1. Thermochemistry
The study and measurement of the heat involed or absorbed during chemical reactions.
Heat  Temperature
Temperature = average kinetic energy
which does depend on heat, but is not the same
Energy

2. Heat vs. Temperature
Heat = Energy that flows between two samples of matter due to differences in temperature.
¿Which has more heat?
a cup of coffee or a bathtub filled with coffee
both at 170 oF? (which takes longer to heat up?)
¿Carpet or hard wood floor in the same room?
Energy

3. Calorimetry Calorimetry: calor (L) + metry (Gr)
The measurement of heat changes.
Calorimetry

4. (e.g., the chemical bonds broke when wood burns
Calorimetry
Essential Terms:
Specific Heat (s)
 amount of heat required to raise the temperature of 1 gram of substance by 1oC.
intensive property
The higher the specific heat, the more energy it takes to heat up the sample. And, the slower it loses heat.
(N.B., Temperature = average kinetic energy is affected by heat. But also includes potential energy
(e.g., the chemical bonds broke when wood burns
Calorimetry

5. Specific Heat (s) q = m s DT m = mass (g)
Calorimetry
Essential Equations:
Specific Heat (s) q = m s DT
m = mass (g)
s = specific heat ( amt. heat required to raise the temperature of 1 gram of a substance by 1oC (J/g oC)
DT = change in temperature (oC)
q = heat associated with reaction (‘ – q’ = exothermic
‘+ q’ = endothermic) ‘
SI Units: Joule(J) 1 J = calories
food Calorie is 1,000 calories
Calorimetry

6. Two Types of Calorimeters:
Calorimetry
Two Types of Calorimeters:
Constant-Volume Calorimeter
used for combustion reaction (e.g., Calories in food)
Constant-Pressure Calorimeter
used for other reactions
(e.g., heat pack: CaCl2 + H2O)
we use this in class
Calorimetry

7. A 394-g sample of water is heated from 10.75°C to 83.20°C.
Specific Heat & Temperature of a Single Substance
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
Check
Sign (+) means heat was absorbed; water was heated; Reaction was endothermic so ‘q’ should be ‘+’
Calorimetry

8. Measuring Heat Change:
qsys = qcal + qrxn = 0
(conservation of energy)
qcal = heat capacity of calorimeter
(assume all q is transferred to water)
qrxn = –qcal
 mA sA DTA = –mB sB DTB
Calorimetry

9. Constant-Pressure Calorimetry Example 6.7
A lead (Pb) pellet having a mass of g at 89.98°C was placed in a constant-pressure calorimeter of negligible heat capacity containing 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!)
Calorimetry

10. mPb sPb DTPb = – mH2O sH2O DTH2O
Constant-Pressure Calorimetry
Summary
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.
mPb sPb DTPb = – mH2O sH2O DTH2O
Calorimetry

11. Constant-Pressure Calorimetry
Solution
Calorimetry

12. Fin
Calorimetry