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Lecture # 2

Lecture # 2. Cassandra Paul Physics 7A Summer Session II 2008. Last time: Made Energy Diagrams Drew Three Phase Diagrams This Time: Lets get quantitative!. Example: two substances.

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Lecture # 2

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  1. Lecture # 2 Cassandra Paul Physics 7A Summer Session II 2008

  2. Last time: • Made Energy Diagrams • Drew Three Phase Diagrams • This Time: • Lets get quantitative!

  3. Example: two substances. Example: 1kg of Copper at 95°C is dropped into an insulated bucket of 1kg of water at 5°C. What is the temperature of the water after the system comes to equilibrium? To set up the problem: • Define system • Define interval • What indicators are changing? • What energies are changing? • How are they changing? • Is this an open or closed system? • What (if anything) is entering/leaving the system? • Write Energy Conservation Equation

  4. Initial Initial Copper Temperature H2O Temperature 2839°C BP 100°C BP 1085°C MP 0°C MP Energy added Energy added (not to scale with each other)

  5. Example: Copper at 95 degrees C is dropped into a cup of water at 5 degrees C. How can we model the Energy Transfer? System: Copper and Water Initial: Copper at 95°C; Water at 5°C Final: Equilibrium (same temp: 5°C<Tequl<95°C) Copper Water Ethermal Ethermal   T  T  I: T=95°C F: T=?°C I: T=5°C F: T=?°C − + ΔEthcopper + ΔEthwater = 0

  6. Important lines: always show these on the board and on exams. ΔEthcopper + ΔEthwater = 0 − + mcΔTcopper + mcΔTwater = 0 Look up in table (1kg)c(Tf-Ti)copper + (1kg)c(Tf-Ti)water = 0 (1kg)(0.386kJ/kgC)(Tf-95°C)copper + (1kg)(4.18kJ/kgC)(Tf-5°C)water = 0 Solve for Tf: Tf = 12.6°C

  7. Initial Initial Final Final Temperature Think about it: Why did the water only increase ~8°C, when the copper decreased 82°C !???? Copper H2O Temperature (2839°C)BP 100°CBP 1085°CMP 0°C MP Energy added Energy added (Freezing point of copper: 1085°) (not to scale with each other)

  8. Why did the water only increase ~8°C,when the copper decreased 82°C !???? • The amount of heat leaving the copper was greater than the amount of heat leaving the water • The amount of heat leaving the water was greater than the amount of heat leaving the copper • The specific heat of the water is greater than the copper • Some of the heat leaving the copper went to the environment • Magic

  9. Heat capacity in Three-phase Model of Matter Temperature (K) C = [C] = J/K gas liquid ∆T solid ∆E Energy added (J)

  10. Which object has the higher heat capacity? B ∆T A ∆T ∆E ∆E

  11. Heat capacity in Three-phase Model of Matter Temperature (K) C = 0 gas Tb liquid solid ∆E Energy added (J)

  12. Heat of vaporazation : ∆H the amount of energy per unit mass (or unit mole) required for a substance to change its phase from liquid to gas or vice versa Temperature (K) gas Tb liquid solid ∆E Energy added (J)

  13. Heat of melting : ∆H the amount of energy per unit mass ( or unit mole) required for a substance to change its phase from solid to liquid or vice versa Temperature (K) gas liquid Tm solid ∆E Energy added (J)

  14. Let’s try a quantitative example: • How much energy does it take to completely melt 2.5kg of ice initially at 0°C? • Draw a three-phase diagram • Draw an energy System Diagram • Calculate!

  15. Which Energy System?How much energy does it take to completely melt 2.5kg of ice initially at 0°C? B) Heat Heat A) Ebond Ebond   ml ml   D) Ebond Heat C) Ebond   ml  ml 

  16. Which Energy System?How much energy does it take to completely melt 2.5kg of ice initially at 0°C? B) Heat Heat A) Ebond Ebond FREEZING!   ml ml   + + - - ΔEbond = Q ΔEbond = Q D) Ebond Heat C) Ebond  IMPOSSIBLE! IMPOSSIBLE!  ml  ml  + - + ΔEbond = Q ΔEbond = 0

  17. B) Was the correct one, let’s solve it. How much energy does it take to completely melt 2.5kg of ice initially at 0°C? Heat Ebond + + ΔEbond = Q  ml  ±lΔmΔHml = Q Look up in table Choose the plus! (2.5kg)(333.5kJ/kg) = Q 833.75kJ = Q = Energy needed to melt 2.5kg of ice!

  18. Today in Lab you will do the DREADED… ICED TEA PROBLEM!!!

  19. Let’s get ready. Let’s not talk Tea yet, let’s start easy: Example: Lets say you have a 3kg of mercury that you want to add 20kJ of heat to. The substance starts in the solid phase at a temperature of 200°K. Find the final temperature of the mercury. • Draw a three-phase diagram • Draw an energy interaction diagram

  20. Temperature (K) Tb Ebond Ethermal Ethermal Ebond Ethermal      ml mg   T  T  T  Energy added (J)

  21. Q5 Q4 Q3 Q2 Q1 Temperature (K) Tb Ethermal Ebond Ethermal Ebond Ethermal  T      ml mg   T  T  ΔEth =Q3 ΔEbond =Q2 ΔEbond =Q4 ΔEth =Q5 ΔEth =Q1 Energy added (J)

  22. Ethermal  T  ΔEth =Q3 ΔEbond =Q2 ΔEbond =Q4 ΔEth =Q5 ΔEth =Q1 Example: You have exactly 20kJ of heat to add to a 3kg block of mercury initially at 200°K. You want to know what the final temperature will be when you have used up all of your energy. Melting point =234°K; Boiling Point=630°K Specific heats: (s)=.141 kJ/kgK, (l)=.140, (g)=.103 Heat of melting: 11.3kJ/kg Heat of Vaporization: 296kJ/kg Ebond Ethermal Ebond Ethermal     ml  mg  T  T 

  23. Start with Q1 Melting point =234°K; Boiling Point=630°K Specific heats: (s)=.141 kJ/kgK, (l)=.140, (g)=.103 Ethermal Temperature (K)  T  630°K 234°K ΔEth =Q1 Energy added (J) mcΔT=Q1 (3)(.141kJ/kgK)(234°K-200°K)=Q1 14.4kJ=Q1

  24. Once we figure out the phase we can then draw the energy system diagram… I figured out the Qs for you: Q1=14.4kJ, Q2=0.423kJ, Q3=166kJ, Q4=0.42kJ Can you tell me what the final state is (remember we only have 20kJ)???? The mercury has enough energy to get into the liquid state but not through it! Temperature (K) Q5 Q4 Q3 Q2 Tb Q1 Energy added (J)

  25. Correct energy system diagram: Ethermal Ethermal   T  T  Ebond  ΔEth+ ΔEbond + ΔEth = Q ml 

  26. Other stuff you will see soon…

  27. Power • Power is a rate of energy transfer. • Amount of energy a system receives per unit time • Power is measured in Watts=Joules/Second Chemical Systems • You will be applying energy interaction diagrams to chemical processes • Breaking bonds: bond energy increases

  28. DL sections • Swapno: 11:00AM Everson Section 1 • Amandeep: 11:00AM Roesller Section 2 • Yi: 1:40PM Everson Section 3 • Chun-Yen: 1:40PM Roesller Section 4

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