1. Thermal Properties, Heat Capacity, Specific Heat.

2. Thermal Properties of Matter When objects absorb heat (Q) their T rises. When objects release/lose (Q) heat their T decreases. Different materials undergo different  T for same amount of heat absorbed/released.

3. Thermal Capacity (C)= amt of heat E (Q) needed to raise an object’s T 1 o C (rather than mass of pure substance). Water has high heat C, takes lots of E to change its T. Metals have low capacities. A little heat E gives a lot of  T

4. Land heats and cools (  T) more quickly than water

5. Why? Temp is measure of average KE of particles. 1 kg of different materials contain different: # of particles, KE must be spread over varying numbers of particles. Weights of particles, it takes more E to move heavier particles. Bond strengths, some materials have particles that are easy to move others are more locked.

6. Why does water have such a high heat capacity? Water molecules form strong intermolecular bonds with each other; Metals have weak intermolecular bonds. Water metal

7. Specific Heat Capacity, c. Amt of E required to raise T of 1kg mass of pure substance 1 o C, or K. Q = mc  Tm = mass in kg c = spec heat const J/kg o C  T = T f – T i. Q = energy in Joules  T is positive if substance absorbs E (T f > T i ), negative for E released (T f < T i ),.

8. Both Iron same c. Same c

9. Specific heats some substances. Easy to heat Way hard to heat Very Easy to heat Way hard to heat

10. 1. A 45-g piece of aluminum is heated from 25 – 55 o C. How much E is required? Q = mc  T Q = (0.045 kg)(897 J/kg o C)(55-25) o C = 1200 J.

11. 2. The SH of water is ~4200 J/kg o C. How much heat will be required to raise the temperature of 300 g from 20 – 60 o C? 50.4 kJ

12. 3. A metal block has a mass of 1.5 kg loses 20kJ of heat. Its temperature drops from 60 – 45 o. What is the specific heat capacity of the metal? 888.9 ~ 900 J/kg o C.

13. Calorimeters – mix substances (usu. w/water) observe Temp change. Use consv E to calculate. Experimental Methods to Determine SH.

14. Mixing substances obeys E conservation : Q lost by one substance = Q gained by another - m 1 c 1  T 1 = m 2 c 2  T 2. or m 1 c 1  T 1 + m 2 c 2  T 2 = 0.

15. Throw hot metal into cold water then Q lost by metal = Q gained by water (+calorimeter). For solids with unknown c, liq with known c.

16. Q lost bolt = Q gained water. -mc  T b = mc  T w. 45  T b = 2520 w  T b = 56 so since T i is more that T f for bolt: 81 o C 4. A 0.05 kg metal bolt is heated to an unknown temperature. It is then dropped into a calorimeter containing 0.15 kg of water of 21.0 o C. The bolt and the water reach a final temperature of 25.0 o C. If the SH of the metal is 899 J/kg o C, find the initial temperature of the metal. Ignore the heat lost to the cup.

17. Heating Coil in Liquid How can we use E conservation to find c liquid?. Electric E lost by coil = E gained by liquid. VIt = mc  T + the cup

18. 5. A heating coil with a voltage potential of 6 V and current of 3 A is immersed in.7 kg of water. How long will it take to heat the water from 20 o C to 35 o C? Assume the cup absorbs no heat. 2442 sec 41 min

19. Feynman https://www.youtube.com/watch?v=ITpDrd tGAmohttps://www.youtube.com/watch?v=ITpDrd tGAmo

20. Phase Changes Melt some ice.

21. The Kinetic Particle Theory of Matter 1.All matter is composed of small particles (atoms, molecules, or ions). 2. They are in constant, random motion. 3.The particles are so small they can be treated as if they are points with no volume.

22. No fixed shape. Minimal intermolecular forces. No fixed shape. Strong intermolecular forces. Fixed shape. Very strong intermolecular forces.

23. Molecular Distance & Type of Motion

24. Phase Changes of Matter Q absorbed - phase change to higher E: –Vaporization: liquid  vapor –Melting: solid  liquid –Sublimation: solid  vapor Q released by phase changes to lower E: –Condensation: vapor  liquid –Fusion: liquid  solid –Deposition: vapor  solid

25. When substances absorb or lose heat E the T changes until: –It reaches “critical” T, T stops changing substance undergoes phase change. –gain/loss of additional heat energy results in the phase transformation of matter from one phase to another: Solid  liquid  gas Substance absorbs E to break molecular bonds. Heat & Phase Change

26. Ex: During the melting process, a solid cube of ice changes phase by absorbing E. Which statement is NOT true about the ice? Its internal energy is increasing. The average KE of the molecules is increasing. The PE of the molecules is increasing. The ice is undergoing a physical change.

27. Consider melting: as heat is absorbed, T goes up particles vibrate more. The KE increases. At the melting point, particles vibrate enough to slip from their fixed positions breaking intermolecular bonds. More and more particles break solid bonds until all are slipping past each other in the liquid phase. The PE of the system increases but the av KE stays constant at the melting (ice) & boiling points (steam).

