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# CH104 Chapter 2: Energy & Matter

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1. CH104 Chapter 2: Energy & Matter Temperature Energy in Reactions Specific Heat Energy from Food States of Matter Heating & Cooling Curves

2. Chapter 2 Energy and Matter 2.2Temperature

3. Units of Measurement meter (m) 1 m = 1.09 yd liter (L) 1 L = 1.06 qt gram (g) 1 kg = 2.2 lb Celsius (oC) C = (F-32)/1.8 Kelvin (K) K = C + 273

4. Temperature 180o 100o

5. Learning Check A. What is the temperature of freezing water? 1) 0 °F 2) 0 °C 3) 0 K B. What is the temperature of boiling water? 1) 100 °F 2) 32 °F 3) 373 K C. How many Celsius units are between the boiling and freezing points of water? 1) 100 2) 180 3) 273

6. Solution A. What is the temperature of freezing water? 1) 0 °F 2) 0 °C 3) 0 K B. What is the temperature of boiling water? 1) 100 °F 2) 32 °F 3) 373 K C. How many Celsius units are between the boiling and freezing points of water? 1) 100 2) 180 3) 273

7. Trivial Pursuit Question • At what temperature does oF = oC? - 40 oF = - 40 oC - 40 o

8. Temperature conversion • Common scales used • Fahrenheit,CelsiusandKelvin. oF = 1.8 oC + 32 oC = (oF - 32) 1.8 K = oC + 273 SI unit

9. Solving a Temperature Problem A person with hypothermia has a body temperature of 34.8 °C. What Is that temperature in °F? F = 1.8(C) + 32 ° F = (1.8)(34.8 °C) + 32 ° exact tenth’s exact = 62.6 ° + 32 ° = 94.6 °F tenth’s

10. Practice: Temp Conversion • What is 75.0 º F in ºC? • ºC = (75.0 º F -32 º) = 23.9 ºC 1.8 • What is -12 º F in ºC? • º F = 1.8 (-12) + 32 º F = 10 º C

11. Learning Check The normal temperature of a chickadee is 105.8 °F. What is that temperature on the Celsius scale? 1) 73.8 °C 2) 58.8 °C 3) 41.0 °C

12. Solution The normal temperature of a chickadee is 105.8 °F. What is that temperature on the Celsius scale? 1) 73.8 °C 2) 58.8 °C 3) 41.0 °C TC = TF – 32 ° 1.8 = (105.8 – 32 °) 1.8 = 73.8 °F = 41.0 °C 1.8 °tenth’s place

13. Learning Check A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale? 1) 423 °C 2) 235 °C 3) 221 °C

14. Solution A pepperoni pizza is baked at 455 °F. What temperature is needed on the Celsius scale? 1) 423 °C 2) 235 °C 3) 221 °C TF – 32 ° = TC 1.8 (455 – 32 °) = 235 °C 1.8one’s place

15. Learning Check On a cold winter day, the temperature is –15 °C. What is that temperature in °F? 1) 19 °F 2) 59 °F 3) 5 °F

16. Solution On a cold winter day, the temperature is –15 °C. What is that temperature in °F? 1) 19 °F 2) 59 °F 3) 5 °F TF = 1.8TC + 32 ° TF = 1.8(–15 °C) + 32 ° = – 27 + 32 ° = 5 °F one’s place Note: Be sure to use the change sign key on your calculator to enter the minus (–) sign. 1.8 x 15 +/ – = –27

17. Temperatures

18. Learning Check What is normal body temperature of 37 °C in kelvins? 1) 236 K 2) 310 K 3) 342 K

19. Solution What is normal body temperature of 37 °C in kelvins? 1) 236 K 2) 310 K 3) 342 K TK = TC + 273 = 37 °C + 273 = 310. K one’s place

20. Chapter 2 Energy and Matter 2.1Energy

21. Energy Energy = The capacity to cause change Heat Light Wind

22. Potential Energy = stored Energy (Has potential for motion) X Kinetic Energy = Energy in motion (Fulfilling its potential)

23. Learning Check Identify each of the following as potential energy or kinetic energy. A. roller blading B. a peanut butter and jelly sandwich C. mowing the lawn D. gasoline in the gas tank

24. Solution Identify each of the following as potential energy or kinetic energy. A. roller blading kinetic B. a peanut butter and jelly sandwich potential C. mowing the lawn kinetic D. gasoline in the gas tank potential

25. Kinetic Energy KE = 1mv2 2 Which has more E? • Truck moving at 5 mph • Bicycle moving at 5 mph

26. Units of Measurement meter (m) 1 m = 1.09 yd liter (L) 1 L = 1.06 qt gram (g) 1 kg = 2.2 lb Celsius (oC) C = (F-32)/1.8 Kelvin (K) K = C + 273 calorie (cal) 1Kcal = 1000 cal = 1Cal Joule (J) 1 cal = 4.18 J

