Section 7.2—Calorimetry &
Heat Capacity
Why do some things get hot more quickly than others?
Temperature
Temperature – proportional to the
average kinetic energy of the molecules
Energy due to motion
(Related to how fast the
molecules are moving)
As temperature
increases
Molecules move
faster
Heat & Enthalpy
Heat (q)– The flow of energy from
higher temperature particles to lower
temperature particles
Enthalpy (H)– Takes into account the
internal energy of the sample along with
pressure and volume
Under constant pressure (lab-top conditions), heat and enthalpy
are the same…we’ll use the term “enthalpy”
Energy Units
The most common energy units are Joules (J) and
calories (cal)
Energy Equivalents
4.18 J
=
1.00 cal
1000 J
=
=
1 kJ
1000 cal
1 Cal (food calorie)
These equivalents can be used in dimensional analysis to convert units
Heat Capacity
Specific Heat Capacity (C) – The
amount of energy that can be absorbed
before 1 g of a substance’s temperature
has increased by 1°C
Cw (C for liquid water) =
1.00 cal/g°C or 4.18 J/g°C
Heat Capacity
High Heat Capacity
Low Heat Capacity
Takes a large amount of
energy to noticeably
change temp
Small amount of energy can
noticeably change
temperature
Heats up slowly
Cools down slowly
Heats up quickly
Cools down quickly
Maintains temp better with
small condition changes
Quickly readjusts to new
conditions
A pool takes a long time to warm up and remains fairly warm over night.
The air warms quickly on a sunny day, but cools quickly at night
A cast-iron pan stays hot for a long time after removing from oven.
Aluminum foil can be grabbed by your hand from a hot oven because it
cools so quickly
What things affect temperature change?
 Heat Capacity of substance
The higher the heat capacity, the slower the
temperature change
 Mass of sample
The larger the mass, the more molecules there are to
absorb energy, so the slower the temperature change
Specific heat capacity of
substance
H  m  C p  T
Energy added
or removed
Mass of sample
Change in temperature
Positive & Negative T
 Change in temperature (T) is always final
temperature – initial temperature
If temperature increases, T will be positive
 A substance goes from 15°C to 25°C.
 25°C - 15°C = 10°C
 This is an increase of 10°C
If temperature decreases, T will be negative
 A substance goes from 50°C to 35°C
 35°C – 50°C = -15°C
 This is a decrease of 15°C
Positive & Negative H
Energy must be put in for temperature to
increase
A “+” T will have a “+” H
Energy must be removed for temperature
to decrease
A “-” T will have a “-” H
Let’s Practice #1
Example:
How many joules must be
removed from 25 g of water
at 75°C to drop the
temperature to 30°? Cw =
4.18 J/g°C
Let’s Practice #1
Example:
How many joules must be
removed from 25.0 g of water
at 75.0°C to drop the
temperature to 30.0°? Cp
water = 4.18 J/g°C
H  m  Cw  T

H = change in energy
m = mass
Cw = heat capacity of water
T = change in temperature (Tf - Ti)



J

H  25.0 g   4.18

30
.
0

75
.
0
C
 
g C

H = - 4703J
Let’s Practice #2
If the specific heat capacity of
aluminum is 0.900 J/g°C, how
much heat must be added to a
30.0 g sample at 15°C to
increase its temperature to
30°C ?
Calorimetry
Conservation of Energy
1st Law of Thermodynamics – Energy
cannot be created nor destroyed in physical
or chemical changes
This is also referred to as the Law of
Conservation of Energy
If energy cannot be created nor destroyed,
then energy lost by the system must be
gained by the surroundings and vice versa
Calorimetry
Calorimetry – Uses the energy change
measured in the surroundings to find energy
change of the system
Because of the Law of Conservation of Energy,
the energy lost/gained by the surroundings is
equal to but opposite of the energy lost/gained by
the system.
Calorimetry
Hsurroundings = - Hsystem
(m×C×T)surroundings = - (m×C×T)system
Calorimetry
Calorimetry also uses the principle of
thermal equilibrium.
Thermal Equilibrium – Two objects at
different temperatures placed together
will come to the same temperature
An example of Calorimetry
Example:
A 23.8 g piece of unknown metal is heated to
100.0°C and is placed in 50.0 g of water at
24°C water. If the final temperature of the
water is 32.5°,what is the specific capacity of
the metal?
An example of Calorimetry
Example:
A 23.8 g piece of unknown metal is heated to
100.0°C and is placed in 50.0 g of water at
24°C water. If the final temperature of the
water is 32.5°,what is the heat capacity of
the metal?
Metal:
m = 23.8 g
Ti = 100.0°C
Tf = 32.5°C
Cm = ?
Water:
m = 50.0 g
Ti = 24°C
Tf = 32.5°C
Cw = 4.184 J/g°C
Step 1: Find the Heat (ΔH) of the water
ΔH = mCwΔT
ΔH = (50)(4.184)(32.5 – 24)
ΔH = 1778.2 J
Step 2: Remember that the water is the
“surroundings”. So, the heat gained by
the water is the same as the heat lost by
the metal but opposite in sign!!
Step 3: Since you know ΔH of the metal
system (-1778.2 J), use this the find Cm.
ΔH = mCmΔT
Now use the values for the metal!
-1778.2 = (23.8)Cm(32.5 – 100)
-1778.2 = (23.8)(-67.5)Cm
(23.8)(-67.5)
Cm = 1.11 J/goC
Let’s Practice
Example:
A 10.0 g of aluminum at 95.0°C is placed in
a container of 100.0 g of water (C = 4.184
J/g°C) at 25.0°. The metal and water reach
a final temperature of 26.5 oC. What is the
specific heat of the metal?