Energy
Unit 8, Chapter 10
Energy,
Temperature, and
Heat
Section 1
Energy
• Energy: the ability to do work or produce
heat.
• Work: force acting over a distance.
• 2 major types:
• Potential energy: energy due to position
or composition.
• Kinetic energy: energy due to motion.
The Law of Conservation of Energy
Energy can be converted from one form to
another but can be neither created nor
destroyed.
Temperature, Heat, and
Thermal Energy
• Temperature: a measure of the random
motions (average kinetic energy) of the
components of a substance.
• H2O molecules in warm water move
faster than H2O molecules in cold water.
Classzone Animation
Temperature, Heat, and
Thermal Energy
• In the animation, energy from the faster
moving particles is transferred through the
wall to the slower moving particles.
• The two samples will eventually reach the
same temperature.
• The amount of energy lost by the hot
water must be equal to the amount of
energy gained by the cold water. (law of
conservation of energy)
Temperature, Heat, and
Thermal Energy
• Heat: a flow of energy due to a
temperature difference; a transfer of
thermal energy.
• Thermal Energy: energy due to the
random motions of the components of an
object.
• Most chemical reactions involve changes
in energy.
• In order to study this, the universe is
divided into 2 parts: the system and the
surroundings.
• The system is the part of the universe
on which we wish to focus attention.
• The surroundings include everything
else in the universe.
SURROUNDINGS
SYSTEM
• Exothermic – refers to a process in which
energy (as heat) flows out of the system
into the surroundings.
- “exo” = out of
- the system releases heat
• Endothermic – refers to a process in which
energy (as heat) flows from the
surroundings into the system.
- “endo” = into
- the system absorbs heat
Exothermic
2H2 (g) + O2 (g)
H2O (g)
2H2O (l) + energy
H2O (l) + energy
Endothermic
energy + 2HgO (s)
energy + H2O (s)
2Hg (l) + O2 (g)
H2O (l)
The energy gained by the surroundings in an
exothermic reaction must be equal to the
energy lost by the system.
(Law of Conservation of Energy)
In any exothermic reaction, some of the
potential energy stored in the chemical
bonds is converted to thermal energy
(random kinetic energy) via heat.
Exothermic or Endothermic?
(Remember to think about the direction!)
1. Your hand gets cold when you touch ice.
exothermic
2. Ice melts when you touch it.
endothermic
3. Your hand gets colder when you touch ice.
exothermic
4. Propane burns in a propane torch.
exothermic
5. A beaker feels cold when two chemicals are
mixed in it.
endothermic
The Flow of Energy
Section 2
Thermodynamics
• Thermodynamics is the study of energy.
• The first law of thermodynamics states
that the energy of the universe is constant.
• Energy is neither created nor destroyed.
• Energy can flow into (or out of) a
system, but the amount gained (or lost)
by the system is equal to the amount
lost (or gained) by the surroundings.
Thermodynamics
• Energy is a state function, or a property of
a system that does not depend on the
pathway
• Example: elevation vs. distance traveled
• Application: ball down a hill
• Work and heat are not state functions
Internal Energy (E)
• The internal energy of a system is the sum
of the kinetic and potential energies of all
particles in the system.
• The internal energy of a system can be
changed by a flow of work, heat, or both.
• Thus,
DE = q + w
the energy change of the system is equal to
the heat plus the work
Thermodynamic Quantities
• Thermodynamic quantities are made up of
2 parts:
• The number tells the magnitude of the
energy change (the amount).
• The sign tells the direction of the heat
flow.
• The direction of heat flow is always from
the system’s point of view.
• Positive (+) = heat into the system
• Negative (-) = heat out of the system
•
•
•
•
Exothermic
Heat flows out of
the system into the
surroundings.
q = -x
system’s energy
decreases
DE < 0
•
•
•
•
Endothermic
Heat flows into the
system from the
surroundings.
q = +x
system’s energy
increases
DE > 0
If you understand the difference between exothermic
and endothermic processes, you just need to
remember that heat flow is put in terms of the system
(from the system’s POV) and DE will make sense.
Measuring Energy Changes
• 2 common units of energy:
• joule
• calorie
• A calorie is the amount of energy (heat)
required to raise the temperature of one
gram of water by one Celsius degree.
• The Calories that you count in your food
(with a capital C) are actually kilocalories,
or 1000 calories.
Measuring Energy Changes
1 Calorie (C) = 1000 calories (cal)
1 calorie (cal) = 4.184 joules (J)
You will need to be able to convert between
Calories, calories, and joules.
Practice Problems
1. A small package of fruit snacks has only
80 Calories (kilocalories). How many
calories (cal) are you consuming if you
eat all of the fruit snacks in the package?
2. Express 60.1 cal of energy in units of
joules.
3. How many calories of energy correspond
to 28.4 J?
Practice Problems
Your McDonald’s Meal
Using the information provided, calculate the
amount of energy in joules that you
consume when you eat your favorite
McDonalds meal. Don’t forget to include
the fries and drink!
Calculating Energy
Requirements
Section 3
Calculating Energy Requirements
• Remember that 1 calorie is the amount of
energy (heat) required to raise the
temperature of 1 gram of water by 1oC.
