1. Dr. Joseph W. Howard ©Summer 2006 Kinetic & Potential Energy on the Nanoscale

2. Dr. Joseph W. Howard ©Summer 2006 Temperature Cool Block Warm Block

3. Dr. Joseph W. Howard ©Summer 2006 Thermal Energy Heat vs. Temperature. Heat is a form of Energy! Temperature is the measure of “degrees of hotness”

4. Dr. Joseph W. Howard ©Summer 2006 Energy Reminders Hmmmm… What kind of energy(ies) have we already talked about? Ah Ha! Potential energy (gravitational) Kinetic Energy (energy of motion)

5. Dr. Joseph W. Howard ©Summer 2006 Temperature All matter is composed of continuously jiggling atoms and molecules (solids, liquids, gases) The average jiggling (avg. KE) of all the individual particles of an object is related to “How Hot.”

6. Dr. Joseph W. Howard ©Summer 2006 Kinetic Energy on the Nanoscale thermal energy Baseball Looking at a tiny piece within the baseball.

7. Dr. Joseph W. Howard ©Summer 2006 Temperature Scales Based ONLY on atomic and molecular motion (Avg. KE) Fast jigglingHot! No upper limit Slow jigglingCooler NO jiggling Coldest Possible A lower limit Shocking!

8. Dr. Joseph W. Howard ©Summer 2006 Fahrenheit Temperature Scale Weatherman? 32 o F Water Freezes/Melts 212 o F Water Boils / Water Vapor Condenses

9. Dr. Joseph W. Howard ©Summer 2006 Celsius Temperature Scale 0oC0oC Water Freezes/Melts 100 o C Water Boils / Water Vapor Condenses

10. Dr. Joseph W. Howard ©Summer 2006 273 o K Water Freezes/Melts 373 o K Water Boils / Water Vapor Condenses Kelvin Temperature Scale 0oK0oK Absolute Zero No KE anywhere 32 o F 0oC0oC 100 o C 212 o F -273 o C -460 o F

11. Dr. Joseph W. Howard ©Summer 2006 TemperatureTemperature Fahrenheit 212 o F 32 o F Celsius 100 o C 0oC0oC Kelvin 373 K 273 K o F = (9/5 × o C) + 32 o C = 5/9 × ( o F – 32) o F = (9/5 × o C) + 32 o C = 5/9 × ( o F – 32)

12. Dr. Joseph W. Howard ©Summer 2006 Time to move Heat (energy) around. Reminder: Total Amount of Energy is always conserved. We can only transform types of energy PE KE Heat

13. Dr. Joseph W. Howard ©Summer 2006 How to “Heat” something? To change its temperature To change its phase!

14. Dr. Joseph W. Howard ©Summer 2006 Specific Heat Specific Heat – “heat-ability” is different for different substances. (Specific Heat Capacity) The amount of heat (energy) required to increase the temperature of 1 gram of a substance by one degree Celsius. Anyone heard of calories?

15. Dr. Joseph W. Howard ©Summer 2006 The “calorie” One calorie is the amount of heat (energy) to raise the temperature of 1 gram of water by 1 o C. Food Calorie? = 1 Calorie = 1000 calories 1 calorie = 4.184 Joules

16. Dr. Joseph W. Howard ©Summer 2006 Heat: Situation #1 Using heat to ONLY change the temperature of a material. “Heating” something? Add Heat! Result? Change the temperature

17. Dr. Joseph W. Howard ©Summer 2006 Conceptual Pitfall Kinetic Energy on the Nanoscale Heat: Situation # 1 Temperature

18. Dr. Joseph W. Howard ©Summer 2006 Heat to change the temperature Heat = (mass)×(specific heat) ×(change in temperature) How much stuff is there to “heat” “Heat ability” of substance Change in temperature (T final – T initial ) Q = m × c ×(  t)

19. Dr. Joseph W. Howard ©Summer 2006 ExampleExample If I add 418.4 Joules of heat to 20 grams of liquid H 2 0 at 20 o C, how much will the temperature of the H 2 O change? What is the final temperature of the H 2 O? Q = m c  T 418.4 J = (20 grams) (4.184 J/g o C)  T  T = (418.4 J) / (20g)(4.184 J/g o C)  T = 5 o C  T = T final – T initial 5 o C = T final – 20 o C T final = 25 o C

20. Dr. Joseph W. Howard ©Summer 2006 Example Con’t What if the situation was the same, except the heat of 418.4 J were added to Iron and not water? First, iron is a different material and has a different Specific heat of 0.451 J/g o C Q = m c  T 418.4 J = (20 grams) (0.451 J/g o C)  T  T = (418.4 J) / (20g)(0.451 J/g o C)  T = 46 o C Much easier to change the temp of iron.

