DO NOW: Hand in specific heat lab & answer the following question on “Do Now” sheet:

1. DO NOW: Hand in specific heat lab &answer the following question on “Do Now” sheet: 56 grams of hot copper are added to 140 grams of water in a 92 g aluminum calorimeter causing the temperature of the water to rise from 22.0oC to 29.5oC. What was the initial temperature of the hot copper? Cwater=4180 J/(kg K) CAl=897 J/(kg K) CCu=385 J/(kg K)

2. 56 g Cu Ti= ? Q1=(mcDT)water Q2=(mcDT)Al Q3=(mcDT)Cu Q1+Q2+Q3=0 29.5oC 22.0oC 140 g water 92 g Al

3. Agenda 3.6.2014 • Describe states of matter • Explore what happens when a substance changes from one state of matter to another. • Label and explain “heating curves” • Use heat of vaporization and heat of fusion to calculate energy absorbed or released when a substance undergoes a change of state. Announcements • Quiz tomorrow on Chapter 12.1 • HW – 12.1 Review Worksheet • List and Photo of Materials – Extended to Monday (3/10)

4. State of matter with fixed volume and shape. Molecules are in motion but are tightly packed and maintain their locations relative to neighboring molecules. • Solid • Liquid • Gas • Plasma

5. Measure of the average kinetic energy per randomly wiggling particle • Heat • Thermal Energy • Temperature • Specific Heat

6. Measure of the total amount of kinetic energy associated with the random motion of particles plus the total potential energy associated with the intermolecular bonds • Heat • Thermal Energy • Temperature • Specific Heat

7. State of matter that takes on the shape of its container but has a constant volume. Although molecules are tightly packed bonds are continuously changing as they change their locations relative to neighboring molecules. • Solid • Liquid • Gas • Plasma

8. In this state of matter molecules do not form lasting bonds with neighbors and spread out to occupy all the space available. With plenty of extra room between molecules this state of matter can be compressed into a smaller volume. • Solid • Liquid • Gas • Plasma

9. At high enough temperatures particles in a gas have enough kinetic energy to knock electrons free from neutral atoms resulting in this “hot soup” of charged ions. • Solid • Liquid • Gas • Plasma

10. Heat is added at a uniform rate to a pot of water that starts off at room temperature. The temperature of the water is shown for the first few minutes. Sketch what you think the rest of the graph look like if the pot is left on the stove for 20 minutes

11. Take 3 minutes to discuss in small groups what is going on here. Would the formula Q=mcDT work for the entire 20 minutes? Why or why not?

12. Why does boiling water stay at 100oC when you continue to add heat? • In liquid state molecules are “stuck” to neighbors by temporary bonds • Addition of heat at the boiling point “un-sticks” molecules from neighboring particles WITHOUT increasing kinetic energy. BIG IDEA – At the boiling point adding heat results in a change of state. As H2O changes from liquid to gas thermal energy increases but temperature stays the same!

13. Changes of State Phase Diagram - Water Melting Point Boiling Point

14. DO NOW Take out your homework and answer the following question (3 min). How much heat is required to raise the temperaure of 1 kg liquid water (C = 4180 J/kg °C) from 0 °C to 100 °C ? Q = mCwaterΔT

15. Heating curves • http://www.kentchemistry.com/links/Matter/HeatingCurve.htm HEAT

16. Heat of fusion (Hf) Amount of heat required to melt 1 kg at the melting point  Heat required to melt something: Q=mHf • Heat required to freeze something: Q=m(-Hf) Example: Adding 668 kJ to a block of ice at 0oC causes 2.0 kg to melt. What is the heat of fusion of ice? Q= 6.68x105 J Hf= Q/m = 3.34x105 J/kg m=2.0 kg

17. Heat of Vaporization (Hv) Amount of heat required to change 1 kg of liquid to a gas at the boiling point.  Heat required to vaporize something: Q=mHv • Heat required to condense something: Q=m(-Hv) Example: The boiling point of ethanol is 78oC. How much heat must be added to 750 g of ethanol at 78oC to turn it into a gas if Hv=846 kJ/kg ? m=0.75 kg  Q=mHv= 635 kJ Hv=846 kJ/kg

18. DO NOW – Have your photo and list of materials ready for me to check and match the definitions with the terms E Definition Term Melting Point Boiling Point Heat of Vaporization Vaporization Melting Heat of Fusion • Change from solid to liquid • Energy required to convert 1 kg from a solid to liquid at the melting point • Liquid phase and gaseous phase can exist in equilibrium • Change from liquid to gas • Energy required to convert 1 kg from a liquid to a gas at the boiling point. • Solid Phase and liquid phase can exist in equililbrium. F B D C A

19. Heating Curve for Water 130 Q = mHv Q = mCsteamΔT Temperature of H20 (°C) 100 Q = mHf Q = mCwaterΔT 0 -40 Q = mCiceΔT Heat added 

20. Drawing Heat Diagrams 130°C Q = mCsteamΔT Heating Curve for Water 100°C Q = mHv 130 Q = mHv Q = mCwaterΔT Q = mCsteamΔT Temperature of H20 (°C) 100 Q = mHf Q = mCwaterΔT 0 0°C Q = mHf -40 Q = mCiceΔT Q = mCiceΔT Heat added  -40°C

21. What is the total amount of heat required to raise the temperature 1 kg of water from -40 °C to 20 °C? Cice= 2060 J/kg°CCwater = 4180 J/kg°CHf = 3.35x105 J/kg 20°C Q = mCiceΔT Q = mHf Q = mCwaterΔT 0°C Qtotal = mCiceΔT+mHf+ mCwaterΔT -40°C Qtotal = 82400 J +3.35x105 J+ 83600 J Qtotal = 5.01x105 J

22. Homework Change of State Problems – p. 337 #61, 62, 64