Transcript Slide 1
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. 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 Copper Temperature H2O Temperature Initial 2839°C BP Initial 100°C BP 1085°C MP 0°C MP Energy added (not to scale with each other) Energy added 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 Etherm Etherm I: T=95°C F: T=?°C − al T T al I: T=5°C F: T=?°C + ΔEthcopper + ΔEthwater = 0 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 Think about it: Why did the water only increase ~8°C, Temperature when the copper decreased 82°C !???? Copper H2O Temperature Initial Fina l 100°CBP (2839°C)BP Initial Fina l 1085°CMP 0°C MP Energy added (not to scale with each other) Energy added (Freezing point of copper: 1085°) Why did the water only increase ~8°C, when the copper decreased 82°C !???? A. The amount of heat leaving the copper was greater than the amount of heat leaving the water B. The amount of heat leaving the water was greater than the amount of heat leaving the copper C. The specific heat of the water is greater than the copper D. Some of the heat leaving the copper went to the environment E. Magic Heat capacity in Three-phase Model of Matter Temperature (K) Q C= T gas [C] = J/K liquid ∆T solid ∆E Energy added (J) Which object has the higher heat capacity? B A ∆T ∆T ∆E ∆E Heat capacity in Three-phase Model of Matter Temperature (K) Tb Q C= T 0 gas liquid solid ∆E Energy added (J) 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) 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) 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! Which Energy System? How much energy does it take to completely melt 2.5kg of ice initially at 0°C? Heat A) B) Ebond Ebond ml ml Heat C) Ebond ml Heat D) Ebond ml Which Energy System? How much energy does it take to completely melt 2.5kg of ice initially at 0°C? B) Heat A) Ebond Ebond FREEZING! ml Heat - ml - + ΔEbond = Q ΔEbond = Q D) Heat C) + Ebond IMPOSSIBLE! m Ebond IMPOSSIBLE! ml l + - ΔEbond = Q + ΔEbond = 0 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! Today in Lab you will do the DREADED… ICED TEA PROBLEM!!! 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 Temperature (K) Tb Energy added (J) Etherm al T Ebond ml Etherm al T Ebond mg Etherm al T Temperature (K) Q5 Q4 Tb Q2 Q1 Q3 Energy added (J) Etherm al T ΔEth =Q1 Ebond ml Etherm al T ΔEbond =Q2 ΔEth =Q3 Ebond mg ΔEbond =Q4 Etherm al T ΔEth =Q5 Etherm al T ΔEth =Q1 Ebond ml ΔEbond =Q2 Etherm al T ΔEth =Q3 Ebond mg ΔEbond =Q4 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 Etherm al T ΔEth =Q5 Start with Q1 Melting point =234°K; Boiling Point=630°K Specific heats: (s)=.141 kJ/kgK, (l)=.140, (g)=.103 Temperature (K) Etherm al T 630°K 234°K ΔEth =Q1 Energy added (J) mcΔT=Q1 (3)(.141kJ/kgK)(234°K-200°K)=Q1 14.4kJ=Q1 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 Tb Q1 Q2 Q3 Energy added (J) Once we figure out the phase we can then draw the energy system diagram… Correct energy system diagram: Etherm al T Ebond ml Etherm al T ΔEth+ ΔEbond + ΔEth = Q Other stuff you will see soon… 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 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