THERMOCHEMISTRY:
HEAT and CHANGE
When a material is heated (or
cooled), it can undergo one of these
changes:

Its temperature changes
OR

Its physical state changes
Type I : Changes In Temperature
Heating and Cooling
Heating or Cooling Only


Involves only an increase or
decrease in temperature
No change in state involved
Formula: Q = m C T
Where q = heat, in cal, J, or kcal,kJ
m = mass, in g or kg
c = specific heat capacity , (value
depends on the substance)
 T = temperature change (final
temperature - initial temperature)
The amount of heat energy that
must be supplied so as to warm a
material depends on three things:
Mass
Specific heat capacity
Temperature change
m
C
T
Which will need more heat in order
to boil?
A cup of water
at room temperature?
Or a bucket of water
at room temperature?
You will need more heat energy
to warm an object with a
bigger mass.
(assuming you have the same material
and the same temperature change)
Which needs more heat to warm up
to 75 degrees Celsius?
1 lb. of water
at room temp?
Or 1 lb. of iron
at room temp?
Different materials have different
Specific Heat capacities (C)

Iron ( C= 0.449)
To warm 1 gram of iron from 10 to 11 deg
Celsius, you must supply 0. 449 joules

Water (C = 4.2)
To warm 1 gram of water from 10 to 11 deg
Celsius, you must supply 4.2 joules

Specific Heat Capacity is measured in
joules / g deg Celsius
Some materials just warm up faster
than others.
You will need more heat to warm a material
which has a high specific heat capacity
(assuming the two materials have the same
mass and same temperature change)
Which will need more heat energy to
warm?
A pound of water from
20˚C to 30˚C
Or a pound of water from
20˚C to 130˚C
The bigger the difference in
temperature, the more the heat energy
needed to warm the material
Difference in temperature is represented by
the symbol ΔT. This is calculated by :
ΔT = Tf - Ti
where Tf represents the final temperature
and Ti represents the initial temperature
Calculate the amount of heat needed
to raise the temperature of 1.2g of
water from 100C to 200C. Specific heat
of water is 4.2 J/g ºC.
Q = mCT
= 1. 2 g ( 4.2 J/g oC ) ( 20 oC – 10 oC)
= 1.2 g ( 4.2 J/g oC) (10 oC)
= 50.4 J
Type 2 : Changes In State
* Freezing/Melting
* Vaporization/Condensation
Changes in State (Phase)
• Most substances can exist in three states—
solid, liquid, and gas—depending on the
temperature and pressure.
When energy is added to or taken
away from a system, one phase can
change into another.
Some phase changes NEED energy ( + Q)
* melting (fusion)
* evaporation (vaporization)
Some phase changes RELEASE energy (-Q)
* freezing (solidification)
* condensation
To simplify this graphic:
Melting
(Fusion)
SOLID ↔
Vaporization
(Evaporation)
LIQUID
Freezing
Condensation
(Solidification)
↔
GAS
What are the terms for each of the heat change
(ΔH) associated with each process?
Heat of Fusion
Heat of Vaporization
Melting
(Fusion)
SOLID ↔ LIQUID
Freezing
( Solidification)
Condensation
Heat of Solidification
Vaporization
(Evaporation)
↔
GAS
Condensation
Heat of
The processes going to the right (melting and
_________) are endothermic. They _______
energy. ( Q has a _____ sign )
The processes going to the left (_______
and condensation) are exothermic. They
________ energy. ( Q has a ______ sign )
The processes going to the right (melting and
vaporization) are endothermic. They
need energy. ( Q has a positive sign )
The processes going to the left (freezing
and condensation) are exothermic. They
release energy. ( Q has a negative
sign )
Formulas for Heat Problems
involving Changes of State
#1 Melting (or fusion)
: solid  liquid
Freezing (or solidification) : liquid solid
Formula :
Q = n ΔHf
Q = amount of heat absorbed or released
n = number of moles
ΔHf = molar heat of fusion
Molar Heat of Fusion ΔHf
Molar Heat of Fusion- amount of energy needed to
change 1 mole of solid to liquid at its melting
temperature
 Different materials have different ΔHf
Ex: Molar Heat of Fusion of H2O = 6.01 kJ/mol
Molar Heat of Fusion of Lead = 4.77 kJ/mol
 Use ΔHf for melting (fusion) and freezing
problems . Remember ΔHf is positive for melting
but negative for freezing.

