1. 6-1 Due: Nuclear Extra Credit Enthalpy Stoichiometry WS Today: 1. Intro to Calorimetry HW Connect Problems 6.3 Calorimetry Prep Lab Notebook: Investigating Commercial Handwarmers

2. 6-2 Warm Up 1. Two solid chemical compounds are mixed together in a beaker. After one minute, ice crystals are observed on the outside of the beaker. What is the best description for the reaction in the beaker? (A) Endothermic (B) Exothermic (C) Exergonic (D) Sublimation 2. A chemical reaction occurs inside a balloon. Before the reaction begins, the volume of the balloon is 2.3 L. After the reaction is complete, the volume of the balloon is 1.7 L. Which of the following describes this system? (A)The internal energy increases because work is done on the system. (B)The internal energy increases because work is done by the system. (C)The internal energy decreases because work is done on the system. (D)The internal energy decreases because work is done by the system.

3. 6-3 Calorimetry q = mc  T q = c = m =  T = 1.What does the symbols stand for with units? 2.Rearrange equation solving for C 1.What is C? Write a sentence looking at the variables that 2. C is equal to 3.Predict the units

4. 6-4 Calorimetry q = mc  T q = heat lost or gained c = specific heat capacity m = mass in g  T = T final – T initial The specific heat capacity (c) of a substance is the quantity of heat required to change the temperature of 1 gram of the substance by 1 K.

5. 6-5 Calorimetry calorimeter calorimeter The amount of heat absorbed or released during a physical or chemical change can be measured, usually by the change in temperature of a known quantity of water in a calorimeter. calorimeter

6. 6-6 Calorimetry Science of measuring heat Calorimeter – device used to measure heat change Heat Capacity (C) – C = heat absorbed increase in temperature The ∆T is DEPENDENT on the substance because KE for substances will vary (mass will be different)

7. 6-7 Table 6.2 Specific Heat Capacities (c) of Some Elements, Compounds, and Materials Specific Heat Capacity (J/g∙K)* SubstanceSpecific Heat Capacity (J/g∙K) Substance Compounds Water, H 2 O ( l ) Ethyl alcohol, C 2 H 5 OH ( l ) Ethylene glycol, (CH 2 OH) 2 ( l ) Carbon tetrachloride, CCl 4 ( l ) 4.184 2.46 2.42 0.862 Elements Aluminum, Al Graphite,C Iron, Fe Copper, Cu Gold, Au 0.900 0.711 0.450 0.387 0.129 Wood Cement Glass Granite Steel Solid materials 1.76 0.88 0.84 0.79 0.45 *At 298 K (25ºC)

8. 6-8 Sample Problem 6.4 Finding the Quantity of Heat from a Temperature Change PROBLEM:A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25ºC to 300.ºC? The specific heat capacity (c) of Cu is 0.387 J/g∙K. SOLUTION: PLAN:We know the mass (125 g) and c (0.387 J/g∙K) of Cu and can find  T in ºC, which equals  T in K. We can use the equation q = cm  T to calculate the heat. q = cm  T = 1.33 x 10 4 J  T = T final – T initial = 300. – 25 = 275ºC = 275 K 0.387 J g∙Kg∙K x 125 gx 275 K=

9. 6-9 Constant-Pressure vs. Constant Volume Calorimetry Constant-PressureConstant-Volume heat of reaction is equal to the enthalpy change heat of reaction is equal to the internal energy change –ΔH = q = specific heat x mass x Δtemperature ΔE = q = Δ temperature x heat capacity of calorimeter q = mcΔT q = CΔT Calorimeter “coffee cup” Bomb Calorimeter

10. 6-10 Figure 6.9Coffee-cup calorimeter. This device measures the heat transferred at constant pressure (q P ).

11. 6-11 Sample Problem 6.5 Determining the Specific Heat Capacity of a Solid PROBLEM:A 22.05 g solid is heated in a test-tube to 100.00ºC and added to 50.00 g of water in a coffee-cup calorimeter. The water temperature changes from 25.10ºC to 28.49ºC. Find the specific heat capacity of the solid. PLAN:Since the water and the solid are in contact, heat is transferred from the solid to the water until they reach the same T final. In addition, the heat given out by the solid (–q solid ) is equal to the heat absorbed by the water (q water ). SOLUTION:  T water = T final – T initial = (28.49ºC – 25.10ºC) = 3.39ºC = 3.39 K  T solid = T final – T initial = (28.49ºC – 100.00ºC) = –71.51ºC = –71.51 K

12. 6-12 Sample Problem 6.5 c solid = c x mass x  T H2OH2OH2OH2OH2OH2O mass solid x  T solid 4.184 J/g∙K x 50.00 g x 3.39 K 22.05 g x (–71.51 K) = – = 0.450 J/g∙K

13. 6-13 Sample Problem 6.6 Determining the Enthalpy Change of an Aqueous Reaction PROBLEM:50.0 mL of 0.500 M NaOH is placed in a coffee-cup calorimeter at 25.00ºC and 25.0 mL of 0.500 M HCl is carefully added, also at 25.00ºC. After stirring, the final temperature is 27.21ºC. Calculate q soln (in J) and the change in enthalpy,  H, (in kJ/mol of H 2 O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL, and that c = 4.184 J/g∙K PLAN:Heat flows from the reaction (the system) to its surroundings (the solution). Since –q rxn = q soln, we can find the heat of the reaction by calculating the heat absorbed by the solution.

14. 6-14 Sample Problem 6.6 SOLUTION: Total mass (g) of the solution = (25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g  T soln = 27.21ºC – 25.00ºC = 2.21ºC = 2.21 K q soln = c soln x mass soln x  T soln = (4.184 J/g∙K)(75.0 g)(2.21 K) = 693 J (a) To find q soln : (b) To find  H rxn we first need a balanced equation: HCl (aq) + NaOH (aq) → NaCl (aq) + H 2 O (l)

15. 6-15 Sample Problem 6.6 For HCl: 25.0 mL HCl x 1 L 10 3 mL 0.500 mol 1 L xx 1 mol H 2 O 1 mol HCl = 0.0125 mol H 2 O For NaOH: 50.0 mL NaOH x 1 L 10 3 mL xx 1 mol H 2 O 1 mol NaOH = 0.0250 mol H 2 O 0.500 mol 1 L HCl is limiting, and the amount of H 2 O formed is 0.0125 mol.  H rxn = q rxn mol H 2 O 0.0125 mol –693 J = 1 kJ 10 3 J x = –55.4 kJ/mol H 2 O

16. 6-16 Figure 6.10A bomb calorimeter. This device measures the heat released at constant volume (q V ).

17. 6-17 Sample Problem 6.7 Calculating the Heat of a Combustion Reaction PROBLEM:A manufacturer claims that its new dietetic dessert has “fewer than 10 Calories per serving.” To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O 2. The initial temperature is 21.862ºC and the temperature rises to 26.799ºC. If the heat capacity of the calorimeter is 8.151 kJ/K, is the manufacturer’s claim correct? PLAN:When the dessert (system) burns, the heat released is absorbed by the calorimeter: –q system = q calorimeter To verify the energy provided by the dessert, we calculate q calorimeter.

18. 6-18 Sample Problem 6.7 SOLUTION: 40.24 kJ x kcal 4.184 kJ 9.62 kcal or Calories The manufacturer’s claim is true, since the heat produced is less than 10 Calories.  T calorimeter = T final – T initial = 26.799ºC – 21.862ºC = 4.937ºC = 4.937 K q calorimeter = heat capacity x  T = 8.151 kJ/K x 4.937 K = 40.24 kJ =