• Remember that temperature only depends
on the motion of the particles in a
substance, but heat depends on the
temperature and the mass of the
substance.
Calculating Energy Requirements
• Knowing these facts allows us to calculate
the amount of energy required to raise the
temperature of a given mass of water a
certain number of degrees.
Calculating Energy Requirements
Practice Problem 1
Determine the amount of energy (heat) in
joules required to raise the temperature of
7.40 g of water from 29.0oC to 46.0oC.
Calculating Energy Requirements
Practice Problem 1
What do we know?
• Mass of water = 7.40 g
• Initial temperature = 29.0oC
• Final temperature = 46.0oC
• 1 cal = 4.184 joules
Calculating Energy Requirements
Practice Problem 1
Where do we want to go?
• 7.40 g water at 29.0oC a 7.40 g water at
46.0oC takes how much energy?
Calculating Energy Requirements
Practice Problem 1
How do we get there?
• 1.00 g water at 29.0oC a 1.00 g water at
30oC would take 4.184 J of energy.
• Because we have 7.40 g of water instead
of 1.00 g, it will take 7.40 x 4.184 J to raise
the temperature by 1 degree.
• Thus, 7.40 g at 29oC a 7.40 g at 30oC
would take 7.40 x 4.184 J of energy.
Calculating Energy Requirements
Practice Problem 1
• However, we want to raise the
temperature by more than 1oC. A change
from 29.0oC to 46.0oC is a total change of
17.0oC
• Final temp. – Initial temp = Total change
in temp
• Thus, it will take 17.0 times the energy
necessary to raise the temperature by
1oC.
Calculating Energy Requirements
Practice Problem 1
• In summary, the total amount of energy
needed to raise the temperature of 7.40 g
of water from 29.0oC to 46.0oC is equal to
17.0 x 7.40 x 4.184
• What are the units?
17.0oC = temperature change
7.40 g = mass of water
4.184 J/g oC = energy per gram of water
per degree of temperature
Calculating Energy Requirements
Practice Problem 1
So, how much energy (heat) in joules is
required to raise the temperature of 7.40 g
of water from 29.0oC to 46.0oC?
17.0oC x 7.40 g x 4.184 J/g oC = 526 J
Calculating Energy Requirements
Review
• The energy (heat) required to change the
temperature of a substance depends on:
• The amount of substance being heated
(# of grams)
• The temperature change (# of degrees)
• And one other factor we haven’t yet
pointed out:
the identity of the substance.
Specific Heat Capacity
• Specific heat capacity is the amount of
energy needed to raise the temperature of
one gram of a substance by one Celsius
degree.
• For water, this amount is 4.184 J because
1 calorie = 4.184 J and it takes exactly 1
calorie of energy to raise the temperature
of 1 gram of water by 1oC.
Specific Heat Capacity
• If 4.184 J of energy were added to 1 gram
of gold, the temperature would increase by
32oC!
• The specific heat capacity of water is very
high, which allows bodies of water to
absorb large amounts of heat with very
small increases in temperature.
Calculating Energy Requirements
• We can calculate the amount of energy
required to raise the temperature of a
given amount of a substance using the
substance’s mass, the substance’s
specific heat capacity, and the
temperature change.
• Set it up like the previous practice problem
for water!
Calculating Energy Requirements
Practice Problem 2
What quantity of energy (in joules) is
required to heat a piece of iron with a
mass of 1.3 g from 25oC to 46oC?
Calculating Energy Requirements
• There is a pattern here! To calculate the
energy (heat) required, we multiply the
sample size in grams times the specific
heat capacity times the change in
temperature in Celsius degrees.
Q = msDT
Calculating Energy Requirements
Practice Problem 3
Determine the amount of energy as heat
that is required to raise the temperature of
a 10.0-g sample of aluminum from 25oC to
58oC. Answer in joules and calories.
Calculating Energy Requirements
Practice Problem 4
A 1.6-g sample of a metal that has the
appearance of gold requires 5.8 J of
energy to change its temperature from
23oC to 41oC. Is the metal pure gold?
Calorimetry
• A calorimeter is a device used to
determine the heat associated with a
chemical reaction or physical change.
• Calorimetry experiments determine the
enthalpy changes of reactions by making
accurate measurements of temperature
changes produced in a calorimeter.
Reactions in a Calorimeter
• Q = msDT is solved for the water (which is
the surroundings)
• The heat (Q) of the reaction (which is the
system) will be equal to the heat of the
surroundings, but the opposite sign
• If the system gains heat, the
surroundings loses the same amount
Energy and Chemical
Reactions
Section 4
Enthalpy
• Enthalpy (H) is used to quantify the heat
flow into or out of a system in a process
that occurs at constant pressure.
DH = heat given off or absorbed during a
reaction at constant pressure
DH = Hproducts – H reactants
Thermochemistry
Hproducts < Hreactants
DH < 0
Hproducts > Hreactants
DH > 0
Enthalpy of Water
http://www.mhhe.com/physsci/chemistry/animations/chang
_7e_esp/enm1s3_4.swf
Thermochemical Equations
Is DH negative or positive?