MELTING
Sample Problem 1: What is the amount of heat
needed to melt 4 moles of ice at its melting point?
(ΔHf of water(ice) = 6.01 kJ/mol)

Q=
n ΔHf
= (4 moles) (6.01 kJ/mole)
= 24.04 moles
MELTING
Sample problem 2: What is the amount of energy
needed to melt 5 grams of ice at its melting point?
(ΔHf of water = 6.01 kJ/mol)


Q = n ΔHf
Number of moles:
5 grams
1 mole
1
18.01 grams
or 0.28 mol
Q = nΔHf = (0.28 mol)(6.01 kJ/mol)
= 1.68 kJ
FREEZING






The reverse process of melting (freezing)
RELEASES energy
Use the same ΔHf as in melting but make the sign
negative
Sample 1: How much energy is released
when 5 moles of water freezes at 0 deg
Celsius? (ΔHf of water = -6.01 kJ/mol)
Q = n ΔHf
= (5 moles) ( -6.01 kJ/mol)
= -30.05 kJ
Sample problem 2: How much energy is
released when 50 grams of water
freezes at 0 degrees Celsius?

Q = nΔHf
n:
50 g
1 mole
1
18.01 g
Q = (2.8 mol)( - 6.01 kJ/mol)
Q = - 16.8 kJ
Formulas for Heat Problems
involving Changes of State
#2
Vaporization :
Condensation:
Formula:

liquid  gas
gas liquid
Q= n ΔHv
Where n = no. of moles
Hv = heat of vaporization
Molar Heat of Vaporization ΔHv

Molar Heat of Vaporization- amount of energy needed
to change 1 mole of liquid to gas at its boiling
temperature

Different Materials have different ΔHv
Molar Heat of vaporization of water= 40.6kJ/mol
Molar Heat of vaporization of ethanol = 38.6kJ/mol

Use ΔHv for vaporization (evaporation) and
condensation problems . Remember ΔHv is positive for
vaporization (evaporation) but negative for
condensation.
VAPORIZATION
Sample problem 1: How much heat in kJ must be
absorbed by 1.5 moles of water in order to
evaporate completely at its boiling point? (ΔHv of
water = 40.6 kJ/mol)



Q = n ΔHv
Q = (1.5 mol)(40.6 kJ/mol)
Q = 60.9 kJ
VAPORIZATION
Sample Problem 2: How much heat in kJ must be
absorbed by 5 grams of water to evaporate
completely at 100 °Celsius?
(ΔHv of water = 40.6 kJ/mol)



Q = nΔHv
Number of moles:
5 grams
1 mole
1
18.01 grams or .28 mol
Q = nΔHv = (0.28 mol)(40.6 kJ/mol)
= 11.37 kJ
CONDENSATION




The reverse process of vaporization (condensation)
RELEASES energy
Use the same ΔHv as in vaporization but make the sign
negative
Example 1: Calculate the amount of heat released
when 2 moles of water vapor change completely to
liquid water at 100°C ?
(ΔHv of water = 40.6 kJ/mol)
Q = nΔHv
Q = (2 mol)( - 40.6 kJ/mol)
Q = - 81.2 kJ

Example 2: Calculate the amount of heat
released when 50g of water vapor
changes completely to liquid water at its
boiling point? (Δ Hv of water = 40.6
kJ/mol)
Q = nΔHv
n: 50 g
1 mole
1
18.01 g
Q = (2.8 mols) ( -40.6 kJ/mol)
= -113.68 kJ