has an efficiency of 45.0%. What are q and q C for one cycle? Since the engine does work on
More informationChapter 3 Thermoelectric Coolers
3- Capter 3 ermoelectric Coolers Contents Capter 3 ermoelectric Coolers... 3- Contents... 3-3. deal Equations... 3-3. Maximum Parameters... 3-7 3.3 Normalized Parameters... 3-8 Example 3. ermoelectric
More information