1. HEATHEAT TEMPERATURE

2. WHAT YOU SHOULD KNOW A WARMER OBJECT CAN WARM A COOLER OBJECT BY CONTACT OR FROM A DISTANCE.

3. SOME MATERIALS CONDUCT HEAT BETTER THAN OTHERS.

4. HEAT IS TRANSFERRED BY THE COLLISION OF ATOMS AGAINST ONE ANOTHER.

5. HEAT FLOWS FROM WARMER OBJECTS TO COOLER OBJECTS UNTIL BOTH REACH THE SAME TEMPERATURE.

6. REMEMBER…. WORK IS FORCE X DISPLACEMENT KINETIC ENERGY IS ENERGY OF MOTION POTENTIAL ENERGY IS ENERGY OF POSITION

7. LAW OF CONSERVATION OF MECHANICAL ENERGY CONSERVATION OF MECHANICAL ENERGY …IN THE ABSENCE OF FRICTION, THE TOTAL MECHANICAL ENERGY OF THE SYSTEM STAYS THE SAME.

8. TEMPERATURE MEASURE OF THE AVERAGE KINETIC ENERGY IN THE OBJECT. TELLS US HOW COLD OR HOT SOMETHING IS BY COMPARISON TO AN ESTABLISHED STANDARD OR SCALE.

9. TEMPERATURE SCALES KELVIN – ABSOLUTE SCALE FAHRENHEIT – USED IN THE U. S. CELSIUS – USED IN COUNTRIES THAT HAVE ADOPTED THE METRIC SYSTEM AND THE SCIENTIFIC COMMUNITY

10. WATER WATER EXPANDS WHEN COOLED TO FREEZING BECAUSE IT FORMS OPEN STRUCTURES AS IT CHANGES STATE. 4 DEGREE WATER IS THE DENSEST WATER THERE IS.

11. HOW DO THERMOMETERS WORK ALL SUBSTANCES EXPAND WHEN HEATED. A SEALED COLUMN WITH A LIQUID IN A VACCUM. AS THE LIQUID GAINS OR LOOSES ENERGY IT EXPANDS OR CONTRACTS IN THE TUBE AGAINS THE SCALE AND WE READ THE TEMPERATURE.

12. TEMPERATURE SCALES CELSIUS SCALE FAHRENHEIT SCALE KELVIN

13. CELSIUS ZERO DEGREES – FREEZING 100 DEGREES – BOILING THE GAP BETWEEN IS DIVIDED INTO 100 EQUAL PARTS CALLED DEGREES. NAMED IN HONOR OF SWEDISH ASTRONOMER ANDERS CELSIUS.

14. FAHRENHEIT SCALE 32 DEGREES – FREEZING 212 DEGREES – BOILING NAMED IN HONOR OF GERMAN PHYSICIST GABRIEL FAHRENHEIT

15. KELVIN ZERO IS THE LOWEST POSSIBLE SCALE CALLED ABSOLUTE ZERO. THE TEMPERATURE AT WHICH ALL ATOMIC MOTION CEASES. THIS IS THE SAME TEMPERATURE AS -273 DEGREE CELSIUS. NAMED IN HONOR OF BRITISH PHYSICIST LORD KELVIN.

16. TEMPERATURE CONVERSION EQUATION °C = 5/9(°F – 32) °F = 9/5 °C + 32 K = °C + 273

17. TEMP AND ENERGY THE MORE KINETIC ENERGY THE GREATER THE TEMPERATURE THE WARMTH YOU FEEL WHEN YOU TOUCH SOMETHING HOT IS KINETIC ENERGY TRANSFERING TO YOUR HAND.

18. 1 LITER OF WATER VS 2 LITERS OF WATER HOW DOES THE TEMPERATURE OF THE TWO CONTAINERS OF WATER COMPARE? HOW DOES THE AMOUNT OF KINETIC ENERGY OF THE TWO CONTAINERS OF WATER COMPARE?

19. HEAT THE ENERGY THAT TRANSFERS FROM ONE OBJECT TO ANOTHER BECAUSE OF A TEMPERATURE DIFFERENCE BETWEEN THEM IS CALLED HEAT.

20. MATTER AND HEAT MATTER DOES NOT CONTAIN HEAT…IT CONTAINS ENERGY. HEAT IS TRANSFER OF ENERGY. ONCE IT IS TRANSFERRED IT IS NO LONGER HEAT.

21. HOW DOES HEAT FLOW? HEAT FLOWS FROM HOT TO COLD.

22. THERMAL EQUILIBRIUM WHEN NO HEAT FLOWS BETWEEN OBJECTS THEY ARE IN THERMAL EQUILIBRIUM.

23. INTERNAL ENERGY THE TOTAL OF ALL ENERGIES WITHIN A SUBSTANCE.

24. COMPONENTS OF INTERNAL ENERGY TRANSLATIONAL KINETIC ENERGY – JOSTLING MOLECUES ROTATIONAL KINETIC ENERGY – MOLECULES SPINNING POTENTIAL ENERGY – DUE TO FORCES BETWEEN MOLECULES

25. TOTAL INTERNAL ENERGY TRANSLATIONAL KINETIC ENERGY + ROTATIONAL KINETIC ENERGY + POTENTIAL ENERGY = INTERNAL ENERGY ANY OF THESE ENERGIES MAY CHANGE WHEN A SUBSTANCE GIVE OFF OR TAKES IN HEAT.

