1. So… about Thermal Energy What’s up with Temperature vs Heat? Temperature is related to the average kinetic energy of the particles in a substance.

2. You know the SI unit for temp is Kelvin K = C + 273 (10C = 283K) C = K – 273 (10K = -263C) Thermal Energy is the total of all the kinetic and potential energy of all the particles in a substance.

3. Thermal energy relationship  As temperature increases, so does thermal energy (because the kinetic energy of the particles increased).

4. Heat The flow of thermal energy from one object to another. Heat always flows from warmer to cooler objects. Ice gets warmer while hand gets cooler Cup gets cooler while hand gets warmer

5. Specific Heat Capacity can be thought of as a measure of how much heat energy is needed to warm the substance up. You will possibly have noticed that it is easier to warm up a saucepan full of oil than it is to warm up one full of water. http://www.cookware- manufacturer.com/photo/418fa6490f24202f2cc 5b5feee0fdde3/Aluminum-Saucepan.jpg

6. Specific Heat (c, sometimes s, but usually c) Things heat up or cool down at different rates. Land heats up and cools down faster than water, and aren’t we lucky for that!?

7. Specific heat is the amount of heat required to raise the temperature of 1 kg (but in Chem we use g) of a material by one degree (C or K, they’re the same size). C water = 4.184 J/(kg*C) (“holds” its heat) C sand = 0.664 J/(kg*C) (less E to change) This is why land heats up quickly during the day and cools quickly at night and why water takes longer.

8. The next table shows how much energy it takes to heat up some different substances. The small values show that not a lot of energy is needed to produce a temperature change, whereas the large values indicate a lot more energy is needed.

9. Approximate values in J / kg °K of the Specific Heat Capacities of some substances are: Air 1000Lead 125 Aluminium 900 Mercury14 Asbestos 840 Nylon 1700 Brass 400 Paraffin 2100 Brick 750 Platinum 135 Concrete 3300 Polythene2200 Cork 2000 Polystyrene1300 Glass 600 Rubber 1600 Gold 130 Silver235 Ice 2100 Steel 450 Iron 500 Water 4200

10. Why does water have such a high specific heat? Water molecules form strong bonds with each other water molecule; (including H-bonds!)so it takes more heat energy to break the bonds. Metals have weak bonds (remember the “sea of e-) and do not need as much energy to break them. water metal

11. WHERE’S THE MATH?! Q = m x Cp x  T Q = change in thermal energy m = mass of substance  T = change in temperature (T f – T i ) C p = specific heat of substance

12. Specific Heat Capacity If 25.0 g of Al cool from 310 o C to 37 o C, how many joules of heat energy are lost by the Al? heat gain/lose = q = (mass) x (c) x (∆T) where ∆T = T final - T initial q = (0.897 J/gK)(25.0 g)(37 - 310)K q = - 6120 J Notice that the negative sign on q signals heat “lost by” or transferred OUT of Al.

13. Heat can be Transferred even if there is No Change in State q transferred = (mass)(c) (∆T)

14. Or… Heat Transfer can cause a Change of State Changes of state involve energy (at constant T) Ice + 333 J/g (heat of fusion) -----> Liquid water Is there an equation? Of course! q = (heat of fusion)(mass)

15. Remember this – it’s the Heating/Cooling Curve for Water! Woa! Note that T is constant as a phase changes Evaporate water

16. So, let’s look at this equation again… q = (heat of fusion)(mass) (There’s also q = (heat of vaporization)(mass), by the way, for when we are talking about vaporization) WHY DO I NEED THIS WHEN I HAVE q transferred = (c)(mass)(∆T)HUH??? Well, when a phase changes THERE IS NO change in temperature… but there is definitely a change in energy!

17. So… if I want the total heat to take ice and turn it to steam I need 3 steps… 1) To melt the ice I need to multiply the heat of fusion with the mass… q = (heat of fusion)(mass)

18. 2) Then, there is moving the temperature from 0 C to 100C… for this there is a change in temperature so we can use… 2) Then, there is moving the temperature from 0 C to 100C… for this there is a change in temperature so we can use… q transferred = (mass)*(c)*(∆T) 3) But wait, that just takes us to 100 C, what about vaporizing the molecules? Well, then we need q = (heat of vaporization)(mass)…

19. Add ‘em all up and there it is! Now, lucky for us, just like there are tables for specific heats, there are also tables for heats of fusion and heats of vaporization. Whew, At least we don’t have to worry about that!

20. Heat & Changes of State What quantity of heat is required to melt 500. g of ice and heat the water to steam at 100 o C? Heat of fusion of ice = 333 J/g Specific heat of water = 4.2 kJ/gK Heat of vaporization = 2260 J/g +333 J/g +2260 J/g

21. And now… More! Heat & Changes of State How much heat is required to melt 500. g of ice and heat the water to steam at 100 o C? 1. To melt ice q = (27.75 mol)(333 J/g) = 9240 J = 9.24 kJ q = (27.75 mol)(333 J/g) = 9240 J = 9.24 kJ 2.To raise water from 0 o C to 100 o C q = (500g)(4.2 J/(gC))(100C – 0C) = 2.1 x 10 5 J q = (500g)(4.2 J/(gC))(100C – 0C) = 2.1 x 10 5 J = 2.1 x 10 2 kJ = 2.1 x 10 2 kJ 3.To evaporate water at 100 o C q = (27.75 mol)(2260 J/g) = 62715 J = 62.715 kJ q = (27.75 mol)(2260 J/g) = 62715 J = 62.715 kJ 4. Total heat energy = 210071.955 kJ = 2.10 x 10 2 kJ

22. 1.How many kJ are required to heat 45.0 g of H 2 O at 25.0 °C and then boil it all away? q = (mass) (C p )( Δ t) q = (moles of water)( Δ H vap ) Givens: Δ t = 75.0 °C The mass = 45.0 g C p = 4.184 kJ*g¯ 1 °C¯ 1 Δ H vap = 40.6 kJ*mol¯ 1 How do we solve this?

