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Energy and Temperature Download Presentation ## Energy and Temperature

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1. Energy and Temperature

2. I. The Nature of Energy • The ability to do work • 3 categories -Radiant energy -Kinetic energy -Potential energy

3. Kinetic Energy • Energy of motion • Types 1. Mechanical - moving parts of a machine 2. Thermal (heat) - energy caused by the random internal motion of particles of matter

4. Potential Energy • Stored energy • Types 1. Electrical - ex. battery 2. Gravitational - ex. water behind dam - used for electricity 3. Chemical - ex. chemical bonds in food

5. Law of Conservation of Energy • In any process, energy is neither created nor destroyed. ex. 1 hitting a baseball transfers kinetic energy from the bat to the ball ex. 2 igniting a match changes chemical energy into heat and light

6. Temperature A measure of the average kinetic energy of molecules in motion Remember: Kelvin = 273 + Celsius Heat Total amount of energy that flows between matter Flows from matter of higher temperature to matter of lower temperature “Hot” molecules quickly move into areas of slower moving “cold” molecules TEMPERATURE VS. HEAT

7. Visual Concepts Chapter 10 Temperature and the Temperature Scale

8. Heat • The transfer of kinetic energy from a hotter object to a colder object. • Symbol is q -particles are always moving -when you heat water  molecules move faster

9. EXOTHERMIC REACTIONS • Chemical reactions that release thermal energy • Feels hot – temperature rises • Examples: condensation, freezing

10. ENDOTHERMIC REACTIONS • Chemical reactions that absorb thermal energy • Feels cold – temperature drops • Examples: boiling, evaporation, melting

11. Measuring Heat • Kelvin Scale (K) • 0 K - point at which there is no molecular motion (absolute zero) • All Kelvin temperatures are positive • K = °C + 273 What is the boiling point of water in Kelvin? What is the freezing point of water in Kelvin?

12. Measuring Heat • Calorie = amount of heat needed to raise the temperature of 1 gram of water by 1 degree Celsius (1 cal = 1 g x 1 C°) • Energy stored in food = Calorie (Cal) 1 Cal = 1000 cal = 1 kcal Joule (J) : 1 cal = 4.184 J

13. Specific Heat • Specific Heat • Amount of heat required to raise 1 gram of a substance 1°C. • physical property • Liquid water 4.184 J/g°C • Fe 0.449 J/g°C

14. Specific Heat • Water has high heat capacity: • Absorbs a large quantity of heat with only a small increase in temperature • Gives up a large quantity of heat with only a small decrease in temperature • Metals have low heat capacity • Small amount of heat  large temperature change

15. HEAT q = cp• m • ΔT Where q = heat released (-) or heat absorbed (+) cp = specific heat (value given on p. 343) m = mass ΔT (means change in Temperature) = Final Temp – Initial Temp

16. HEAT q = cp• m • ΔT Where q = Joules (J) or calories (cal) cp = J/g•K or J/g•˚C or cal/g •˚C m = grams ΔT = K or °C

17. SAMPLE PROBLEM A If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed?

18. SAMPLE PROBLEM A If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed? Step 1: Outline what you know. q = ? m = 75 g cp = 0.449 J/g•K ΔT = 314 - 274

19. SAMPLE PROBLEM A If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed? Step 1: Outline what you know. q = ? m = 75 g cp = 0.449 J/g•K ΔT = 40 K

20. SAMPLE PROBLEM A If 75 g of iron (cp = 0.449 J/g•K) is heated from 274 K to 314 K, what is the amount of heat that is absorbed? Step 2: Plug-in and solve. q = m • cp•ΔT q = (75) (0.449) (40) q = 1347 J Endothermic

21. SAMPLE PROBLEM B A 4.0 g sample of glass was heated from 274 K to 314 K and was found to have absorbed 32 J of energy as heat. What is the specific heat of this type of glass?

22. SAMPLE PROBLEM B A 4.0 g sample of glass was heated from 274 K to 314 K and was found to have absorbed 32 J of energy as heat. What is the specific heat of this type of glass? Step 1: Outline what you know. q = 32 J m = 4.0 g Cp = ? ΔT = 314 – 274 = 40 K

23. SAMPLE PROBLEM B A 4.0 g sample of glass was heated from 274 K to 314 K and was found to have absorbed 32 J of energy as heat. What is the specific heat of this type of glass? Step 2: Plug-in and solve. q = m • cp•ΔT 32 = (4.0) (x) (40) 32 = 160 x x = 0.20 J / g •K

24. Practice Problems How much heat would be required to raise the temperature of 20.0 g of water from 50.0ºC to 100. ºC. (c of H2O = 4.18J/gºC) q = cmΔT = (4.18J/gºC)(20.0g)(100. ºC–50.0ºC) q = 4,180 J

