1.  Labs due today (place in red box)  Section 7.1 Quiz tomorrow!  Write any questions you have on scrap paper at the front Brainteaser How many words in the English language end in “-gry”?

2. December 7, 2010

3.  Is the transfer of thermal energy constant?  When does a change in state occur?  What is latent heat?  What happens to the temperature during a phase change?

4.  There are two methods of heating and cooling:  Constant Power  This method increases (or decreases) the temperature of an object at a constant rate

5.  Fixed Temperature  This method raises (or lowers) the temperature of an object quickly at first, then slows down until it reaches the temperature of the surrounding area

6.  Physical states, or phases, are determined by the organization of particles in matter. ◦ Solids ◦ Liquids ◦ Gasses  During a phase change, particles get enough energy, they can break the bonds between nearby particles ◦ When this happens, the potential energy increases, but average kinetic energy remains constant

7. http://apollo.lsc.vsc.edu/classes/met130/notes/chapter2/lat_heat3.html

8.  Heating a solid at a constant rate results in the following: http://resources.yesican-science.ca/lpdd/g07/lp/nelson/nel10.html

9.  Latent Heat: The amount of thermal energy needed for a phase change  Specific Latent Heat of Fusion: The amount of thermal energy needed to melt 1 kg of a substance at its melting point  Specific Latent Heat of Vaporization: The amount of thermal energy needed to vaporize 1kg of a substance at its boiling point

10.  Specific Latent Heat of Fusion Where: L F = Specific Latent Heat of Fusion (kJ/kg) Q F = Thermal Energy transferred (kJ) m = mass of substance (kg)  Specific Latent Heat of Vaporization Where: L V = Specific Latent Heat of Vaporization (kJ/kg) Q V = Thermal Energy transferred (kJ) m = mass of substance (kg)

11. Practice Problems #2. pg. 224 The specific latent heat of fusion of silver is 88.0 kJ/kg. How much energy is released when 6.20 μ g of silver solidify? (1 μ g = 1 x 10 -6 g) Given: L F = 88.0 kJ/kg m = 6.20 x 10 -6 g = 6.20 x 10 -9 kg R.T.F.: Q F = ? Solution: L F = Q F m Q F = L F m = (88.0 kJ/kg)(6.20 x 10 -9 kg) = 5.46 x 10 -7 kJ = 5.46 x 10 -4 J

12. Practice Problems #2. pg. 225 The boiling point of Freon-12 is -29.8°C. How much thermal energy is absorbed by 455 g of Freon-12 going from the liquid phase at -29.8°C to the gaseous phase at 20.02°C? The specific latent heat of vaporization of Freon-12 is 143 kJ/kg and its specific heat capacity is 950 J/(kgK). Given: m = 455 g = 0.455 kg T i = -29.8°C T f = 20.02°C ΔT = 20.02°C – (-29.8°C) = 49.82°C or 49.82 K L V = 143 kJ/kg c = 950 J/(kgK) R.T.F.: Q V = ? Q = ? Solution: Q V = L v m = (143 kJ/kg)(0.455 kg) = 65.1 kJ Q = mc ΔT = (0.455 kg)(950 J/kgK)(49.82 K) = 21534.7 J = 21.5 kJ Q TOT = Q V + Q = 65.1 kJ + 21.5 kJ = 86.6 kJ