1. Calorimetry Calorimetry: calor (L) + metry (Gr) The measurement of heat changes. food Calories (C) = 1,000 cal (1 cal = 4.184 J; we use Joules in chemistry) 1 cal = heat required to raise the temperature of 1 gram of water by 1 o C. q sign is the same as before (‘–’ exothermic)

2. Calorimetry Essential Terms: Calorimetry Specific Heat (s)  amount of heat required to raise the temperature of 1 gram of substance by 1 o C. intensive property Heat Capacity (C)  amount of heat required to raise the temperature of a specific system by 1 o C. Comparison: s = heat needed to raise 1 gram of water by 1 o C. C = heat needed to raise a glass of ice, tea, &glass by 1 o C.

3. Calorimetry Essential Equations: (q=heat; m=mass;  Temp) Calorimetry Specific Heat (s)q = m s  T Heat Capacity (C) q = C  T (C = m s) Heat Exchange (2 substances: A & B) m A s A  T A = – m B s B  T B

4. Calorimetry Two Types of Calorimeters: Calorimetry Constant-Volume Calorimeter used for combustion reaction (e.g., Calories in food) Constant-Pressure Calorimeter used for other reactions (e.g., heat pack: CaCl 2 + H 2 O) we use this in class

5. Calorimetry Example 6.5 A 394-g sample of water is heated from 10.75°C to 83.20°C. Calculate the amount of heat absorbed (in kilojoules) by the water. Solution Specific Heat & Temperature of a Single Substance Check Sign (+) means heat was absorbed; water was heated; Reaction was endothermic so ‘q’ should be ‘+’

6. Calorimetry Example 6.6 A quantity of 1.274 g of naphthalene (C 10 H 8 ), a pungent- smelling substance used in moth repellants, was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 21.49°C to 26.52°C. If the heat capacity of the bomb plus water was 10.17 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion. Constant-Volume Calorimetry

7. Calorimetry Strategy Knowing mass,  T, and C: 1.How do we calculate the heat absorbed by the calorimeter (q cal ) ? 2.What is the heat generated by the combustion of 1.274 g of naphthalene (q rxn ) ? 3.What is the conversion factor between grams and moles of naphthalene? Example 6.6 – Problem Summary m =1.274 g of C 10 H 8  T=21.49°C to 26.52°C; C=10.17 kJ/°C; q = ? kJ/mol

8. Calorimetry 1.What is the heat absorbed by the calorimeter (q cal )? Solution The sum of the heat absorbed by the container and the water is equal to the product of the heat capacity and the temperature change. (Assume no heat loss.)

9. Calorimetry 2. What is the heat generated by the combustion of 1.274 g of naphthalene? Solution The heat absorbed by the calorimeter was released by the chemical reaction.

10. Calorimetry 3.What is the conversion factor for naphthalene between grams and moles?

11. Calorimetry Measuring Heat Change: q sys = q cal + q rxn = 0 (conservation of energy) q cal = heat capacity of calorimeter (assume all q is transferred to water) q rxn = –q cal  m A s A  T A = –m B s B  T B

12. Calorimetry Example 6.7 A lead (Pb) pellet having a mass of 26.47 g at 89.98°C was placed in a constant-pressure calorimeter of negligible heat capacity containing 100.0 mL of water. The water temperature rose from 22.50°C to 23.17°C. What is the specific heat of the lead pellet? Strategy Draw the initial and conditions. (It really does help!) Constant-Pressure Calorimetry

13. Calorimetry Strategy Treat the calorimeter as an isolated system. We know the masses of water and the lead pellet as well as the initial and final temperatures. Assuming no heat is lost to the surroundings, we can equate the heat lost by the lead pellet to the heat gained by the water. Knowing the specific heat of water, we can then calculate the specific heat of lead. Summary Constant-Pressure Calorimetry m Pb s Pb  T Pb = – m H2O s H2O  T H2O

14. Calorimetry Constant-Pressure Calorimetry Solution

15. Fin Calorimetry