The document describes different types of calorimeters used to measure heat changes in chemical reactions, including bomb calorimeters and coffee cup calorimeters. Bomb calorimeters allow measurement of combustion reactions under constant volume conditions, while coffee cup calorimeters measure heat changes under constant pressure. Examples are provided demonstrating how to use calorimetry data to determine the heat of reaction for a combustion reaction and a neutralization reaction. Both examples show the reactions are exothermic as heat is released.

1. GROUP MEMBER’S:
CHANG PEI XIAN D 20101037455
CHAN SU FANG D 20101037450
LECTURER :PN. FARIDAH YUSOFF

2. INTRODUCTION OF
CALORIMETER
A calorimeter (from Latin calor, meaning heat) is a
device used for calculate the enthalpy change of
reaction, the science of measuring the heat of
chemical reactions or physical changes as well as heat
capacity.
The most common types of calorimeter is the bomb
calorimeter(constant volume calorimeter) and coffee
cup(constant pressure calorimeter).
The other types include differential scanning
calorimeters, isothermal micro calorimeters, titration
calorimeters and accelerated rate calorimeters.

3. BASIC CONCEPT IN APPLICATION OF
CALORIMETER
Endothermic process occurs in which the
system absorbs heat.
Exothermic process occurs in which a system
loses heat.
The enthalpy change that accompanies a
reaction is called the enthalpy of reaction, or
merely the heat of reaction, and is sometimes
written as ∆Hrxn, where rxn means reaction.

4. Since , ∆H = H final – H initial, the enthalpy change
for a chemical reaction is given by the enthalpy
of the products minus the enthalpy of the
reactants:
∆H = H products – H reactants
Typically, we can determine the magnitude of the
heat flow produces.
The measurement of heat flow is
calorimetry, and the device used to measure heat
flow is a calorimeter.
The heat capacity of an object is the amount of
heat needed to raise the temperature by 1 K (or 1
C).

5. The heat capacity of one gram of a substance is called
its specific heat capacity, or merely is called its
specific heat.
The specific heat, Cs, of a substance can be
determined experimentally by measuring the
temperature change, ∆T, that a known mass, m, of the
substance undergoes when it gains or loses a specific
quantity of heat, q:
Specific heat =
(quantity of heat transferred)
(grams of substance) x (temperature change)
Cs = q
m x ∆T

6. BOMB CALORIMETER(CONSTANT-
VOLUME CALORIMETER)

7. INTRODUCTION/THEORY
Combustion reactions are most conveniently
studied using a bomb calorimeter.
Combustion is a process which an organic
compound reacts completely with excess
oxygen.
Heat is released when combustion occurs. This
heat is absorbed by the calorimeter
contents, causing a rise in the temperature of
the water.

8. To calculate the heat of combustion, we must know the
total heat capacity of the calorimeter, Ccal. This quantity is
determined by measuring the resulting temperature
change.
For example, the combustion of exactly 1 g of benzoic
acid, C6H5COOH in a bomb calorimeter produces 26.38
kJ of heat. It increases the temperature by 4.857 0C. The
heat capacity of the calorimeter is then given by Ccal =
26.38 kJ / 4.8570C = 5.341 kJ/0C. By knowing the value of
Ccal , we can measure temperature changes produced by
other reactions , and then we can calculate the heat
evolved in the reaction, qrxn .

9. qrxn = Ccal x T
Since the reactions in a bomb calorimeter are carried out
under constant-volume conditions, the heat transferred
corresponds to the change in internal energy, E, rather
than the change in enthalpy, H .

10. EXAMPLE 1
Using Bomb Calorimeter Data to Determine a Heat of
Reaction
The combustion of 1.010 g sucrose, C12H22O11, in a bomb
calorimeter causes the temperature to rise from 24.92 to
28.33 C. The heat capacity of the calorimeter assembly is
4.90 kJ/ C.
a) What is the heat of combustion of sucrose expressed
in kilojoules per mole of C12H22O11?
b) Verify the claim of sugar producers that one teaspoon
of sugar (about 4.8 g) contains only 19 Calories.

11. Solution:
Calculate the qCalorimeter with the equation
qCalorimeter = heat capacity of calorimeter x ∆ T
qCalorimeter = 4.90 kJ/ C x (28.33 – 24.92) C
= (4.90 x 3.41) kJ
= 16.7 kJ
qreaction = - qCalorimeter = -16.7 kJ
The heat of combustion of the 1.010 g sample is -16.7kJ.
Per gram C12H22O11:
qreaction = - 16.7 kJ/ 1.010 g C12H22O11
= -16.5 kJ/ g C12H22O11
Per mole C12H22O11:
qreaction = - 16.5 kJ/ g C12H22O11 x ( 342.3 g C12H22O11/ 1 mol
C12H22O11)
= - 5.65 x 103 kJ/ mol C12H22O11

12. COFFEE-CUP(CONSTANT-
PRESSURE CALORIMETR)

13. INTRODUCTION/THEORY
A very simple “ coffee-cup” calorimeter , is
often used in general chemistry labs to
illustrate the principles of calorimeter.
This is because the calorimeter is not
sealed, the reaction occurs under the essentially
constant pressure of the atmosphere.

14. We can assume that the calorimeter itself does not absorb
heat.
For an exothermic reaction, heat is “lost” by the reaction
and “gained” by the solution, so the temperature rises.
For an endothermic reaction, heat is “lost” by the solution
and “gained” by the reaction, so the temperature drops.
The heat gained by the solution, qsoln equal to qrxn.
qsoln = -qrxn
qsoln = -qrxn = mc T

15. EXAMPLE 1
Determining a Heat of Reaction from Calorimetric Data
Two solutions, 25.00 mL of 2.50 M HCI(aq) and
25.00 mL of 2.50 M NaOH(aq), both initially at
21.1 C, are added to a Styrofoam-cup calorimeter and
allowed to react. The temperature rises to 37.8 C.
Determine the heat of the neutralization
reaction, expressed per mole of H2O formed. Is the
reaction endothermic or exothermic?

16. Solution:
Because the reaction is a neutralization reaction, let us
call the heat of reaction qneutralization. Now, according to
equation qneutralization = - q calorimeter
q calorimeter = 50.00 mL x (1.00 g/mL) x (4.18 J/g C) x
(37.8 – 21.1) C
= 3.5 x 103 J
q neutralization = - q calorimeter
= - 3.5 x 103 J
= - 3.5 kJ

17. In 25.00 mL of 2.50 M HCI, the amount of H+ is
? mol H+ = 25.00 mL x (1 L/1000 mL) x (2.50 mol/ 1 L) x (1 mol
H+/ 1mol HCI)
= 0.0625 mol H+
So , in 25.00 ml of 2.50 M NaOH there is 0.0625 mol of OH- .
The H+ and the OH- combine to form 0.0625 mol H2O
The amount of heat produced per mole of H2O is
q neutralization = ( - 3.5 kJ/0.0625 mol H2O)
= - 56 kJ/ mol H2O
b)Because q neutralization is a negative value, the neutralization
reaction is exothermic.