Calorimetry
CP Unit 9
Chapter 17
CALORIMETRY

The enthalpy change associated with a
chemical reaction or process can be
determined experimentally.

Measure the heat gained or lost during a
reaction at CONSTANT pressure
Calorimeter

Device used to
measure the heat
absorbed or released
during a chemical or
Styrofoam
physical process
cup
Example

If you leave your keys and your chemistry
book sitting in the sun on a hot summer
day, which one is hotter?

Why is there a difference in temperature
between the two objects?
Because…

Different substances have different
specific heats (amount of energy needed
to raise the temperature of 1 g of a
substance by 1 degree Celsius).
What happens in a calorimeter

One object will LOSE heat, and the other will
ABSORB the heat

System loses heat to surroundings = EXO = -q

System absorbs heat from surroundings = ENDO = +q
When
a hot chunk of metal is dropped in a cool glass
of water, the metal cools off. Where did the heat from
the metal go?
Did the metal lose more heat then the water gained?
HEAT GAINED = HEAT LOST (ALWAYS!)
The numbers in these two boxes are
always the same, but with different signs
(+/-). What heat one lost, the other
gained.
To do calorimetry problems…

Make a Chart:
Water
Heat
Mass
Specific Heat
Final Temp
Initial Temp
4.184
Object/Reaction
EXAMPLE: A small pebble is heated and placed in a
foam cup calorimeter containing 25.0 g of water at 25.0
C. The water reaches a maximum temperature of 26.4
C. How many joules of heat were released by the pebble?
The specific heat of water is 4.184 J/g C.
Water
Heat
Mass
Specific Heat
25.0 g
4.184
Final Temp
26.4 oC
Initial Temp
25.0 oC
Pebble
(#13)
The numbers
in these two
boxes are
always the
same, but with
different signs
(+/-). What
heat one lost,
the other
gained.

The pebble because the water heated up from
25.0 C to 26.4 C.

Pebble loses heat (-q, exothermic) while water
gains heat (+q, endothermic)

Do you calculation based on water (since the
problem gave all the water’s information)

Water
Pebble
Heat
Mass
Specific Heat
25.0 g
4.184
Final Temp
26.4 oC
Initial Temp
25.0 oC
qwater = mwatercwaterTwater
qwater = (25.0g)(4.184J/goC)(26.4oC-25.0oC)
qwater = 150 J
If the water ABSORBED 150 J of heat,
then the pebble _______ ______ J of heat.
qpebble = - 150 J
Example 2 (LAB type of CALC)

Suppose that 100.00 g of water at 22.4 °C
is placed in a calorimeter. A 75.25 g
sample of Al is removed from boiling
water at a temperature of 99.3 °C and
quickly placed in a calorimeter. The
substances reach a final temperature of
32.9 °C . Determine the SPECIFIC HEAT of
the metal.The specific heat of water is
4.184 J/g C.
 MAKE YOUR CHART
Suppose that 100.00 g of water at 22.4 °C is placed in a calorimeter. A
75.25 g sample of Al is removed from boiling water at a temperature of
99.3 °C and quickly placed in a calorimeter. The substances reach a
final temperature of 32.9 °C . Determine the SPECIFIC HEAT of the
metal.The specific heat of water is 4.184 J/g C.
1. Make chart
Water
Pebble
Heat
Mass
Specific Heat
Final Temp
Initial Temp
100.00 g
75.25 g
4.184
32.9 oC
22.4 oC
32.9 oC
99.3 oC
2. Calculate q for
water
3. Q for water is
the same (but
with different
sign) as q for
metal.
4. Using q metal,
calculate c
metal
0.879 J/g °C
Example 3

A lead mass is heated and placed in a
foam cup calorimeter containing 40.0 g of
water at 17.0C. The water reaches a
temperature of 20.0 C.
How many joules of heat were released by
the lead?
The specific heat of water is 4.184 J/g C.
502 J = 5.0 x 102 J for 2 sig figs