An attempt to take ideas about Energy and Conservation from

1. 1 Defining Energy Across Disciplines An attempt to take ideas about Energy and Conservation from several disciplines and form an integrated organization.

2. 2 What Causes the Confusion About ENERGY?

3. Statements about energy from various sources: 3 1. There are many forms of energy: wind, water, sun, potential, chemical, electrical, nuclear, light, motion, heat, mechanical, internal, external, atomic, nuclear, molecular, gravitational, bond energy, bonding energy, sound energy etc. Energy can change from one form to another. Energy is Conserved 2. PotentialEnergy can be stored in an atom or nucleus or molecule or by gravity or in a battery” 3. Work: Force times Distance= KE= 1/2 mv2 PE= mgh 4. In an exothermic reaction heat is given off by a chemical reaction. The energy comes from substances involved. 5. Electricity often turns into heat 6. A windmill can turn wind energy into electrical energy 7. Nuclear energy is very large and comes from the nucleus! Both fission and fusion produce energy. 8. Potential energy is in a form you often cannot see.

4. Where does the confusion about energy come from? 1. Why is it “work” to push on a wall? We get tired! Books say no work is done! 2. Do these topics have any conceptual similarities as we teach them. Biology (sun energy, cell energy, chemical energy, potential energy), Physics (mechanics:gravitational potential energy and kinetic energy, frictional heat energy), thermodynamics: heat transfer, nuclear and electrical energy), Chemistry (thermochemistry, bond energy, nuclear energy, heat of reaction, heat of fusion and evaporation and in all areas The Law of Conservation of Energy 3. How do you describe how energy is stored in molecules and bonds and the nucleus? 4. In an exothermic reaction where exactly does the heat come from? At the atomic level what is the same for all exothermic reactions? 5. What is the connection between mass and energy? 6. How is electrical energy related to the equations of Mechanics in Physics? 7. How are all the different Forms of Energy..mentioned in most texts …..related? Are there really so many different forms of energy? 8. How is Work done actually related to PE, KE, heat, electricity ? 9. How is Binding Energy related to energy given off in nuclear reactions? Is Binding Energy a form of energy? 10. What do all the forms of Potential Energy have in common? 4

5. …………………………………Some confusing definitions………………………….. Definition: Energy is the capacity of a physical system to perform work. Energy exists in several forms such as heat, kinetic or mechanical energy, light, potential energy, electrical, or other forms. Energy Definition: Energy may be defined as the ability to do work. It is a scalar physical quantity. Although energy is conserved, there are many different types of energy, such as kinetic energy, potential energy, light, sound, and nuclear energy … a form of power such as , electricity, heat or light that is used for moving things around…work. …the power that is present in all physical things and that can be changed into something such as heat, movement or light. Electrical energy is the movement of electrons. That is kinetic energy. The voltage in an electrical circuit is the potential energy that can start electrons moving. Electrical forces cause the movement to occur. Chemical energy is potential energy until the chemical reaction puts atoms and molecules in motion. Heat energy (KE) is often the result of a chemical reaction. Confusion also comes from using “h” instead of h for example in “mgh” and Vlost instead of V in V=IR and energy = VIt etc etc 5

6. 6 The following are examples using consistent concepts of KE and PE throughout science

7. Definitions that prove consistent throughout the curriculum: 7 1. Work is the Transfer of Energy from one form or condition to another. Mass=Energy is always Conserved in all interactions 2. There are 3 forms of energy: kinetic, potential and electromagnetic. (Mass is converted into Energy via E= ΔmC2) 3. Kinetic energy may assume the form of Macroscopic 1/2 MV2 or Microscopic Heat. (Heat in or out can be calculated) 4. Potential energy is associated with the relative positions between bodies. Most common: Since all molecules and atoms and opposite charged ions attract each other …Potential Energy goes UP when attracting bodies move away from each other and goes down when they approach…at least until they get to their “bond distance” in which case repulsion becomes greater than attraction. Ultimately all losses PE involve mass being converted into energy. 5. Electromagnetic energy refers to any of the electromagnetic spectrum.

8. 8 Examples of sources of potential energy in systems 1. Gravitational: Masses attract the earth…when moved further away PE goes up. 2. Electrostatic A. Electrical: When electrons are moved from protons PE goes up; when electrons are moved closer together PE goes up. …as in charging a capacitor etc. B. Chemical: Anytime PE goes up: Sum of attracting bodies get further apart and repelling bodies get closer together. It often is difficult to see all of this but if you analyze simple systems you can see it at work. 4. Nuclear: In any exothermic nuclear reaction…be it fusion or fission it can be shown that potential energy goes down because attracting bodies are getting closer…the nuclear force of attraction is much larger than the electrostatic forces of repulsion. Potential energy can be shown in graphical form for ALL changes….not just chemical reactions.

