Chapter 16
Calorimetry
I
II
III
Calorimetry
 The
study of heat flow
and heat measurement.
Heat Capacity
 The
amount of heat
needed to raise the
temperature of an
object by 1 Celsius
degree.
Specific Heat
 The
heat capacity of 1
gram of a substance
 Specific
Heat of liquid
water is 4.184 J/gºC
 4.184
J = 1 calorie
1
Calorie = 1000 calories = 1
kilocalorie
1
Food Calorie = 1000 calories
Calorimetry Experiments
 Determine
the heats of
reaction (ENTHALPY
CHANGES) by making
accurate measurements
of temperature changes
using a calorimeter.
 Scientists
use q to denote
measurements made in a
calorimeter.
 Heat
transferred in a reaction is
EQUAL, but OPPOSITE in sign
to heat absorbed by the
surroundings.

qrxn = - qsur
HEAT
qsur = m x Cp x (Tf –Ti)
Mass of Water
Specific heat
of Water
Temperature
change
Practice Problem: #1
When a 12.8g sample of KCl dissolves
in 75.0g of water in a calorimeter, the
temperature drops from 31.0ºC to
21.6ºC. Calculate H for the process.
KCl(s) 
K+
(aq)
+ Cl
(aq)
First, calculate qsur and then calculate H.
Practice Problem: #1
When a 12.8g sample of KCl dissolves
in 75.0g of water in a calorimeter, the
temperature drops from 31.0ºC to
21.6ºC. Calculate H for the process.
KCl (s) 
K+
(aq)
+ Cl
(aq)
qsur = m x Cp x (Tf –Ti)
qsur = 75.0g x 4.184 J/gºC x (21.6 ºC –31.0 ºC )
qsur = -2949.72 J
qsur is negative (as expected)
based on the temperature drop of
the water.
qrxn = - qsur = +2949.72 J
qrxn represents the heat absorbed
due to the reaction of 12.8g KCl.
+
KCl  K
+ Cl
(s)
(aq)
(aq)
Now you must convert the KCl to moles.
Convert grams KCl to moles.
12.8 g KCl 1 mol KCl
= 0.173 mol KCl
74 g KCl
Calculate H for the reaction
+
KCl  K
+ Cl
(s)
H =
(aq)
+2949.72 J
(aq)
x 1 mol KCl
0.173 mol KCl
H = +17050 J
H = +17.1 kJ
Coefficient from
balanced
equation
H Must be positive because
it was an endothermic reaction!
Practice Problem: #2
What is the specific heat of aluminum
if the temperature of a 28.4 g sample
of aluminum is increased by 8.1ºC when
207 J of heat is added?
qsur = m x Cp x (Tf –Ti)
207J = 28.4g x Cp x 8.1oC
Cp =
207J
28.4 g x 8.1 oC
= 0.90 J/g oC