28. Critical Temps/ Equilibrium Multiple phases exist at a particular T. Molecules leave and enter phases at equal rates. Water: Ice / liquid water: 0 °C Liquid water / steam: 100 °C Ice water

29. Distinguish between the specific heat and the latent heat of a substance. Specific - Q involved  T, increase av KE. Latent – Q involved with phase. No change av KE, change in PE.

30. Latent heat define, L - the amount of heat needed to change the state of 1-kg of material w NO  T. Latent heat of fusion L f – heat absorbed during melting or released during freezing. Latent heat of vaporization L v – heat absorbed during vaporization, released during condensation. Latent Heat

31. describes heat energy (Q) gained/lost to mass of substance (m) that undergoes a phase change is: Q = m. L m = mass in kg Q – E absorbed/released in J. L - latent heat (the SI units are J/kg) L f or L v.

32. Some Thermal Values

33. Ex: Compare the L f to the L v of the substances. Why do you think the L v for each is so much higher. E is required to completely separate the molecules for vaporization.

34. Phases/States of Matter https://www.youtube.com/watch?v=ndw9X YA4iF0 5 minhttps://www.youtube.com/watch?v=ndw9X YA4iF0 https://www.youtube.com/watch?v=8bGYV 6ypP5o 2 minhttps://www.youtube.com/watch?v=8bGYV 6ypP5o

35. Heating/Cooling Curves

36. Label what’s happening here.

37. What is the slope on this portion of the graph?  T = Q or 1 (Q)The slope is 1 mcmc mc Q = mc  T

38. Ex 1: How much E is released cooling 10g of water from steam at 133 o C to liquid at 53 o C? Q to cool steam 133 o – 100 o mc  T stm Q to condense at 100 o mL v Q to cool liquid 100 – 53 o mc  T wat C stm = 2.01x10 3 J/kg o C L v = 2.26 x 10 6 J/kg C wat = 4186 J/kg o C

39. To cool steam (0.01kg)(2.01x10 3 J/kg o C)(33 o C) = 633 J To condense at 100 o (0.01 kg)(2.26x10 6 J/kg) = 2.26 x 10 4 J To cool water (0.01kg)(4186 J/kg o C)(47 o ) = 1967 J Total E2.52x10 4 J

40. Ex 2. How much energy is released when 1.5 kg of water vapor at 100 o C is placed in a freezer and converted to ice at -7 o C. Specific Heat – c ice = 2.1 x 10 3 J/K kg. – c Wat = 4.2 x 10 3 J/K kg. –. L f = 3.34 x 10 5 J/kg L v = 22.5 x 10 5 J/kg

41. 4.5 x 10 6 J

42. Mixing Method: Add ice known mass & T to water. Let it melt while measuring T. Determining L f for Ice -mc(  T) w = mc  T ice + mL f ice + mc  T liq ice. Thermal E lost by water = thermal E gained by ice.

43. Hwk pg 59 Hamper IB Set Specific, Latent Heat 2014

44. Heating/Cooling Curves show temperature change with absorbed energy for a substance.

45. Temp vs. time graph Heating time related to Q absorbed. Rate of E = Power.

46.  T = Pt mc Relate Power to  T  T = P (t) mc

47. Suggest a method to measure the power output of a hotplate.

48. Example Cooling Curve Hypothetical substance. Phase Changes for ice IB Prob. Kerr pg 91 #3- 6, 8, 11.

49. Evaporation vs. Boiling.

50. Evaporation Molecules leave surface & enter vapor phase at any T.

51. Evaporation The molecules in a liquid have a distribution of speeds. The average speed determines the T of the liquid. The fastest molecules have enough energy to overcome the attraction between the molecules in the liquid. They evaporate and become vapor.

52. Evaporative cooling: as the fastest molecules evaporate, they leave behind the coolest molecules and the average temperature of the liquid decreases. Example: sweat

53. Volatile liquids = rapid evaporation. Evaporation rate depends on: Surface area. Ambient Temperature. Vapor Pressure – hi –Vap press = hi volatility Humidity of Surroundings

54. Boiling Boiling occurs when the average motion of particles is fast enough to overcome the forces holding them close together. This happens evenly throughout a boiling liquid. You will see bubbles form. The temperature is uniform throughout.

55. Boiling vs. Evaporation Both boiling & evaporation, intermolecular forces between particles are present. The greater the space between the particles, the weaker the intermolecular force. Evaporation – only at surface, boiling throughout. Boiling at single T, Evaporation range of Temps. To break the bond between two particles, one particle has to be moving fast enough to overcome the pull of the other, until it gets so far away that pull is diminished.

56. Boiling and Melting Boiling E must break intermolecular bonds and separate molecules very far. Happens at T c. No  T. Melting E mostly for breaking of bonds. Occurs over range of T. Cools liquid by leaving lower E molc’s behind.