24. What if I wanted to know the total Δ H when converting that amount of water to steam?

25. CALORIMETRY Aka… How we Measure Heats of Reaction In a Constant Volume or “Bomb” Calorimeter, we burn a combustible sample. From the heat change, we measure heat evolved in a reaction to get ∆E for reaction.

26. First, some heat from reaction warms the water, which we know the mass of and “c” for… q water = (c)(water mass)(∆T) THEN, some heat from reaction warms “bomb,” which has a known specific heat for the entire apparatus (typically), so we don’t need the mass… q bomb = (heat capacity, J/K)(∆T) Total heat evolved = q total = q water + q bomb BOOM! Combustible material ignited at constant volume! This heats up the “bomb”, which heats up the water surrounding it…

27. The Mathy bit… Measuring Heats of Reaction using… CALORIMETRY Calculate heat of combustion of octane. C 8 H 18 + 25/2 O 2 --> 8 CO 2 + 9 H 2 O C 8 H 18 + 25/2 O 2 --> 8 CO 2 + 9 H 2 O You could burn 1.00 g of octane… or you could just note that… Temp rises from 25.00 to 33.20 o C Calorimeter contains 1200 g water Heat capacity of bomb = 837 J/K

28. Step 1 Calc. heat transferred from reaction to water. q = (4.184 J/gK)(1200 g)(8.20 K) = 41,170 J Step 2 Calc. heat transferred from reaction to bomb. q = (bomb heat capacity)(∆T) = (837 J/K)(8.20 K) = 6860 J = (837 J/K)(8.20 K) = 6860 J Step 3 Total heat evolved 41,170 J + 6860 J = 48,030 J 41,170 J + 6860 J = 48,030 J Heat of combustion of 1.00 g of octane = - 48.0 kJ VIOLA!

29. Hesse’ Law Coal gasification converts coal into a combustible mixture of carbon monoxide and hydrogen, called coal gas, in a gasifier: H 2 O (l) + C (s) → CO (g) + H 2(g) Δ H = ? 2 C (s) + O 2(g) → 2 CO (g) Δ H = -222 kJ 2 H 2(g) + O 2(g) → 2H 2 O (g) Δ H = -484 kJ H 2 O (l) → H 2 O (g) Δ H = +44 kJ How do we solve this?

32. Overview Changes of State Phase transitions Phase Diagrams Liquid State Properties of Liquids; Surface tension and viscosity Intermolecular forces; explaining liquid properties Solid State Classification of Solids by Type of Attraction between Units Crystalline solids; crystal lattices and unit cells Structures of some crystalline solids Calculations Involving Unit-Cell Dimensions Determining the Crystal Structure by X-ray Diffraction John A. Schreifels Chemistry 211 Chapter 11-32

33. Comparison of Gases, Liquids and Solids Gases are compressible fluids. Their molecules are widely separated. Liquids are relatively incompressible fluids. Their molecules are more tightly packed. Solids are nearly incompressible and rigid. Their molecules or ions are in close contact and do not move. John A. Schreifels Chemistry 211Chapter 11-33 Figure 11.2 States of Matter

34. Phase Transitions Melting: change of a solid to a liquid. Freezing: change a liquid to a solid. Vaporization: change of a solid or liquid to a gas. Change of solid to vapor often called sublimation. Condensation: change of a gas to a liquid or solid. Change of a gas to a solid often called deposition. John A. Schreifels Chemistry 211 Chapter 11-34 H 2 O(s)  H 2 O(l) H 2 O(l)  H 2 O(s) H 2 O(l)  H 2 O(g) or H 2 O(s)  H 2 O(g) H 2 O(g)  H 2 O(l) or H 2 O(g)  H 2 O(s)

35. John A. Schreifels Chemistry 211Chapter 11-35 Energy of Heat and Phase Change Heat of vaporization: heat needed for the vaporization of a liquid. H 2 O(l)  H 2 O(g)  H = 40.7 kJ Heat of fusion: heat needed for the melting of a solid. H 2 O(s)  H 2 O(l)  H = 6.01 kJ Temperature does not change during the change from one phase to another. E.g. Start with a solution consisting of 50.0 g of H 2 O(s) and 50.0 g of H 2 O(l) at 0°C. Determine the heat required to heat this mixture to 100.0°C and evaporate half of the water.

36. John A. Schreifels Chemistry 211 Chapter 11-36 Intermolecular Forces Intermolecular forces: attractions and repulsions between molecules that hold them together. Intermolecular forces (van der Waals forces) hold molecules together in liquid and solid phases. Ion-dipole force: interaction between an ion and partial charges in a polar molecule. Dipole-dipole force: attractive force between polar molecules with positive end of one molecule is aligned with negative side of other. London dispersion Forces: interactions between instantaneously formed electric dipoles on neighboring polar or nonpolar molecules. Polarizability: ease with which electron cloud of some substance can be distorted by presence of some electric field (such as another dipolar substance). Related to size of atom or molecule. Small atoms and molecules less easily polarized.

37. John A. Schreifels Chemistry 211Chapter 11-37 Comparisonof Energies for Intermolecular Forces Interaction Forces:Approximate Energy Intermolecular London1 – 10 kJ Dipole-dipole3 – 4 kJ Ion-dipole5 – 50 kJ Hydrogen bonding10– 40 kJ Chemical bonding Ionic100 – 1000 kJ Covalent100 – 1000 kJ