9. 9 Energy Conversions

10. Kinetic PotentialElectromagnetic A and B C D 10 Macroscopic KE A 1/2 M V2 C D E=MC2 For now EM is left out of the scheme. B Heat Microscopic KE Within each category energy can go from one body to another

11. Conversions within a category 1. Potential Energy: a. Chemical PE -----Electrical PE: Battery charging capacitor b. Gravitational PE-----Spring PE: A mass set atop a spring compresses it. c. Chemical PE -----Gravitational PE: Lift a box 2. Kinetic Energy a. Microscopic KE: Heat flows from hot to cold b. Macroscopic KE: Two billiard balls collide. 3. Electromagnetic Energy a. Fluorescence: UV changed to Visible light

12. Kinetic PotentialElectromagnetic A 1/2 M V2 C E=MC2 Within each category energy can go from one body to another Heat B 1. Expanding gases cool: B to C Compress gases: C to B 2. Water evaporates and cools: B to C Water condenses: C to B 3. Wax freezes on your finger: C to B Wax melts: B to C 4. Car brakes to a stop: A to B Engine accelerates car: B to A 5. Exothermic chem reaction: C to B Endothermic chem reaction: B to C 6. Bullet shot from gun: C to B & A Car bounces down on springs: A to C 7. Car coasts up a hill: A to C Car coasts down a hill: C to A 8. Driver brakes down a hill: C to A & B Car brakes going up a hill: A to B & C 9. Pendulum: C to A and A to C Eventually: all ends up as B 10. Drop a book on the table: C to A to B Pour hot water into cold: B to B 11. Lift a rock:Cbody to Cgravity Push on the wall: C to B 12. Pool balls collide: Aball 1 to Aball 2 A mass compresses a spring: Cgravityto Cspring 13. Battery lights light bulb: C to B etc Electric motor: Celectrical to A and B E=MC2

13. 11 The Law of Conservation of Energy Energy is never destroyed; it just changes forms. (JUST CONSIDERPOTENTIAL AND KINETIC ) PE + KE = CONSTANT Example: 5 + 5 = 10 6 + 4 = 10 8 + 2 = 10 If PE gets bigger then KE gets smaller Boils down to: KE + PE = 0

14. 12 A good definition below The bonds between atoms are the source of all chemical or covalent type have relatively low potential chemical energy, as it requires a large amount of outside energy simply to break the bonds. Weaker bonds, like those of the van der Waal type, have more potential chemical energy, as they require relatively little energy to break. Energy is released when these bonds form between atoms, and the energy in chemical reactions is not created or destroyed. This means that chemical reactions may be analyzed like mathematical equations. Since a strong bond requires a large amount of energy to break, this must mean that when that same bond forms, much energy is released. By the same logic, when a weak bond forms, relatively little energy is released.

15. 13 Bond lengths and Reactions

16. 14 Hydrogen gas + chlorine gas ==== hydrogen chloride gas • + 199 > 2 ( 127 ) bond distances pm • 273 > 254 H + H H + H + Cl + Cl Cl + Cl Exo Endo H2 H2 ……………………………………………… ExoEndo PE average PE Cl2 Cl2 …………………………….. Exo Exo Endo Coordinate 2H = H2 + heat 2Cl = Cl2 + heat 2H + 2Cl = 2HCl + heat H2 + Cl2 = 2HCl + heat The Net reaction is Exothermic because the bond distances between hydrogen and chlorine are smaller than between hydrogen-hydrogen or chlorine-chlorine…PE is down.

17. 15 Bond lengths shorten because of stronger bonds

18. 16 PE + KE = 0 A Simple Chemical Reaction C(g) + 02 (g) ------ CO2(g) + heat 121 > 113 PE + KE PE + KE O + O O C O C + O + O PE Atoms are closer together O2 + C CO2 ……………………. Coordinate

19. Combustion of Methane 17 1 CH4 + 2 O2 2 H2O + CO2 +2632 kj/mole ch4 -3338 kj/mole methane 916 > 848 992 kj/mole O2 ENERGY 2 O2 120 pm 1486 kj/mole CO2 1640 kj/mole 116 pm Ch4 1852 kj/mole methane 2 H2O 109 pm bond length 96 pm

20. 19 Some examples of application of a uniform outline to a variety of problems

21. 20 Heat energy going from one object to another.