26. FORMULA INTERNAL ENERGY = PE + KE CHANGE IN PE + CHANGE KE + CHANGE IN INTERNAL ENERGY = 0

27. MEASURING HEAT calorie – THE AMOUNT OF HEAT REQUIRED TO RAISE THE TEMPERATURE OF 1 GRAM OF WATER BY 1 °C

28. calorie VS CALORIE calorie IS THE OFFICIAL SI UNIT. CALORIE IS 1000 calories. CALORIE IS THE CALORIE USED FOR FOOD. IF AN ITEM HAS 200 CALORIES THAT MEANS IT HAS 200,000 calories OF ENERGY.

29. CONVERSION OF calories TO JOULES 1 calorie = 4.184 J 1 CALORIE = 4184 J HOW DO WE DETERMINE THE ENERGY VALUE IN FOOD? WE BURN THE FOOD AND MEASURE THE ENERGY RELEASED AS HEAT.

30. PROBLEM A WOMAN WITH AN AVERAGE DIET CONSUMES AND EXPENDS ABOUT 2000 CALORIES PER DAY. THE ENERGY USED BY HER BODY IS EVENTUALLY GIVEN OFF AS HEAT. HOW MANY JOULES PER SECOND DOES HER BODY GIVE OFF? OR, IN OTHER WORDS, WHAT IS HER AVERAGE THERMAL POWER OUTPUT?

31. ANSWER 2000 CAL. 1 HOUR 4184 J 24 HOUR 3600 S 1 CAL. 96.85 JOULES

32. QUESTION SUPPOSE YOU USE A FLAME TO ADD A CERTAIN QUANTITY OF HEAT TO 1 LITER OF WATER, AND THE WATER TEMPERATURE RISES BY 2 °C. IF YOU ADD THE SAME QUANTITY OF HEAT TO 2 LITERS OF WATER, BY HOW MUCH WILL ITS TEMPERATURE RISE? °C

33. ANSWER 1 °C WHY? BECAUSE THE DOUBLED VOLUME HALVES THE TEMPERATURE INCREASE.

34. THERMAL EXPANSION DIFFERENT SUBSTANCES EXPAND AT DIFFERENT RATES AND THEIR EXPANSION IS DIRECTLY RELATED TO THE AMOUNT OF KINETIC ENERGY IN THE SUBSTANCE.

35. LINEAR EXPANSION WHEN A SOLID OBJECT EXPERIENCES A TEMPERTURE CHANGE, ITS LENGTH WILL INCREASE BY A CERTAIN AMOUNT DEPENDING UPON THE NATURE OF THE MATERIAL.

36. EQUATION CHANGE IN LENGTH = ORIGINAL LENGTH X COEFFICIENT OF EXPANSION X CHANGE IN TEMPERATURE ∆L = L o α∆T

37. COEFFICIENT OF LINEAR EXPANSION A CHARACTERISTIC PROPERTY OF A PARTICULAR MATERIAL THAT TELL YOU HOW READILY A MATERIAL EXPANDS. SI UNIT IS 1/°C OR °C -1

38. AREA EXPANSION AN OBJECT MAY ALSO EXPAND IN AREA WHEN HEATED. CHANGE IN AREA = 2 X ORIGINAL AREA X COEFFICIENT OF EXPANSION X CHANGE IN TEMP. ∆A = 2A o α∆T

39. VOLUME EXPANSION CHANGE IN VOLUME = ORIGINAL VOLUME X COEFFICIENT OF VOLUME EXPANSION X CHANGE IN TEMP. ∆V = V o β∆T Β IS THE COEFFICIENT OF VOLUME EXPANSION

40. QUESTIONS HOW DO WE APPLY THIS INFORMATION IN A THERMOSTAT? …TO A BRIDGE? …TO OPENING A JAR?

41. PROBLEM FOR DIFFERENT LENGTH OF STEEL, EXPANSION WOULD FOLLOW THE SAME PROPORTION. FOR SHORT LENGTH OF STEEL, EXPANSION MAY BE NEGLIGIBLE. BUT CONSIDER THE EXPANSION FOR A SNUGLY FITTING STEEL PIPE THAT COMPLETELY ENCIRCLES THE EARTH. HOW MUCH LONG ER WOULD THIS 40 MILLION METER PIPE BE IF ITS TEMPERATURE INCREASED BY 1°C (STEEL CHANGES 1 PART IN 1000000 FOR EACH °C CHANGE IN TEMP.

