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

Energy added Temperature Energy added Temperature H2OH2O Copper (not to scale with each other) Initial 0°C MP 100°C BP 1085°C MP 2839°C BP

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

ΔEth copper + ΔEth water = 0 −+ mcΔT copper + mcΔT water = 0 (1kg)c(T f -T i ) copper + (1kg)c(T f -T i ) water = 0 (1kg)(0.386kJ/kgC)(T f -95°C) copper + (1kg)(4.18kJ/kgC)(T f -5°C) water = 0 Solve for T f : T f = 12.6°C Important lines: always show these on the board and on exams. Look up in table

Energy added 0°C MP 100°C BP Temperature Energy added 1085°C MP (2839°C) BP Temperature H2OH2O Copper (not to scale with each other) Initial (Freezing point of copper: 1085°) Think about it: Why did the water only increase ~8°C, when the copper decreased 82°C !???? Fina l

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) Energy added (J) solid liquid gas ∆T ∆E C = [C] = J/K

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

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

Temperature (K) Energy added (J) solid liquid gas TbTb ∆E 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) Energy added (J) solid liquid gas TmTm ∆E 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

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? E bond   mlml   mlml   mlml   mlml Heat A) B) C) D)

Which Energy System? How much energy does it take to completely melt 2.5kg of ice initially at 0°C? E bond   mlml   mlml   mlml   mlml Heat A) B) C) D) ΔE bond = Q ΔE bond = FREEZING! IMPOSSIBLE!

B) Was the correct one, let’s solve it. E bond   mlml Heat ++ ΔE bond = Q How much energy does it take to completely melt 2.5kg of ice initially at 0°C? ±l ΔmΔH m l = Q Choose the plus! (2.5kg)(333.5kJ/kg) = Q kJ = Q = Energy needed to melt 2.5kg of ice! Look up in table

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) Energy added (J) TbTb E bond   mgmg E therm al T   T   E bond   mlml E therm al T  

Temperature (K) Energy added (J) TbTb E bond   mgmg E therm al T   T   E bond   mlml E therm al T   ΔEth =Q 1 ΔEbond =Q 2 ΔEth =Q 3 ΔEbond =Q 4 ΔEth =Q 5 Q1Q1 Q2Q2 Q4Q4 Q5Q5 Q3Q3

E bond   mgmg E therm al T   T   E bond   mlml E therm al T   ΔEth =Q 1 ΔEbond =Q 2 ΔEth =Q 3 ΔEbond =Q 4 ΔEth =Q 5 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

Start with Q 1 E therm al T   ΔEth =Q 1 mcΔT=Q1 (3)(.141kJ/kgK)(234°K-200°K)=Q1 Melting point =234°K; Boiling Point=630°K Specific heats: (s)=.141 kJ/kgK, (l)=.140, (g)=.103 Temperature (K) Energy added (J) 630°K 234°K 14.4kJ=Q1

Once we figure out the phase we can then draw the energy system diagram… I figured out the Qs for you: Q 1 =14.4kJ, Q 2 =0.423kJ, Q 3 =166kJ, Q 4 =0.42kJ Can you tell me what the final state is (remember we only have 20kJ)???? Temperature (K) Energy added (J) TbTb Q1Q1 Q2Q2 Q4Q4 Q5Q5 Q3Q3 The mercury has enough energy to get into the liquid state but not through it!

Correct energy system diagram: E therm al E bond   mlml E therm al T   ΔEth+ ΔEbond + ΔEth = Q T  

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 EversonSection 1 Amandeep: 11:00AM Roesller Section 2 Yi: 1:40PM Everson Section 3 Chun-Yen: 1:40PM Roesller Section 4