22. Thermochemistry 21 For example: Suppose 35ml of 80C water is mixed with 19ml of 24C water. What is the resulting temperature? (60 C) Heat energy is transferred from hot object to a cooler one. COLD HOT Energy Conserved Kinetic energy falls Kinetic energy rises 1. Hot water is mixed with cold waterH = MC  T= KE change Δ Hhot water + Δ Hcold water = ZERO [ (35g)(1cal/gC)(T2 - 80C)] + [(19g)(lcal/gC)(T2 - 24C)] = 0 T2= 60 C

23. 22 Determining the Specific Heat of Iron2 Heat a 1.0 kg mass in water to about 80 c. Place 200 ml tap water in a styrafoam cup, measure temp. and insert mass. Measure T2: T1 Water = 22.2C 0 = heat lost by Fe + heat gained by water heat Kg 0 = mc Δ t iron + mc Δ t water 0 = (1000g) (CFe) (T2-81.2C) + (200g) (1cal/gC) (T2-22.2C) 200.0 ml water Insert the final temperature and solve for CFe T1 iron = 81.2 C Notice that in the following problems that delta T can always be interpreted as T2 – T1 and no confusion of what to do.

24. 23 Energy transfers between potential and kinetic (heat)

25. Some examples 24 1. Going from solid oxygen to nucleons. 6 Nucleons Δ PE + Δ KE = 0 up down 5 Ions Exothermic Endothermic 4 PE Getting further apart Atoms Getting closer together 3 Δ PE + Δ KE = 0 down up Gas 2 Liquid 1 solid 1-2. O2(s) + energy --- O2(L) 2-3 . O2(L) + energy ----- O2( g ) 3-4. O2(g) + energy ---- 2 O(g) 4-5 O(atom) + Energy ----- O(ion) + 8 electrons5-6 O(ion) + Energy ---- 8 protons + 8 neutrons

26. 25 Ice…to....nucleons thru heating Ionize atoms O Nucleons Heat atoms O Heat ionz Decompose water * O Heat Vapor Vaporize Water O nuceons ENERGY O Heat Water Heat Ice Melt Ice * O O ions Total Energy * O * atoms gas O * KE liquid solid PE *= phase change

27. 26 Demonstration Have a student insert one finger into 60F liquid wax and another into 60F water. Essentially both will feel the same. Now remove both fingers: the wax gets hotter as it freezes and the water gets cooler as it evaporates. Water Gas Wax Liquid PE PE Liquid solid Heat Absorbed Heat given off

28. Ice in Water 27 ICE Water heat 0 C KE water PE ice Heat from the water is used to (1) melt the ice and (2) warm the resulting ice water.

29. 18 Burn methanol with oxygen to make carbon dioxide and water + heat 96 pm 108 pm 95 pm 4 3 2 2 143 pm = 113 pm 121 pm 1850 total to 1672 total = atoms closer together and PE down 0 = KE + PE

30. 28 Liquid Hot Wax Solid Hot Wax Water Thermochemistry 20g of wax at 80C in mixed with 50ml of water at 25C. What is the final temp? Fd =ΔPE +ΔKE (macroscopic) + Heat 0 = ( - ) + 0 + + ( - ) + ( + ) 0 = (Mass x H fusion wax) + [(Mwax x Cwax T) + (M x Cwater T)] 0= [(20g)( - 20cal/g)] + [(20g)( 0.1cal/gC)(T2- 80C)] + [(50g)(1cal/gC)(T2- 25C)] Heat of fusion is (-) PE goes down cools heats 0 = ( - 400) + 2T2+ ( - 160 ) + ( + 50T2) +( -1250 ) 0 = - 1810 + 52 T2T2 = 35 C no confusion about t

31. 29 Dissolving Some compounds are exothermic when dissolved, some are endothermic when dissolved and some don’t seem to affect water temperature when dissolved

32. 30 Dissolving of Ammonium Chloride Endothermic +1 -1 NH4 +1 (g) + Cl-1 (g) PE HYDRATION PE DISSOCIATION (lattice energy) - 307 kj/mole - 381 kj/mole ENERGY NH4 +1 (aq) + Cl-1 (aq) + 705 kj/mole 0 0 0 0 +1 1 - 0 0 0 0 NH4Cl (s) +1 -1 H = +18 KJ/MOLE + Note: H20 = 0 = 0 -

