Heat and Thermodynamics
Review

What are the three phases of matter?
The Phases
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There are three basic phases of matter: solid, liquid, and
gas.
We can change between these phases:
A.
B.
C.
D.
E.
F.
Solid changing to a gas is called sublimation
Liquid changing to a gas is called evaporation
Gas changing to a liquid is called condensation
Solid changing to a liquid is called melting
Liquid changing to a solid is called freezing
Gas changing to a solid is called deposition
The Phases
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At certain temperatures matter can be in multiple
forms
H2O (l)  H2O (s) is the chemical equation for
water when it is melting or freezing.
Pure water has a melting/freezing point of 0°C.
Above 0°C melting will happen & the equation
goes from right to left.
Below 0°C freezing will happen & the equation
goes from left to right.
The Experiment

You have two beakers on hot plates.
One beaker contains one ice cube.
The other beaker contains eight ice
cubes. For each beaker, the
temperature and the time were
recorded as the ice cubes melted.
The Data
The Graphical Representation
What does this all mean?

Segment 1 (melting point): At the solid
surface melting is occurring as molecules
slip by one another. Energy is used for the
phase change at the solid surface so
temperature stays constant. The type of
particle motion is changing, not the speed of
the particles.
What does this all mean?

Segment 2 (liquid heating): All
molecules are moving past each other.
All the increasing heat energy is used
to increase the velocity of the particles.
Kinetic energy increases so
temperature goes up.
What does this all mean?

Segment 3 (boiling point): Energy is
used to make more & more of the
particles escape the liquid. Escaping
particles form water vapor. Energy is
used for the phase change at the liquid
surface so temperature stays constant.
What does this all mean?

The Slope: Why is the slope of the curve
larger for the beaker with 1 ice cube than for
the beaker with 8 ice cubes? There are
more particles in the beaker so it will take
longer for them to melt & heat.

Temperature: The temperatures at melting
& boiling were the same for both beakers.
Temperature is a measure of the average
kinetic energy of the particles in a substance.
What does this all mean?

Heat: The beaker with 8 ice cubes
needed more heat to reach the melting
& boiling points because it had more
particles. Heat is a measure of the
total amount of energy transferred
from an object of high temperature to
low temperature.
Kinetic Energy
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KE = ½m x v2
M stands for mass
V stands for velocity
KE = ½m x v2 = (kg) x (m/s2) = Joules
Joules (kg•m/s2) are the unit for
energy.
Heat Capacity

Specific heat capacity is the amount of
energy needed to raise the temperature of
1 gram of a substance by 1°C.

The lower the specific heat of a substance,
the faster the object heats up.

Specific heat capacity is abbreviated Cp &
has units of J/g°C.
What does a joule of energy look like?
1 gram of pure ethyl alcohol produces
29,700 joules of energy
The average 18 year old creates 14,000
joules of energy per day. This energy
keeps your body at 98.6º F.
How do I calculate how much heat
an object holds?

The Greek letter delta, Δ, means change in.
ΔT means change in temperature.

Change in temperature is usually found by
taking the final T & subtracting the initial T

ΔT = Tfinal – Tinitial = T2 – T1
Cont..
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The formula for calculating change in heat of
a certain mass of a substance in a given
phase (s, l, g) through a temperature range is:
H = mCpΔT
H= heat; m= mass; Cp = specific heat;
ΔT = change in temperature
The units are (Grams) (°C) (J/g°C) = Joules
(g) (°C) ( J ) = J
g°C
Example 1
Calculate the joules of energy required to heat
454 g of water from 5.4°C to 98.6°C.
The specific heat of liquid water is 4.184 J/gºC.
H = mCpΔT
H = 454 g * 4.184 J * (98.6 ºC – 5.4ºC) = 93.2ºC
1
gºC
1
H = 177036.76 J
Example 2
A 2.80 g sample of pure metal requires
10.1 J of energy to change temperature
from 21ºC to 36ºC. What is the metal?
Example 2- solution
H = mCpΔT
10.1 J = 2.80 g * Cp * (36ºC – 21ºC)
10.1 J = 2.80 g * Cp * 15ºC
0.240= Cp
Cp = 0.240 J/g°C
SILVER
SPECIFIC HEAT VALUES
Substance
Specific Heat
Capacity (J/g°C)
Aluminum (s)
Iron (s)
Mercury (l)
Carbon (s)
Silver (s)
Gold (s)
Tungsten (s)
Copper (s)
0.89
0.45
0.14
0.71
0.24
0.13
0.132
0.385

Define the term exothermic.

Heat or energy is released- EXiting

Define the term endothermic.

Heat or energy is absorbed- ENtering
ΔH


Heat of fusion (ΔHfus) is the energy
needed to melt 1 mole of a substance or
joules needed to melt 1 gram.
Heat of vaporization (ΔHvap) is the energy
in kJ needed to vaporize 1 mole of a
substance or joules needed to vaporize 1
gram.
Equations for Heat
FILL in your diagram
-
H in J
condensation
evaporation
freezing
melting
+
H in J
Write this on the back of your diagram

Words to watch for:








