Thermochemistry
Part 2: Calorimetry
Questions to ponder…

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).
Questions to ponder…
 How
much heat does it take to melt
an ice-cube?
Questions to ponder…
 How
much heat does it take to melt
an ice-cube?
Q=mc∆T,
but ∆T=0
Questions to ponder…
 How
much heat does it take to melt
an ice-cube?
But I
KNOW
q≠ 0
Questions to ponder…
 How
much heat does it take to melt
an ice-cube?
How can
I solve
this???
Calorimetry!!!
Heat required to melt ice (a.k.a. latent
heat of fusion) cannot be measured
directly, but calorimetry provides an
experimental method allowing this
heat transfer to be measured
indirectly.
Calorimetry!!!
Calorimetry: measurement of the
amount of heat evolved or absorbed
in a chemical reaction, change of
state or formation of a solution.
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
A calorimeter is a device used to
measure the heat absorbed or released
during a chemical or physical process
Coffee Cup Calorimeter
The cup is filled with water, which
absorbs the heat evolved by the reaction,
so:
qice = -qwater
OR
qrxn = -qcal
Coffee Cup Calorimeter

A more “high tech”
drawing…
Styrofoam
cup
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?
Magnitude of 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:
measurement
Heat (q)
Mass (m)
Specific Heat (c)
Final Temp (Tf )
Initial Temp(Ti)
“Cal”
(often water)
Object/”Rxn”
EXAMPLE 1: 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
(cal)
Heat
Mass
25.0 g
Specific Heat
4.184
Final Temp
26.4 oC
Initial Temp
25.0 oC
Pebble
(rxn)
The numbers
in these two
boxes are
always the
same, but with
different signs
(+/-). What
heat one lost,
the other
gained.
Hints:

The pebble lost heat 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
(cal)
Pebble
(rxn)
Heat
Mass
25.0 g
Specific Heat
4.184
Final Temp
26.4 oC
Initial Temp
25.0 oC
qcal = mwatercwaterTwater
qcal = (25.0g)(4.184J/goC)(26.4oC-25.0oC)
qcal = 150 J
qrxn = - 150 J
If the water ABSORBED 150 J of heat,
then the pebble RELEASED 150 J of heat.
Example 2: When 1.00 g of ammonium nitrate, NH4NO3, is added to 50.0 g of
water in a coffee cup calorimeter, it dissolves, NH4NO3 (s)  NH4+(aq) + NO3(aq), and the temperature of the water drops from 25.00C to 23.32C. Calculate
q for the reaction system.
cal
rxn
50.0 g
1.00 g
q
m
c
4.184
Tf
23.32 oC
Ti
25.00 oC
qcal = mwcwt
qcal = (50.0g)(4.18 J/gC)(-1.68C)
qcal = - 351 J
qrxn = - qcal
qrxn = 351 J (endothermic)
When one substance absorbs heat, the other substance
releases heat (energy)
Example 3
(similar to what you will do in tomorrow’s lab)

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
Example 3: 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.
cal (H2O)
rxn (Al)
2. Calculate q for water
q
m
c
100.00 g
1. Make chart
75.25 g
3. Q for Al is the same (but
with different sign) as q
for metal.
4. Using q metal, calculate c
metal
4.184
Tf
32.9 oC
32.9 oC
Ti
22.4 oC
99.3 oC
Example 3 cont’d: specific heat of aluminum…
cal (H2O)
rxn (Al)
qcal = mwcwT
qcal = (100.00g)(4.184J/gC)(10.5C)
q
4,393.2 J
m
100.00 g
c
4.184 J/gC
Tf
32.9 C
32.9 C
qrxn = - qcal
Ti
22.4 C
99.3 C
qrxn = -4,393.2 J
-4,393.2 J
75.25 g
qcal = 4,393.2 J
qrxn = mAlcAlT
-4,393.2 J = (75.25 g)(cAl)(-66.4C)
cAl = 0.879 J/gC

