2. CHAPTER 10 Reaction Energy

3. Visual Concepts Heat Chapter 10

4. TEMPERATURE VS. HEAT 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

5. Visual Concepts Temperature and the Temperature Scale Chapter 10

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

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

8. CALCULATING HEAT

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

10. HEAT q = c p m ΔT Where q = Joules (J) or calories (cal) c p = J /gK or J/g˚C or cal/g ˚C m = grams ΔT = K or °C

11. SPECIFIC HEAT The amount of energy required to raise 1 g of substance 1 K or 1 ˚C

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

13. SAMPLE PROBLEM A If 75 g of iron (c p = 0.449 J/gK) 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 c p = 0.449 J/gK ΔT = 314 - 274

14. SAMPLE PROBLEM A If 75 g of iron (c p = 0.449 J/gK) 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 c p = 0.449 J/gK ΔT = 40 K

15. SAMPLE PROBLEM A If 75 g of iron (cp = 0.449 J/gK) 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 c p ΔT q = (75) (0.449) (40) q = 1347 J Endothermic

16. 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?

17. 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 C p = ? ΔT = 314 – 274 = 40 K

18. 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 c p ΔT 32 = (4.0) (x) (40) 32 = 160 x x = 0.20 J / g K

19. Molar Heat of Fusion / Vaporization

21.  How much heat does it take to melt / evaporate something.  Expressed in kJ / mol Energy (kJ) mol Molar Heat =

22. 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)

23. 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

24. 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. Energy (kJ) 2. 61 mol 6.009 =

25. 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 (kJ) 2. 61 mol 6.009 = (2.61 mol)

26. 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