Thermochemistry
Problems
1
Dealing with Heat TCIS Style
A GUIDE TO SOLVING OUR
THERMOCHEMISTRY PROBLEMS.
Introduction
2
We want to solve several
types of Heat Problems.
Remember to use the 4
steps of problem solving
that we have stressed all
year.
Problem Solving Steps Reviewed
3
Step 1: Write down the given
information
Step 2: Find the formula.
Step 3: Substitute the
information into the formula.
Step 4: Solve
Problem Type 1:
4
Using Q = m ● ΔT ● Cp
Q = heat in Joules or calories
m = mass in grams
ΔT= change in Temp. (°C)
Cp= Specific Heat (J/g°C)
or (cal/g°C)
Solving Problem Type 1:
5
 How much heat is needed to raise the
temperature of 50 grams of water from
18°C to 30°C ?
 YOU NEED to know that the Specific
heat of water is 4.18 J/g°C (or
1cal/g°C). You don’t need to memorize
the specific heat values for every
material – just water !
Solving Problem 1
6
 Q=?
ΔT = 12°C
Mass = 50 grams
Cp = 4.18 J/g°C
 Q = m ● ΔT ● Cp
 Q = 50 ● 12 ● 4.18
 Q = 2508 J
Solving Problem 2
7
 Similar Problem except that we are
going to look for a different variable.
 What is the Cp of a metal if 900 Joules
are needed to raise 50 grams of it
48°C ?
 Q = m ● ΔT ● Cp
 Substitute and Solve
 Cp = .375 J / g °C
Solving Problem Type 3
8

A more difficult type of this problem
is to find the final temperature.
Remember that ΔT is change in
temperature and not the initial or
final temperature. These type of
problems are easier if you remember
to solve for ΔT first and then work
out the final temperature.
Solving Problem Type 3
9
 What is the final temperature of a metal
if 2750 Joules are added to 150 grams
of it at an initial temperature of 20°C ?
It has a Cp of .4 J / g°C.
 Q = m ● ΔT ● Cp
 Solve for ΔT
ΔT = 45.83 °C
 Since it started at 20°C the final temp.
is 20°C + 45.83 °C = 65.83 °C
Solving Problem Type 4
10
A more difficult type of problem is one where
one material is added to another.
 Remember that HEAT LOST = HEAT GAINED
(Law of Conservation of Energy) so we can adjust our
formula as follows:

QLost = QGained
and therefore
mLost ● ΔTLost ● Cp = mGain ● ΔTGain ● Cp
Problem Type 4
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 The Law of Conservation of Energy states that in any
chemical or physical process, energy is neither
created nor destroyed. Heat Energy can be
transferred from one object to another.
 We will be using the formula from the previous slide

mL ● ΔTL ● Cp = mG ● ΔTG ● Cp
 If 400 grams of Aluminum (Cp = .9 J/g°C) is
warmed to 95°C and then added to 1,000
grams of 20°C water, what is the final
temperature ?
Solving Problem Type 4
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 Heat Lost = Heat Gained
QLost = QGained
 mL ● ΔTL ● Cp = mG ● ΔTG ● C
 400 ● (95°C – Tf ) ● .9 =
1,000 ● (Tf – 20°C ) ● 4.18
 360 (95°C – Tf ) = 4,180 (Tf – 20°C )
 34200 – 360Tf = 4,180Tf – 83,600
 + 83,600 + 360Tf = + 360Tf + 83,600
117,800 = 4540 Tf
 Tf = 25.95 °C
Solving Problem Type 5
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 This type of problem is simple stoichiometry.
 If you have a reaction and you know the Heat of
Reaction (ΔHrxn) then you have a lot of information.
 Example: In the equation below the ΔH = -393.5 kJ
 C + O2  CO2
 So one mole of CO2 produced gives off -393.5 kJ of
heat.
 However, as we have stated numerous times
(remember our campfire example) that the more
material reacted the more heat you have.
 2 moles of CO2 produced will give off twice as
much heat : 2 x -393.5 kJ = -787 kJ of heat.
Solving Problem Type 5
14
 How much heat is given off when 14
grams of CO2 is produced in the rxn. ?
C + O2  CO2
ΔH = -393.5 kJ
u It is best to simply set up a proportion: 1
mole of C gives off -393.5 kJ of heat.
Therefore:
u
u
1 n / -393.5 kJ = x moles / y heat
WHAT VALUE DO WE HAVE ?
Solving Problem Type 5
15
We have the number of grams of Carbon
Dioxide (14 grams). Converting to
moles gives us .318 moles.
Our proportion looks like this now:
1 n / - 393.5 kJ = .318 moles / y heat
y = - 125.13 kJ of heat
Hess’s Law
Germain Henri Hess
(1802-1850)
(developed in 1840)
16
 If you add two or more
chemical equations to give a final
equation, then you can also add
the heats of reaction to give the
final heat of reaction.
Called Hess’s Law of Heat Summation
How do we use Hess’s Law # 1
17
1) If you turn an equation around, you
change the sign:
If H2(g) + 1/2 O2(g)  H2O(g)
ΔH= - 285.5 kJ
then the reverse is:
H2O(g)  H2(g) + 1/2 O2(g)
ΔH = + 285.5 kJ
Did you notice that the reactants are now the products and
the products are now the reactants ?
Also notice that the heat has the same value but the sign is
changed.
How do we use Hess’s Law # 2
18
2) If you multiply the equation by a number, you
multiply the heat by that number:
2 H2O(g) 2 H2(g) + O2(g)
DH =+ 571.0 kJ
The equation above is double the amount of both
reactants and products so the heat is also
doubled.
How do we use Hess’s Law # 3
19
 Our third option is that the equation remains
unchanged and you use the Heat of Reaction
given.
 Please note that you can use more than one
‘option’ at a time. For example you can
reverse a reaction and double the reactants
and products together.
Why is Hess’s Law important ?
20
 The reason Hess’s Law is important is because there
are millions of compounds in the world and naturally
there are countless ways to combine them in
chemical reactions. It would take way too much
time to do every conceivable chemical reaction to
see the heat change in them all.
 Hess’s Law enables this to take already known heats
of reactions and manipulate them to get the heats of
reactions of previously unknown reactions without
having to do the actual experiment.
Solved Hess’s Law Problem
21
 Time to dive in and solve one a Hess’s Law problem.
 We want to find the Heat of Reaction. Let’s look at
the following reaction:
C + 2 H2  CH4
 We want to find the Heat of Reaction and to find it
we are going to use information from a few other
reactions that already have the heat of reaction
calculated.
Solved Hess’s Law Problem
22
 Rxn. 1
H2(g) + 1/2 O2(g)  H2O(l)
• ΔHrxn = - 285.8 kJ
 Rxn. 2
•
 Rxn. 3
C(graphite) + O2(g)  CO2(g)
ΔHrxn = - 293.5 kJ
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
• ΔHrxn = - 890.4 kJ
Solved Hess’s Law Problem
23
 We want to look for ‘clues’ to decide which equation
to manipulate and how to manipulate it.
 So for our reaction: C + 2H2  CH4
 I notice that CH4 is only found in equation 3 and it is on the
reactant side so I have to reverse equation 3. Doing this is
the only way to get CH4 on the product side which is where it
is at in the problem. This gives us the following:
 CO2(g)