What if you wanted to measure the heat of a
reaction or process that couldn’t be measured in a
simple coffee cup calorimeter (e.g., heat of
combustion of Mg)?
You would need something like this…
Bomb Calorimeter
Another type of calorimeter is a
Bomb Calorimeter
NOTE: In a bomb calorimeter, heat is transferred from the
sample to the oxygen-enriched chamber, to the metal that
makes up the chamber, to the water… thus we cannot just
use the specific heat of water; instead heat capacity of
the calorimeter, Ccal, can be used or calculated.
If Heat Capacity (
C) is known
It is possible to calculate the amount of
heat absorbed or evolved by the reaction if
you know the heat capacity, Ccal, and the
temp. change, ΔT, of the calorimeter:
qcal =
C ΔT
cal
Everything else is the same (remember, the
heat lost from the reaction goes into the
calorimeter)
EXAMPLE 4: The reaction between hydrogen and chlorine, H2 + Cl2  2HCl, can be
studied in a bomb calorimeter. It is found that when a 1.00 g sample of H2 reacts completely,
the temp. rises from 20.00C to 29.82C. Taking the heat capacity of the calorimeter to be
9.33 kJ/C, calculate the amount of heat evolved in the reaction.
Cal
q
c
9.33 kJ/oC
Tf
29.82 oC
Ti
20.00 oC
Rxn
qcal = CcalΔT
qcal = (9.33 kJ/C)(9.82 C)
qcal = 91.6 kJ
qrxn = - 91.6 kJ
*** must have a negative sign if “heat
evolved in the reaction” = released
EXAMPLE 5: When 1.00 mol of caffeine (C8H10N4O2) is burned in air, 4.96 x 103 kJ of
heat is evolved. Five grams of caffeine is burned in a bomb calorimeter. The temperature is
observed to increase by 11.37C. What is the heat capacity of the calorimeter in J/C?
4.96 x 103 kJ is for 1.00 mol of caffeine. We are burning only 5.00 g of
caffeine to increase the temp by 11.37 oC. We FIRST need to figure out
how much heat energy is given off by just 5 grams.
 4.96  10 3 kJ
x

194.0g C 8H10N4 O2 5.00 g
Cal
q
+128 kJ
c
?
T
11.37 oC
Rxn
x  128 kJ
 128kJ  (Ccal )  (11.37C)
-128 kJ
C cal  11.3
kJ

C
 11,300
J

C
EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0.714 g/mL) is burned
in a bomb calorimeter, the temperature rises from 24.7C to 88.9C. The calorimeter heat
capacity is 10.34 kJ/C.
(a) What is q for the calorimeter?
(b) What is q when 20.0 mL of ether is burned?
(c) What is q for the combustion of one mole of ethyl ether?
Cal
q
c
10.34 kJ/oC
Tf
88.9 oC
Ti
24.7 oC
Rxn
qcal = CcalΔt
qcal = (10.34 kJ/C)(64.2 C)
qcal = 664 kJ
EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0.714 g/mL) is burned
in a bomb calorimeter, the temperature rises from 24.7C to 88.9C. The calorimeter heat
capacity is 10.34 kJ/C.
(a) What is q for the calorimeter?
(b) What is q when 20.0 mL of ether is burned?
(c) What is q for the combustion of one mole of ethyl ether?
qrxn = -qcal
qrxn = -664 kJ
EXAMPLE 6: When twenty milliliters of ethyl ether, C4H10O. (d=0.714
g/mL) is burned in a bomb calorimeter, the temperature rises from 24.7C to
88.9C. The calorimeter heat capacity is 10.34 kJ/C.
(a) What is q for the calorimeter?
(b) What is q when 20.0 mL of ether is burned?
(c) What is q for the combustion of one mole of ethyl ether?
664kJ 1 mL 74.0 g
3 kJ


 3.44  10 mol
20.0 mL 0.714 g 1 mol