21. Dr. Joseph W. Howard ©Summer 2006 Using “Heat” Hmmmm…. What will happen to our 20 g water sample if we keep adding “heat”? Another 418.4 J 30 o C Another 418.4 J 35 o C Another 418.4 J 40 o C Another 418.4 J 100 o C Temp keeps going up! Water boils

22. Dr. Joseph W. Howard ©Summer 2006 Boiling?Boiling? What is boiling? A liquid begins to change into a gas!! Phase Change

23. Dr. Joseph W. Howard ©Summer 2006 Heat: Situation #2 Heat ONLY to change the phase of a material. Add heat to liquid (water) temperature goes up, but when it reaches the “boiling point”, add heat and NO CHANGE IN TEMPERATURE until all the liquid converts to gas! It takes some energy (heat) just to convert phase! This is true for any change of phase !

24. Dr. Joseph W. Howard ©Summer 2006 Conceptual Pitfall Potential Energy on the Nanoscale Heat: Situation # 2 Phase Change Case 1 Case 2 Different position means different potential energy & different phase

25. Dr. Joseph W. Howard ©Summer 2006 Phase changes! Heat of vaporization – how much “heat” (energy) is removed/added when a gas condenses / when liquid vaporizes. Heat of fusion – how much “heat” (energy) is removed/added when a liquid freezes / when solid melts. Q=m×L Q = (mass)× (Heat needed for phase change)

26. Dr. Joseph W. Howard ©Summer 2006 Phase Changes Description of Phase Change Name of Change Solid  Liquid Melting or Fusion Liquid  Solid Freezing Liquid  Gas Vaporization Solid  Gas (directly without changing to liquid first, dry ice does this) Sublimation Gas  Solid (directly without changing to liquid first) Condensation or Deposition

27. Dr. Joseph W. Howard ©Summer 2006 Ice Bath Activity Was there a temperature change? Was there a flow of heat? Was energy conserved?

28. Dr. Joseph W. Howard ©Summer 2006 Example Calculate how much heat should be removed to change 100 g of steam at 100 o C to water at 100 o C. 1 g steam1 g water2260 J/g Q=m×L=(2260J/g)(100g)= 226000J Notice!! Heat was removed! Only the phase changed! Not the temperature!

29. Dr. Joseph W. Howard ©Summer 2006 Example Con’t Suppose this heat removed from the steam is used to heat a 2000 g rock from 25 o C to 79 o C. What is the specific heat of the rock? Q = m c  T 226000 J = (2000 g) c (79 o C-25 o C) c = (226000 J) / (2000g)(54 o C) c = 2.09 J/g o C

30. Dr. Joseph W. Howard ©Summer 2006 Heat & Energy Heat can change the kinetic energy of a particles in which case the temperature will change. Heat can change the potential energy of particles (as during a phase change) Heat can change the kinetic energy of a particles in which case the temperature will change. Heat can change the potential energy of particles (as during a phase change)

31. Dr. Joseph W. Howard ©Summer 2006 Follow the Heat A 50.0g piece of copper at 90 o C is placed into 100.0g of water at 15 o C. The final temperature of the water and copper is 18 o C. The specific heat of water is 4.184 J/(g o C). How much did the temperature of the copper change? How much did the temperature of the water change? Copper?  T = T final – T initial  T = 18 o C – 90 o C = - 72 o C How much heat was gained by the water? Water?  T = T final – T initial  T = 18 o C – 15 o C = + 3 o C Q = m c  T Q = (100g)(4.184 J/g o C)(3 o C) = 1255 Joules

32. Dr. Joseph W. Howard ©Summer 2006 Follow the Heat A 50.0g piece of copper at 90 o C is placed into 100.0g of water at 15 o C. The final temperature of the water and copper is 18 o C. The specific heat of water is 4.184 J/(g o C). Calculate the specific heat of copper. If the mass of the copper were doubled how much would the specific heat of the copper change? How much heat was lost by the copper? -1255 Joules !! Q = m c  T c = Q m  T c = (-1255 J) (50g)(-72 o C) = 0.35 J/g o C It would not change!