25. II.A. Calorimetry • Calorimeter - an insulated device to measure temperature changes • Can determine enthalpy changes (heat) of reactions

26. Calorimetry • Exothermic process  releases heat  temperature of surroundings increases • Endothermic process  absorbs heat  temperature of surroundings decreases

27. Calorimeter • qrxn = - qsur Transferred Surroundings Put hot iron ring into cool water Leave until the temp Is the same for both How does heat lost by the iron compare to heat gained by the water? Heat lost  Iron Heat gained Water

28. LAB: Heat and Molecular Motion Purpose: To illustrate what happens when a substance is heated. Procedure: • Fill 2 250 mL beakers ¾ full of tap water. • Heat 1 to near boiling on a hot plate. • Gently put 1 drop of food coloring in each.

29. Quiz 12-9 A 10.0 g sample of metal X requires 25.0 J of heat to raise its temperature from 17ºC to 67ºC. What is the specific heat of the metal?

30. Quiz 12-8 How much heat is required to raise the temperature of 30.0 g of H2O from 18.0°C to 28.0°C? c = 4.18 J/gºC

31. THERMOCHEMISTRY- • Study of the changes in heat in chemical reactions • All chemical reactions involve changes in energy. • Heat energy is either absorbed or released.

32. Enthalpy • Enthalpy (H)-heat content of a substance • Depends on temperature, physical state, and composition • Enthalpy Change - the amount of heat absorbed or released during a chemical reaction; ΔH ΔH = Hproducts – Hreactants

33. Exothermic Reactions • Reactions that release heat to their surroundings • Combustion C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g) + heat - Heat is produced because the energy released as new bonds are formed (products) > energy required to break the old bonds (reactants)

34. Thermite Reaction Al(s) +Fe2O3(s) Al2O3(s) +Fe(l) +heat

35. energyreleased Exothermic Reaction • Products have lower potential energy than reactants. Al(s) +Fe2O3(s) Al2O3(s) +Fe(l) +heat

36. If heat is released, • Hproducts < Hreactants DH is negative (exothermic)

37. Endothermic Reactions • Reactions that absorb heat from surroundings • Ammonium nitrate in water NH4NO3(s) + heat  NH4+(aq) + NO3-(aq) • Energy released as new bonds are formed (products) < energy required to break bonds (reactants) • This energy must be supplied by surroundings and is stored in the chemical bonds of the products

38. energyabsorbed Endothermic Reaction • Reactants have lower potential energy than products NH4NO3(s) + heat  NH4+(aq) + NO3-(aq)

39. If heat is absorbed, Hproducts > Hreactants DH is positive (endothermic)

40. Calculating Heat of Reaction How much heat will be released when 34.0g of hydrogen peroxide decomposes? 2 H2O2(l)  2 H2O(l) + O2(g) ΔH° = -190 kJ -190 kJ 2 mol H2O2 34.0 g H2O2 1 mol H2O2 - 190 kJ 34.0 g H2O2 2 molH2O2 = - 95.0 kJ Draw an energy diagram for this reaction. Ratio of coefficients

41. 2. How much heat will be released when 184 g of NO2 is formed at STP? N2(g) + 2 O2(g) 2 NO2(g) ΔH° = +68.0 kJ 184 g NO2 1 mol NO2 68.0 kJ 46 g NO2 2 mol NO2 = 136 kJ Draw an energy diagram for this reaction.

42. What type of reaction is this? 2. Would the value of ΔH be positive or negative?

43. What type of reaction is this? 2. Would the value of ΔH be positive or negative?

44. Changes of State • Melting= s  l ΔH= • Freezing= l  s ΔH= • Condensation= g  l ΔH= • Evaporation/Boiling= l  g ΔH= • Deposition= g  s ΔH= • Sublimation= s  g ΔH=

45. Changes of State • Heat of vaporization- The amount of heat required to convert a unit mass of a liquid at its boiling point into vapor without an increase in temperature. • Heat of fusion- the amount of heat required to convert a unit mass of a solid at its melting point into a liquid without an increase in temperature.

46. Energy (kJ) mol Molar Heat = Molar Heat of Fusion / Vaporization • How much heat does it take to melt / evaporate something. • Expressed in kJ / mol

47. Sample Problem A How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol) Step 1: Outline what you know. Molar Heat = 6.009 kJ / mol Energy = ? kJ mol = 2.61 mol

48. Energy (kJ) 2. 61 mol 6.009 = Sample Problem A How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol) Step 2: Plug into the equation.

49. Energy (kJ) 2. 61 mol 6.009 = Sample Problem A How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol) Step 3: Solve. (2.61 mol) (2.61 mol)

50. Sample Problem A How much energy is absorbed when 2.61 mol of ice melts? (Molar Heat of Fusion for ice = 6.009 kJ / mol) Step 3: Solve. Energy = 15.7 kJ