This means:
Is heated


Is cooled

“ from ______

“ to ________

“ through _______
Heat energy is released 
Heat energy is absorbed
Up the stairs = +J, Exo
Down the stairs = -J Endo
Start here
End here
INCLUDE this step
Exo, + J
Endo, - J
Example 3
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
Find the heat released as 10.5g H2O cools
from 130°C to -15°C.
Step 1- Fill in MP and BP for H2O
100°C
0°C
-15°C
130°C
Given information:
MP = 0°C
BP = 100°C
Hfus = 334 J/g
Hvap = 2260J/g
Cp solid = 2.07 J/g°C
Cp liquid = 4.18 J/g°C
Cp gas = 2.04 J/g°C
Example 3: Solution
Step 1: Find the heat needed to be released
from gaseous (steam) H2O to reduce the temperature
of the gas.
m * Cp(gas) * ΔT ; ΔT= T final – T inital
10.5 g * 2.04 J * (100°C – 130°C)
1
gºC
10.5 g * 2.04 J * -30°C = -642.6 J
1
gºC
Example 3: Solution
Step 2: Find the amount of heat that needs to
be released to condense the steam into a liquid.
m * Hvap
10.5 g * 2260 J = - 23730 J
1
g
Hvap is – because we
are doing DOWN the
stairs
Example 3: Solution
Step 3: Calculate how much heat needs to
be released to decrease the temperature of
the liquid water.
m* Cp(liquid) * ΔT
10.5 g * 4.18 J * (0°C-100°C)
1
gºC
10.5 g * 4.18 J * -100°C = -4389 J
1
gºC
Example 3: Solution
Step 4: Calculate the heat that needs to be
released to freeze all of the water.
m * Hfus
10.5 g * 334 J = - 3507 J
1
g
Hfus is – because we are
doing DOWN the stairs
Step 5: Calculate how much heat must be
released to decrease the temperature of the
solid water from100 ºC to -15ºC.
m * Cp(solid) * ΔT
10.5g * 2.07 J * (-15ºC - 0ºC)
1
gºC
10.5 g * 2.07 J * -15ºC = -326.025 J
1
gºC
Example 3: Solution
Step 6: Add up all of the products to get a total
- 642.6 J
- 23730 J
- 4389 J
- 3507 J
+ - 326.025 J
- 32594.63 J
This heat needs to be released to decrease
water from 130ºC to -15ºC.

When changing a
Cooling Curve
substance from an
vs Heating Curve
ice to gas, then a
heating curve is
used instead of a
H in J
cooling curve.
Uses the same
process as a
cooling curve, but
backwards and the
answer is positive.
-

+
H in J
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
Determine the final temperature of a 15.5 g
block of solid copper originally at 30°C if
you apply 680 J of energy to it.
Hint- Use the equation:
H = m Cp ∆ T
∆ T = (T final – T initial)
143.95°C
Calorimetry
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Calorimeter: a very insulated device used to
measure heat energy. A reaction happens in the
container & the heat given off raises the
temperature of the water.
(mass) (ΔT) (Cp water liquid) = heat (joules)
Calorimeters are used to measure food energy.
Nutritional information is determined in a lab.
A calorie is the energy needed to raise 1 gram of
H2O by 1°C.
1 calorie = 4.18 Joules
heat gained bywater
specific heat of metal 
(mass of metal)( T of metal)
Enthalpy
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Enthalpy is just another word for heat.
The symbol for enthalpy is ΔH.
Enthalpy is the total energy a substance contains.
ΔHrxn is the symbol for heat of reaction.
Heat of reaction is the amount of energy (kJ)
released or absorbed during a reaction.
ΔHf is the symbol for heat of formation.
Heat of formation is the amount of energy absorbed
or released during the synthesis of 1 mole of a
compound.
Endothermic or Exothermic?
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This graph shows an
exothermic reaction.
Energy is released or
given off to the
surroundings.
Enthalpy or ΔH is negative
during an exothermic
reaction because heat is
going out of the substance.
If you could feel the
reaction happening, it
would feel warm/hot.
Endothermic or Exothermic?
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This graph shows an
endothermic reaction.
Energy is absorbed or
taken in from the
surroundings.
Enthalpy or ΔH is positive
during an endothermic
reaction because heat is
going into the substance.
If you could feel the
reaction happening, it
would feel cold.
Activation Energy
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Activation energy is the amount of energy needed
to start a reaction.
Catalysts are used to lower activation energy.
How does this help?
1.
2.

Don’t need as much energy to get the reaction started.
The reaction will take less time, speeding it up!
Example: Enzymes (amylase, lactase, lipase)
Hess’s Law

The total enthalpy change for a chemical or
physical change is the same no matter how
many steps it takes to occur.
Entropy
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Entropy is the degree of disorder or
randomness of a system.
Entropy is represented by the symbol ΔS.


More disorder has a positive ΔS.
Less disorder has a negative ΔS.

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If we go from solid liquid gas
entropy is positive because the molecules
are more disorganized
If we go from gas liquid solid
entropy is negative because the molecules
more organized.
Free Energy

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Free energy a quantity that combines both
enthalpy and entropy. It is the amount of
potential or stored energy available to do
work or cause a change.
The symbol for free energy is ΔG.
Example: diffusion

Spray perfume or cologne and the molecules will slowly
travel through out the room naturally.
Free Energy

By comparing enthalpy and entropy as
well as temperature it can be
determined how much free energy a
substance has.
Free Energy

Free energy can also tell us if a reaction is
spontaneous.
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Spontaneous means that a reaction will occur
automatically, (not necessarily quickly).
If ΔG is negative the reaction is spontaneous.
If ΔG is positive the reaction is
nonspontaneous.
How do I know what the sign of ΔG is?

Use the equation: ΔG = ΔH – TΔS
Example 4
If ΔH is +12, ΔS is -4.10, & T is 100 K, what is
ΔG? Is this reaction spontaneous or
nonspontaneous?
Example 4 - Solution
ΔG = ΔH – TΔS
ΔG = +12 – (100° x -4.10)
ΔG = +12 – -410
ΔG = 422
Spontaneous or Nonspontaneous?
The End!
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