1. 1 CH110 Chapter 2: Energy & Matter Potential & Kinetic Energy Energy from Food Temperature Specific Heat States of Matter Heating & Cooling Curves

2. 2 Energy Energy Energy = The capacity to cause change Heat Light

3. 3 Potential Energy stored Energy Potential Energy = stored Energy (Has potential for motion) Kinetic Energy Energy in motion Kinetic Energy = Energy in motion (Fulfilling its potential) X

4. 4 MetricSICommonConversions Length Volume Mass Energy Units of Measurement meter (m) 1 m = 1.09 yd liter (L) 1 L = 1.06 qt gram (g) 1 kg = 2.2 lb calorie (cal) 1Kcal = 1000 cal = 1Cal Joule (J) 1 cal = 4.18 J

5. 5 Energy Units of Energy = caloriecal kilocaloriekcal 1000 cal = 1 kcal CalorieCal Cal = kcal jouleJ 4.18 J = 1 cal British Thermal Unit BTU

6. 6 Energy calorie: E to raise 1 g H 2 0 by 1 o C H2OH2O 1 1 cal 1g 1 o C 1 1 cal 1g 1 o C Specific Heat 1oC1oC

7. 7 Energy from Food Carbohydrate Fat Protein 4 kcal 1 g carb 4 kcal 1 g carb 9 kcal 1 g fat 9 kcal 1 g fat 4 kcal 1 g protein 4 kcal 1 g protein

8. 8 = 196 Cal 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 16 g fat 4 kcal 1 g carb 4 kcal 1 g carb 9 kcal 1 g fat 9 kcal 1 g fat 4 kcal 1 g protein 4 kcal 1 g protein 7 g protein = 24 kcal = 144 kcal = 28 kcal = 196 kcal = 200 Cal

9. 9 MetricSICommonConversions Length Volume Mass Energy Temperature 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 ( o C) C = (F-32)/1.8 Kelvin (K) K = C + 273 calorie (cal) 1Kcal = 1000 cal = 1Cal Joule (J) 1 cal = 4.18 J

10. 10 180 o 100 o Temperature

11. 11 Temperature conversion Common scales used Fahrenheit, Celsius and Kelvin.Fahrenheit, Celsius and Kelvin. o F= 1.8 o C + 32 o C = ( o F - 32) 1.8 K = o C + 273 K = o C + 273 SI unit

12. 12 At what temperature doesAt what temperature does o F = o C? Trivial Pursuit Question - 40 o F = - 40 o C - 40 o

13. 13 Practice: Temp Conversion What is 75.0 º F in ºC? ºC = (75.0 º F -32 º) / 1.8 = 23.9 ºC What is -12 º F in ºC? º F = 1.8 (-12) + 32 º F = 10 º C

14. 14 Energy calorie: E to raise 1 g H 2 0 by 1 o C H2OH2O 1 1 cal 1g 1 o C 1 1 cal 1g 1 o C Specific Heat 1oC1oC

15. 15 Specific Heat E to raise Temp of 1g substance by 1 o C Fe0.11 Cu 0.093 H 2 O 1.00 Ag 0.057 Au 0.031 Sand 0.19 Al 0.22 cal 1g 1 o C cal 1g 1 o C 0 o C = start Add 1 cal

16. 16 Specific Heat Fe0.11 Cu 0.093 Ag 0.057 Au 0.031 Sand 0.19 Al 0.22 0 o C = start H 2 O 1.00 E to raise Temp of 1g substance by 1 o C cal g o C cal g o C 1o1o 10 o 30 o 20 o Add 1 cal

17. 17 High SpHt; Resists change Specific Heat Fe0.11 Cu 0.093 Ag 0.057 Au 0.031 Sand 0.19 Al 0.22 H 2 O 1.00 1o1o 10 o 30 o 20 o Low SpHt; Heats quickly

18. 18 Specific Heat Sand 0.19 H 2 O 1.00 Resists change Stays cold Heats quickly Gets hot Hydrated person Resists change in body temp Dehydrated person Body temp rises quickly

19. 19 75 o C Specific Heat Sample Problem: How much energy does is take to heat 50 g’s of water from 75 o C to 87 o C? 87 o C  T = 12 o C E = m  T SpHt 1 1 cal 1g 1 o C 1 1 cal 1g 1 o C m = SpHt = 50g H 2 O

20. 20 50g H 2 O Specific Heat Sample Problem: How much energy does is take to heat 50 g’s of water from 75 o C to 87 o C? 12 o C E = m  T SpHt 1 1 cal 1g 1 o C 1 1 cal 1g 1 o C m m SpHt TT TT = cal to heat water 600

21. 21 Elemental states at 25 o C He Rn XeI KrBrSe ArClS NeFO P NC H Li Na Cs Rb K TlHgAuHfLsBa Fr PtIrOsReWTaPoBiPb Be Mg Sr Ca CdAgZrYPdRhRuTcMoNb AcRa ZnCuTiScNiCoFeMnCrV InSbSn GaGe Al Gd Cm Tb Bk Sm Pu Eu Am Nd U Pm Np Ce Th Pr Pa Yb No Lu Lr Er Fm Tm Md Dy Cf Ho Es At Te As Si B Solid Liquid Gas

22. 22 Changes of State Melting Pt = Freezing Pt Boiling Pt Solid Liquid Vapor CondenseCondense FreezeFreeze MeltMelt VaporizeVaporize Slow, close, Fixed arrangement Moderate, close, Random arrangement Fast, far apart, Random

23. 23 Solid Liquid Vapor Changes of State Frost DepositDeposit SublimeSublime Freeze Dry

24. 24 80 cal to melt ice g 1.00 cal to heat water g o C g o C 540 cal to vaporize g water g water Heat of Vaporization Specific Heat of H 2 O Heat of Fusion Heating Curve

25. 25 BeginBegin EndEnd Example: Example: Calculate the total heat needed to change 500. g of ice at 0 o C into 500. g of steam at 100 o C. 500g 80 cal g = 40,000 cal to melt ice 500g 100 o C 1.00 cal g o C g o C = 50,000 cal to heat water 500g 540 cal g = 270,000 cal to vaporize water 360,000 cal = 3.60 x 10 5 cal