27. Examples of Energy Values in Joules

28. Learning Check How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized?

29. Solution How many calories are obtained from a pat of butter if it provides 150 J of energy when metabolized? Given Need 150 J 1 cal 4.184 J = cal 36 Conversion Factors: 1 cal and 4.184 J 4.184 J 1 cal

30. Chapter 2 Energy and Matter 2.3Specific Heat

31. Energy Units of Energy = calorie cal kilocalorie kcal 1000 cal = 1 kcal Calorie Cal Cal = kcal joule J 4.18 J = 1 cal British Thermal Unit BTU

32. H2O Energy • calorie: • E to raise 1 g H20 by 1 oC Specific Heat 1 cal 1g 1oC 1oC

33. Au 0.031 Ag 0.057 Cu 0.093 Fe 0.11 Sand 0.19 Al 0.22 H2O 1.00 Add 1 cal Specific Heat E to raise Temp of 1g substance by 1 oC cal g oC 0oC = start

34. Au 0.031 Ag 0.057 Cu 0.093 Fe 0.11 Sand 0.19 Al 0.22 H2O 1.00 Specific Heat E to raise Temp of 1g substance by 1 oC 30o cal g oC 20o 10o 1o Add 1 cal 0oC = start

35. Au 0.031 Ag 0.057 Cu 0.093 Fe 0.11 Sand 0.19 Al 0.22 H2O 1.00 Specific Heat 30o Low SpHt; Heats quickly 20o 10o 1o High SpHt; Resists change

36. Sand 0.19 H2O 1.00 Specific Heat Dehydrated person Body temp rises quickly Hydrated person Resists change in body temp Heats quickly Gets hot Resists change Stays cold

37. Learning Check A. For the same amount of heat added, a substance with a large specific heat 1) has a smaller increase in temperature 2) has a greater increase in temperature B. When ocean water cools, the surrounding air 1) cools 2) warms 3) stays the same C. Sand in the desert is hot in the day and cool at night. Sand must have a 1) high specific heat 2) low specific heat

38. Solution A. For the same amount of heat added, a substance with a large specific heat 1) has a smaller increase in temperature 2) has a greater increase in temperature B. When ocean water cools, the surrounding air 1) cools 2) warms 3) stays the same C. Sand in the desert is hot in the day and cool at night. Sand must have a 1) high specific heat 2) low specific heat

39. 24.8g Specific Heat Sample Problem: What is the specific heat of a metal if 24.8 g absorbs 65.7 cal of energy and the temp rises from 20.2 C to 24.5 C? 65.7 0.62 cal g oC = cal 24.8g 4.3oC = cal 1g 1oC SpHt = 24.5oC DT = 4.3 Co m = 20.2oC

40. 50g H2O Specific Heat E = m DT SpHt Sample Problem: How much energy does is take to heat 50 g’s of water from 75oC to 87oC? 87oC DT = 12 Co m = 75oC 1 cal 1g 1oC SpHt =

41. 50g H2O Specific Heat E = m DT SpHt Sample Problem: How much energy does is take to heat 50 g’s of water from 75oC to 87oC? m DT SpHt 12 Co 1 cal 1g 1oC = cal to heat water 600

42. 750g H2O Specific Heat E = m DT SpHt Sample Problem: A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temp (37 °C), how many calories of heat could be transferred to sore muscles? 65oC DT = 28 Co m = 37oC 1 cal 1g 1oC SpHt =

43. 750g H2O Specific Heat E = m DT SpHt Sample Problem: A hot-water bottle contains 750 g of water at 65 °C. If the water cools to body temp (37 °C), how many calories of heat could be transferred to sore muscles? m DT SpHt 28 Co 1 cal 1g 1oC = cal from cool water 21000

44. Learning Check How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C?

45. 325g H2O Solution E = m DT SpHt How many kilojoules are needed to raise the temperature of 325 g of water from 15.0 °C to 77.0 °C? m DT SpHt = KJ 84.3 62 Co 4.184 J 1g 1oC 1 KJ 1000 J 77oC DT = 62 Co 15oC

46. Chapter 2 Energy and Matter 2.4Energy and Nutrition 1 Cal = 1000 calories 1 Cal = 1 kcal 1 Cal = 4184 J 1 Cal = 4.184 kJ

47. Energy Units of Energy = 1 Cal 1000 cal = 1 kcal 1 cal = 4.18 J 1 Cal 4184 J = 4.184 kJ

48. Calorimeters A calorimeter • A reaction chamber & thermometer in H2O used to measure heat transfer • indicates the heat lost by a sample • indicates the heat gained by water

49. Energy from Food 4 kcal g 17 kJ g Carbohydrate 9 kcal g 38 kcal g Fat Protein 4 kcal g 17 kcal g

50. Energy from Food Sample Problem: How much energy in kcal (= Cal) would be obtained from 2 Tbl peanut butter containing 6 g carb, 16 g fat, & 7 g protein? 6 g carb 4 kcal 1 g carb = 24 kcal = 196 kcal = 196 Cal 16 g fat 9 kcal 1 g fat = 144 kcal = 200 Cal 7 g protein 4 kcal 1 g protein = 28 kcal