System absorbs heat =
Endothermic
DH > 0
6.01 kJ of energy are absorbed for
every 1 mole of ice that melts at 0oC
and 1 atm.
H2O (s)
H2O (l) DH = 6.01 kJ
Thermochemical Equations
• The stoichiometric coefficients always refer to the
number of moles of a substance
H2O (s)
H2O (l)
DH = 6.01 kJ
• If you reverse a reaction, the sign of DH changes
H2O (l)
H2O (s)
DH = -6.01 kJ
• If you multiply both sides of the equation by a
factor n, then DH must change by the same factor
n.
2H2O (s)
2H2O (l) DH = 2 x 6.01 = 12.0 kJ
Thermochemical Equations
The physical states of all reactants and products
must be specified in thermochemical equations.
H2O (s)
H2O (l)
DH = 6.01 kJ
H2O (l)
H2O (g)
DH = 44.0 kJ
Thermochemical Equations
The amount of heat absorbed or released in a
chemical reaction depends on the # of moles
reacted.
How much heat is evolved when 266 g of white
phosphorus (P4) burn in air?
P4 (s) + 5O2 (g)
266 g P4 x
P4O10 (s)
1 mol P4
123.9 g P4
x
DH = -3013 kJ
3013 kJ
= 6470 kJ
1 mol P4
Practice Problems
When 1 mol of methane is burned at
constant pressure, 890 kJ of energy is
released as heat. Calculate DH for a
process in which a 5.8-g sample of
methane is burned at constant pressure.
Practice Problems
How much heat will be released if 1.0 g of
hydrogen peroxide (H2O2) decomposes to
form water and oxygen in a bombardier
beetle to produce a steam spray against
predators?
DH = -190 kJ
Practice Problems
The reaction that occurs in the heat packs
used to treat sports injuries is
4Fe(s) + 3O2(g) g 2Fe2O3(s)
DH = -1652 kJ
How much heat is released when 1.00 g of
Fe(s) is reacted with excess O2?
And what if…
• Two reactant masses are given in the
problem?
• Find the limiting reactant by determining
which transfers the smaller amount of energy.
• The reaction is done in a calorimeter?
• Calculate qsurroundings (the water)
• Reverse the sign for the qsystem (the reaction)
• Divide the qsystem by the number of moles of
reactant used in the reaction
Hess’s Law
Section 5
Hess’s Law
• States that if a series of reactions are
added together, the enthalpy change for
the net reaction will be the sum of the
enthalpy changes for the individual steps.
• In other words, in going from a particular
set of reactants to a particular set of
products, the change in enthalpy is the
same whether the reaction takes place in
one step or a series of steps.
• Energy is a state function!
Hess’s Law
• The oxidation of nitrogen to produce
nitrogen dioxide can occur in one step:
N2(g) + 2O2(g) g 2NO2(g)
DH = 68 kJ
• The same net reaction can also be carried
out in two distinct steps:
N2(g) + O2(g) g 2NO(g)
DH = 180 kJ
2NO(g) + O2(g) g 2NO2(g)
DH = -112 kJ
N2(g) + 2O2(g) g 2NO2(g)
DH = 68 kJ
Hess’s Law
• Hess’s Law is useful because it allows us
to calculate heats of reaction that might be
difficult or inconvenient to measure directly
in a calorimeter.
Characteristics of DH
• In order to use Hess’s Law, we must
review and understand two characteristics
of enthalpy changes:
1. If a reaction is reversed, the sign of DH
is also reversed.
2. If the coefficients in a balanced
equation are multiplied by an integer,
the value of DH is multiplied by the
same integer.
Applying Hess’s Law
• Verify that the reactants and products of
the net reaction are on the correct sides
of the step reactions.
If a step reaction must be reversed,
reverse the sign of DH for that reaction
as well.
Applying Hess’s Law
• Verify that the coefficients in the net
reaction match those in the step
reactions.
If a reaction must be multiplied (or
divided) by a factor, multiply (or divide)
the DH value by that factor as well.
Applying Hess’s Law
• Cancel any elements or compounds that
occur as a product in one step reaction
and a reactant in another step.
Coefficients must match to cancel
completely, otherwise some will be left
over
*Note: These general steps are not always
done in this order, but should all be
applied to Hess’s Law Problems
Practice Problem
The combustion of sulfur can produce SO2
as well as SO3 depending upon the supply
of oxygen. From the following reactions
and their enthalpy changes,
2SO2(g) + O2(g) g 2SO3 (g) DH = -196 kJ
2S(g) + 3O2(g) g 2SO3(g)
DH = -790 kJ
calculate the enthalpy change for the
combustion of sulfur to produce SO2.
S(g) + O2(g) g SO2(g)
Practice Problem
Given the following data:
2O3(g) g 3O2 (g) DH = -427 kJ
O2(g) g 2O(g)
DH = +495 kJ
NO(g) + O3(g) g NO2(g) + O2(g)
DH = -199 kJ
calculate the enthalpy change for the
reaction
NO(g) + O(g) g NO2(g)