42. ANSWER 1 = X 1 X 10 6 40 X 10 6 X = 40 METERS

43. TEMPERATURE INCREASE WHEN TEMPERATURE INCREASES IT IS A CHANGE IN THE TRANSLATIONAL ENERGY ONLY.

44. SPECIFIC HEAT CAPACITY SOME OBJECTS REMAIN HOT LONGER THAN OTHER OBJECTS BECAUSE THEY CAN STORE MORE HEAT THAN OTHER OBJECTS. A MEASURE OF THE AMOUNT OF HEAT NEEDED TO RAISE THE TEMPERATURE OF 1 KG OF A SUBSTANCE BY 1°C

45. SPECIFIC HEAT SI UNIT – J/Kg°C CHANGE IN HEAT = MASS X SPECIFIC HEAT X CHANGE IN TEMP. ∆Q = mc ∆T c IS THE SPECIFIC HEAT

46. CONSERVATION OF ENERGY HEAT LOST = HEAT GAINED mc ∆T = mc ∆T c OF WATER IS 4187 J/Kg°C c OF ICE IS 2090 J/Kg°C

47. LATENT HEAT THE AMOUNT OF ENERGY THAT IS NEEDED TO CAUSE A CHANGE IN PHASE. Q = MASS X LATENT HEAT LATENT HEAT WILL BE HEAT OF VAPORIZATION OR HEAT OF FUSION.

48. HEAT OF FUSION THE QUNATITY OF HEAT NEEDED PER KILOGRAM TO MELT A SOLID AT A CONSTANT TEMPERATURE AND ATMOSPHERIC PRESSURE. CHANGE IN HEAT = MASS X HEAT OF FUSION ∆Q = mh f SI UNIT – J/Kg

49. HEAT OF FUSION WATER HEAT OF FUSION IS 3.35 X 10 5 J/Kg. THAT IS THE AMOUNT OF HEAT 1 KG OF ICE MUST ABSORB IN ORDER TO TURN TO WATER.

50. HEAT OF VAPORIZATION THE QUANTITY OF HEAT NEEDED PER KILOGRAM TO VAPORIZE A LIQUID AT A CONSTANT TEMPERATURE AND ATMOSPHERIC PRESSURE. CHANGE IN HEAT = MASS X HEAT OF VAPORIZATION ∆Q = mh V

51. HEAT OF VAPORIZATION WATER 2.26 X 10 6 J/Kg THIS IS THE AMOUNT OF HEAT THAT 1 KG OF WATER MUST ABSORB TO CHANGE TO WATER VAPOR.

52. PROBLEM SUPPOSE WE KNOW THE NUMBER OF CALORIES NEEDED TO RAISE THE TEMPERATURE OF 1 LITER OF WATER BY 15°C. THE SPECIFIC HEAT CAPACITY FOR WATER, c, IS 1 cal/g°C, AND THE MASS OF 1 LITER OF WATE IS 1 KILOGRAM, WHICH IS 1000 GRAMS. REMEMBER 1 GRAM OF WATER HAS VOLUME OF 1mL. HOW MUCH ENERGY IS USED?

53. ANSWER Q = mc ∆T Q = 1000G X 1 cal/g°C X 15°C Q = 15000 CAL X4.184 Q = 62760 J SUPPOSE WE DELIVER THIS ENERGY TO THE WATER WITH A 1000 WATT BURNER. HOW LONG WILL IT TAKE TO HEAT THE WATER.

54. ANSWER WE KNOW THAT 1000 WATTS DELIVERS ENERGY AT THE RATE OF 1000 JOULES PER SECOND. ENERGY = POWER X TIME 62760 J= 1000 W/S X T 62.76 S = T

55. WATER BECAUSE WATER HAS SUCH A HIGH SPECIFIC HEAT CAPACITY, IT CAN ABSORB A GREAT AMOUNT OF ENERGY. THIS MAKES IT A WONDERFUL COOLING AGENT.

56. STEAM V.S. WATER VAPOR WATER VAPOR IS AN INVISIBLE GAS THAT RESULTS WHEN WATER BOILS OR EVAPORATES STEAM IS WHAT YOU SEE WHEN WATER VAPOR IS COOLED AND CONDENSES BACK INTO WATER DROPLETS.

57. EXAMPLE AN IGLOO IS MADE OF 224 BLOCKS OF ICE AT 0 °C, EACH WITH A MASS OF 12.0 KG. HOW MUCH HEAT MUST BE GAINED BY THE ICE TO MELT THE ENTIRE IGLOO?

58. SOLUTION M = 2690 KG h f = 3.35 x 10 5 J/kg ∆Q = mh f ∆Q = 2690 X 3.35 x 10 5 J/kg ∆Q = 9.01 X 10 8 J

59. EXTENSION HOW MUCH HEAT IS NECESSARY FOR THE IGLOO TO INCREASE ITS TEMPERATURE BY 2 °C? Q = mc∆T Q = 2690 X 2.09 X10 3 X 2 °C Q = 1.12 X 10 7 J

60. FURTHER WHAT WAS THE TOTAL ENERGY NEEDED TO CHANGE THE TEMPERATURE 2 °C AND MELT THE ICE? Q T = ∆Q + Q Q T = 9.01 X 10 8 J+ 1.12 X 10 7 J Q T = 9.12 X 10 8 J