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
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
Calorimetry
• Calorimetry:
• calor (L) + metry (Gr)
• The measurement of heat changes.
Calorimetry
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
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:
Calorimetry
Joule(J)
1 J = 4.184 calories
food Calorie is 1,000 calories
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
• 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
q  m s DT
J
1kJ
o
q = (394g)(4.184 o )(83.20 -10.75 C)(
)
g× C
1000J
q =119kJ
Check
Sign (+) means heat was absorbed; water was heated;
Reaction was endothermic so ‘q’ should be ‘+’
Calorimetry
• 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
• Constant-Pressure 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!)
Calorimetry
• 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.
m s DT = – m s DT
Pb Pb
Calorimetry
Pb
H2O H2O
H2O
• Constant-Pressure Calorimetry
Solution
Calorimetry
Fin
Calorimetry