33. Dissolving of Sodium Hydroxide exothermic 31 -1 +1 Na +1 (g) + OH -1 (g) PE PE DISSOCIATION ENERGY HYDRATION NaOH (S) +1 -1 Na+1 (aq) + OH-1 (aq) o o o H = - 45 kj/mol o -1 o +1 o o o o o o o

34. 32 Dissolving of Sodium Chloride very slightly exothermic -1 +1 Na+1 (g) + Cl-1 (g) PE -406 = - (406 +363) = -788 kj/mol +787 kj/mol HYDRATION Energy DISSOCIATION o o o o -1 o -1 +1 +1 o o o o o o NaCl (s) Na+1 (aq) + Cl-1 (aq) Delta H = -1 kj/mol

35. Thermochemistry 33 Example: Add 5.0 g of ice at 0C to 100ml of water at 25C. What is the final temp? Δ H = mC(specific heat) Δ T = ΔKE ΔH = m (Hf) ….heat of fusion = ΔPE H25C water=(100g)(1cal/gC)(T2-25C) Δ Hice =(5.0g) (+80 cal/g) Hice water=(5.0g)(1cal/gC)(T2-0C) (Use +80 when PE rises and -80 when PE falls; when a substance freezes the attracting particles get closer together and PE goes down. This also happens with water but because of hydrogen bonding its less obvious to see….ice floats) warm water cools cool water heats ice melts Δ H total =(100g)(1cal/gC)(T2-25C) + 5.0g)(1cal/gC)(T2-0C)+ 5.0g) (+80 cal/g) = 0 KE goes down + KE goes up + PE goes up = 0 ΔKE + Δ KE + Δ PE = 0 ( - ) ( + ) ( + ) Conservation of Energy 100T2 - 2500 + 5T2 - 0 + 400 = 0 Final Temp = 20C

36. Exothermic Reactions 34 A. 2.0g (0.050mol) of Na0H are put into 300ml of water at 22C and the temperature rises 2.5C. What happened to the Potential energy of the Na0H? Fd = ΔPE + Δ KE (macroscopic) + Heat 0 = ΔPE + 0 + heat 0= Δ PE + mCΔT 0= Δ PE + (300g)(1cal/gC)(+2.5C) chemical PE is lost and KE transferred to the water Δ PE= -750 calories The reaction is EXOTHERMIC! B. 0.10mol of Ammonium Chloride dissolves in 250ml of water and the temp drops from 24C to 21C. What is the H of the reaction? Fd = Δ PE + Δ KE (macroscopic) + Heat 0 = ΔPE + 0 + mC Δ T 0= Δ PE + (5g)(1cal/gC)(21C-24C) KE is lost by the water and converted to PE in the chemical system 0= Δ PE + (-15cal) Δ H = + 15cal or + 150cal/mol ENDOTHERMIC!

37. 35 Other examples of energy transformations and the conservation of energy found in various science topics

38. Line Spectra : ΔPE + Δ Electromagnetic energy = 0 down up Energy levels in a hydrogen atom 36 0= ΔPE + Δ EM energy “ As the attracting electron gets further from the nucleus the PE goes up” Ionization N=4 PE IR N=3 PE Visible light PE Average distance from nucleus N=2 UV N=1 0 PE

39. 37 Ionization Energy Calcium 2, 8, 2 Ionization Level Ionization energy increases as electrons closer to the nucleus are reached. Rise in PE corresponds to more work needed to be done to remove it. 1st e Potential Energy 1st Electron …Outermost 2nd Electron 3rd Electron

40. Mechanics in Physics 38 External Work Done on a Body changes to Kinetic Energy (macroscopic) Kinetic Energy (microscopic) Potential Energy

41. Mechanics 39 1. Work in mechanics (F d) is an usually is an instance where external energy is transferred to a system. 2. Most texts show equations like: Fd = KE or Fd=PE Fd=mgh Fd=1/2mv2 This leads to an incomplete understanding of just what is going on. The equations are really: External work Fd= mg Δh and Fd= Δ KE = 1/2mv22 -1/2mv12 3. A system that consistently works in mechanics is: Work = change in PE + change in KE + heat generated Fd = Δ PE + Δ KE (macroscopic) + Heat F Δ d = mg Δ h + (1/2 mv22 -1/2 mv12) + heat F Δ d = mg Δ h + (1/2 mv22 -1/2 mv12) + Ffrictiond Which is the same as 0 = F d + Δ PE + Δ KE + heat