+ 2H2O(l)  CH4(g) + 2O2(g)
ΔH = + 890.4 kJ
Notice that I always include a sign with heat even if it is positive.
Solved Hess’s Law Problem
24
 Next I notice that C is needed on the reactant side
and equation 2 already has it on the reactant side so
in this case I can just leave the equation as is.

C(graphite) + O2(g)  CO2(g)
ΔHrxn = - 393.5 kJ
Solved Hess’s Law Problem
25
O2 into the equation which is not in the
final desired equation and I also have to get H2 in the equation.
 I have now included
One clue for these types of equations is that if you are given an
equation it will be used somewhere. This leaves us equation 1.
 H2(g) + 1/2 O2(g)  H2O(l)
ΔHrxn = - 285.8 kJ
 I notice that in our desired equation that I need 2 moles
of H2 as a product. If I double the H2 in this equation, I
have 2 moles and I also create 1 mole of O2 that will
cancel out the mole of O2 as a product that I had to
include when I used equation 3. This gives me.
 2 H2(g) + O2(g)  2 H2O(l)
ΔHrxn = - 571.6 kJ
Solved Hess’s Law Problem
26
 Now I combine the three equations that I manipulated.
CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) ΔH = + 890.4 kJ
 C(graphite) + O2(g)  CO2(g)
ΔH = - 393.5 kJ
 2 H2(g) + O2(g)  2 H2O(l)
ΔH = - 571.6 kJ


Canceling out (for instance the oxygen gas) and combining the
three equations gives me the following:
C
I
+ 2H2  CH4
also combine the heat term to give me -74.7 kJ
 So ΔHrxn = - 74.7 kJ
Specific Heat
27
 For water, Cp = 4.18 J/g oC (Joules),
and Cp = 1.00 cal/g oC (calories).
 Water
takes a long time to heat up AND
 Water takes a long time to cool down
 Water is used as a coolant in cars for
this very reason. It is also why the
climate in cities next to large bodies
of water is so different than the
climate in cities that are inland.
What is Calorimetry ?
28
 Calorimetry - the measurement of the
heat into or out of a system for chemical
and physical processes.
 Based on the fact that the heat
released = the heat absorbed
 The device used to measure the
absorption or release of heat in chemical
or physical processes is called a
calorimeter
Calorimetry
29
 Ordinary foam cups are excellent heat
insulators, and are commonly used as
simple calorimeters.
 This is why they are used for hot drinks
like coffee and hot chocolate.
A simple Coffee-cup calorimeter
30
Simple coffee – cup
calorimeter:
2 cups are nested
together for better
insulation
Calorimetry
31
 Calorimetry experiments can be
performed at a constant volume
using a device called a “bomb
calorimeter” - a closed system
Used
by nutritionists to measure
energy content of food.
A bomb calorimeter
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A Bomb Calorimeter
http://www.chm.davidson.edu/ronutt/che115/Bomb/Bomb.htm
How to determine Exothermic v. Endothermic
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 FOR EXOTHERMIC REACTIONS:


DH is negative: DH = -100 kJ
DH is written as a product such as:
A + B  C + heat

Look at the next slide to see how the
energy diagram appears for an
exothermic reaction. Notice that the
energy of the products are lower than
the energy of the reactants.
C + O2 → CO2 + 393.5 kJ
Energy
34
C + O2
393.5kJ
given off
CO2
Reactants

Products
How to determine Exothermic v. Endothermic
35
 FOR ENDOTHERMIC REACTIONS:


DH is positive: DH = + 400 kJ
DH is written as a product such as:
A + B + heat  C + D

Look at the next slide to see how the
energy diagram appears for an
exothermic reaction. Notice that the
energy of the products are higher than
the energy of the reactants.
Energy
CaCO3 + 176 kJ → CaO + CO2
CaO + CO2
176 kJ
absorbed
CaCO3
Reactants

36
Products
Heat in Changes of State: Boiling and Freezing
37
 When heat is added to a sample one of two
things can happen.
 1. The temperature changes as we saw with
the formula Q = m ● ΔT ● Cp OR
 2. The phase of the material can change.
The next slide summarizes these changes
and gives the formula that we will use.
Heat in Changes of State: Freezing
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Molar Heat of Fusion (DHfus.) = the heat absorbed
by one mole of a substance in melting from a
solid to a liquid
q = mol x DHfus. (no temperature change)
Molar Heat of Solidification (DHsolid.) = the heat
lost when one mole of liquid solidifies (or freezes)
to a solid
q = mol x DHsolid. (no temperature change)
Heat in Changes of State: Freezing
39
 Note: You may also have the
value of these equations as:
q = mass x DH
This is because some textbooks
give the value of DH as kJ/gram,
instead of kJ/mol
Heat in Changes of State: Freezing
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 Heat absorbed by a melting
solid equals the heat lost when
a liquid solidifies.

DHfus. = - DHsolid.
Heat in Changes of State: Boiling
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 When liquids absorb heat at their
boiling points, they become a vapor.
Molar Heat of Vaporization (DHvap.) =
the amount of heat necessary to
vaporize one mole of a given liquid.
q = mol x DHvap. (no temperature change)
Heat in Changes of State: Boiling
42
 Condensation is the opposite of vaporization.
Molar Heat of Condensation (DHcond.) =
amount of heat released when one mole of
vapor condenses to a liquid
q = mol x DHcond. (no temperature change)

DHvap. = - DHcond.
Heat in Changes of State: Boiling
43
 The large values for water DHvap. and
DHcond. is the reason hot vapors such as
steam are very dangerous!
You can receive a scalding burn from
steam when the heat of condensation is
released!
H2O(g)  H2O(l) DHcond. = - 40.7kJ/mol
Heating Curves
44
 The following slides shows a heating curve.
 Notice that there are various parts of the curve.
Some are showing a change in temperature and
others are showing a change in phase.
 The parts of the curve showing a change in phase
are the parts that are flat lines. These parts use the
formulas found in the previous few slides concerning
Heats of vaporization and heats of fusion, etc.
 The next several slides will break down a heating
curve for you. This is a heating curve for water. We
will assume a 50 gram sample of water.
Leg ‘A’ of a 50 gram sample of water
45
←
The solid temperature is rising from -20
to 0 oC (use q = mass x ΔT x C) and we
get 2100 Joules of heat needed. (note that
the Cp of Ice is 2.1 J/g C and would be a given)
Leg ‘B’ of a 50 gram sample of water
46
←
The solid is melting at 0o C; no
temperature change (use q = mass x
ΔHfus.) and we get 16,700 J (Note that
the ΔHfus. for water is 334 J/g )
Leg ‘C’ of a 50 gram sample of water
47
←
The liquid temperature is rising from
0 to 100 oC (use q = mass x ΔT x
C) and we get 20,920 Joules.
Leg ‘D’ of a 50 gram sample of water
48
↑
The liquid is boiling at 100o C;
no temperature change (use q
= mass x ΔHvap.) we get 113,000
Joules. The ΔHvap. of water is
2260 J/g
Leg ‘E’ of a 50 gram sample of water
49
↑
The gas temperature is rising from 100
to 120 oC (use q = mol x ΔT x C) so we
get 2,000 Joules. Note that the Cp of
water vapor is 2.0 J/g °C
Let’s finish the heating curve problem.
50
 We need to add up all of the heat terms from each
part of
 Heat
 Heat
 Heat
 Heat
 Heat
the heating curve.
change from -20 to 0.
change during melting
change from 0 to 100
change during boiling
change from 100 to 120
 TOTAL
HEAT NEEDED
2100 J
16700 J
20920 J
113000 J
2000 J
154720 J
(154.72 kJ)
Special Note about Heating Curve Problems
51
 The above example was a complete
heating curve. A problem doesn’t have
to have all 5 parts. For example, the
water can start at 30 degrees in which
you will only need the last 3 parts.
 It all comes down to reading the
problem carefully.