42. Mechanics 40 Suppose a 2.0 kg block at rest slides down a frictionless ramp 25 cm high and out on a horizontal surface upon which there was a friction force of 4.0N. Where does the block stop? No external work is done. F Δ d = mg Δ h + (1/2 mv22- 1/2 mv12) + Ffrictiond 0 = ( -) + (0) - (0) + ( +) Suppose a 2.0 kg block at rest slides down a frictionless ramp 25 cm high and out on a horizontal surface upon which there was a friction force of 4.0N. Where does the block stop? No external work is done. Zero = 0 = [(2.0kg)(9.8N/kg)(-0.25m)] + [0 - 0] + [4.0N)(d)] d=+1.23m

43. 41 Mechanics Suppose YOU lifted a 3.0kg block vertically upward 9.0 meters and let it slide down an incline upon which the constant force of friction was 15N. What would be the speed of the block after it had slid a distance of 4.0 meters 15m down the incline and dropped a vertical distance of 3.0 meters? Only consider the starting and ending points involved F Δ d = mg Δ h + (1/2 mv22- 1/2 mv12) + Ffrictiond (+) (-) zero (+) (3.0)(9.8)(+9.0) = (3.0)(9.8)(-3.0) + ( 1/2 (3.0)( v22 ) - ( 0 ) + (15)(4.0) +265 J =- 88.2J + 1.5 V22+ +60J V2 = 14 m/s

44. 42 The curious example of taking energy from something by pushing on it. KE1 = 40 J Applied Force 4N Frictionless Motion 10 m F•D = Δ PE + Δ KE + heat From the dot product we get negative work…which \ means we are taking energy out instead of putting it in. 0 = - F•D + Δ PE + Δ KE + heat 0 = -[-FD] + 0 + Δ KE + 0 0= +(4)(10) + KE2 - KE1 0= +40 J + 0 - 40J 40 J of some form of energy must be created from the lost KE; we know we did “work” …in fact as much work is done stopping it as was done getting it going.

45. 43 The amount of work we did took energy from the block and therefore that amount of energy must be vented via heat from our bodies. This term does not appear in the equation for the same reason as when we push on a wall and get tired we are actually working on the air….its gaining energy…but this equation can’t be used to “measure” the energy gained by the air. We can calculate the energy gained by the air in the above case with reason. If you did 40 J of work lifting a rock you would do 40 J of work setting it gently back down. In setting it down the rock loses 40 J of PE and you lose 40 J of heat….explaining your sense of getting tired.

46. Electricity in Physics 44 Potential Energy changes to KE (macroscopic) KE (microscopic) Other form of PE

47. 45 Electricity: ΔV=IR and EL = IR Ohm’s Law Electric Field PE V1Q Δ VQ= Δ VIt ELQ=VIt=  PE PE V2Q KE or Heat - + R L=length of resistance: HEAT formed

48. 46 Suppose you touch a charged van der graaf at 50,000 volts And 2 x 10 -6 coulomb of charge flows thru your body. How many joules of heat is generated? V x q = 50,000 j/c x 2 x 10 -6 coul = 1 x 10 -1 j same as a kg falling 1 cm Electrical PE changes to electron KE which changes into heat.  PE electrical = PE gravitational = same amt of heat formedl • PE + KE electrons = 0 KE electrons + KE atoms = 0

49. 47 Electricity A 1.5v battery delivers 0.30amps to a toy car for 5.0 s and the 1.0kg car goes from rest to 2.0 m/s in that time. How much energy is “wasted” as heat? Fd = Δ PE + Δ KE (macroscopic) + Heat Change of electrical potential energy is: ΔVIt or( Power x time) ΔV is really Vlost in all equations using ohm’s law: V=IR F Δ d =ΔVIt + (1/2 mv22- 1/2 mv12) + Ffrictiond zero = Vlost I t + 1/2mV22 - 0 + heat 0 = (-1.5)(2.0)(5.0) + (0.5)(1.0)(3.0)2 - 0 + heat 0 = -15J + +4.5J + heat battery lost 15J car gained 4.5J gained heat 10.5J

50. 48 The Role of Mass “The Rest Mass particles have is simply the work done in separating them against their mutual attraction after they are created” Richard Feynman This implies that when particles that attract each other are moved closer and PE goes down…energy must be given off and MASS MUST BE LOST. In the case of burning a mol of carbon to carbon dioxide the amount of KE (= to loss of PE) is equivalent to 5x10-9g. In normal chemical reactions this mass loss is not measurable.