Vann - Chemistry ch. 6.1

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Chapter 6.1-6.2
Thermochemistry
THERMODYNAMICS
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Chapter 6
Table of Contents
•
•
•
•
•
•
6.1
6.2
6.3
6.4
6.5
6.6
The Nature of Energy
Enthalpy and Calorimetry
Hess’s Law
Standard Enthalpies of Formation
Present Sources of Energy
New Energy Sources
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Chapter 6
Table of Contents
• Finishing ch. 5 test - on computer
• Notes with computer for section 6.1-6.2 including
problems (note to me: assigned as HW last year incorporated into notes this year)
• HW: Pre-lab Calorimetry
• Turn in part 2 ch. 5 take home Test
• Turn in previous lab reports
• Turn in ch. 5 Interactive Notes packet for grade.
• Turn in any missing or late assignments - check
Pinnacle for grades
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4
Section 6.1
The Nature of Energy
Energy
•
•
Capacity to do work or to produce heat.
That which is needed to oppose natural
attractions.
• Law of conservation of energy – energy
can be converted from one form to
another but can be neither created nor
destroyed.
• The First Law of Thermodynamics
 The total energy content of the
universe is constant.
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Section 6.1
The Nature of Energy
Thermodynamics
• Thermodynamics is the study of energy and its
interconversions.
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6
Definitions #1
Energy: The capacity to do work or produce heat
Potential Energy: Energy due to position or
composition
Kinetic Energy: Energy due to the motion of the
object
Section 6.1
The Nature of Energy
Energy
•
•
Potential energy – energy due to position
or composition.
Kinetic energy – energy due to motion of
the object and depends on the mass of
the object and its velocity.
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Section 6.1
The Nature of Energy
Initial Position
•
In the initial position, ball A has a higher
potential energy than ball B.
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Section 6.1
The Nature of Energy
Final Position
•
After A has rolled down the hill, the potential
energy lost by A has been converted to random
motions of the components of the hill (frictional
heating) and to the increase in the potential
energy of B.
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Section 10.1
The Nature of Energy
Energy
Transformations
Types of Energy
1. Kinetic energy
2. Potential energy
3. Chemical energy
4. Heat energy
5. Electric energy
6. Radiant energy
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Section 6.1
The Nature of Energy
What are 2 ways energy is transferred?
1) By doing work
2) by Heat transfer - thermal energy is transferred
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Section 6.1
Energy Transfer by Doing work
The Nature of Energy
•
Energy can be transferred from one object to another by doing work. When work is done on an object, it
results in a change in the object's motion (more specifically, a change in the object's kinetic energy).
•
Energy is often defined as the ability to do work. Work equals force multiplied by distance.
•
An illustration of how doing work is an example of energy transfer
•
Suppose that a person exerts a force on the wheelbarrow that is initially at rest, causing it to move over a
certain distance. Recall that the work done on the wheelbarrow by the person is equal to the product of the
person's force multiplied by the distance traveled by the wheelbarrow. Notice that when the force is
exerted on the wheelbarrow, there's a change in its motion. Its kinetic energy increases. But where did the
wheelbarrow get its kinetic energy? It came from the person exerting the force, who used chemical energy
stored in the food they ate to move the wheelbarrow. In other words, when the person did work on the
wheelbarrow, they transferred a certain amount of chemical energy stored in the person was transferred to
the wheelbarrow, causing its kinetic energy to increase. As a result, the person's store of chemical energy
decreases and the wheelbarrow's kinetic energy increases.
•
Wherever you look, you can see examples of energy transfers. When you turn on a light, you see result of
energy being transferred from the sun to the plants to the coal to electricity and finally to light you see.
During each of these transfers, energy changes form. There are two main forms of energy, kinetic energyReturn to TOC
(motion) and potential energy (position). To further classify energy, these forms are sometimes further
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Section 6.1
The Nature of Energy
Energy
•
Heat involves the transfer of energy between
two objects due to a temperature difference.
• Work – force acting over a distance.
• Energy is a state function; work and heat are
not:
 State Function – property that does not
depend in any way on the system’s past
or future (only depends on present state).
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State Functions depend ONLY on the
present state of the system
ENERGY IS A STATE
FUNCTION
A person standing at the
top of Mt. Everest has the
same potential energy
whether they got there by
hiking up, or by falling down
from a plane 
WORK IS NOT A
STATE FUNCTION
WHY NOT???
Section 6.1
The Nature of Energy
Chemical Energy
•
•
System – part of the universe on which we
wish to focus attention.
Surroundings – include everything else in
the universe.
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Section 6.1
The Nature of Energy
Chemical Energy
•
Exothermic Reaction:
 Energy flows out of the system.
• Energy gained by the surroundings must
be equal to the energy lost by the system.
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Section 6.1
The Nature of Energy
Chemical Energy
•
Endothermic Reaction:
 Heat flow is into a system.
 Absorb energy from the surroundings.
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Section 6.1
The Nature of Energy
Concept Check
•Is the freezing of water an endothermic or
exothermic process? Explain.
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Section 6.1
The Nature of Energy
Concept Check
•Classify each process as exothermic or
endothermic. Explain. The system is
underlined in each example.
a)
b)
c)
d)
e)
Your hand gets cold when you touch
ice.
The ice gets warmer when you touch it.
Water boils in a kettle being heated on a
stove.
Water vapor condenses on a cold pipe.
Ice cream melts.
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Return to TOC
Section 6.1
The Nature of Energy
Concept Check
•Classify each process as exothermic or
endothermic. Explain. The system is
underlined in each example.
Exo
a)
Endo
Endo
Exo
Endo
b)
c)
d)
e)
Your hand gets cold when you touch
ice.
The ice gets warmer when you touch it.
Water boils in a kettle being heated on a
stove.
Water vapor condenses on a cold pipe.
Ice cream melts.
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Return to TOC
Section 6.1
The Nature of Energy
Concept Check
•For each of the following, define a system
and its surroundings and give the direction
of energy transfer.
a)
b)
Methane is burning in a Bunsen burner in
a laboratory.
Water drops, sitting on your skin after
swimming, evaporate.
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22
Section 6.1
The Nature of Energy
Concept Check
•Hydrogen gas and oxygen gas react
violently to form water. Explain.

Which is lower in energy: a mixture
of hydrogen and oxygen gases, or
water?
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23
Section 6.1
The Nature of Energy
Internal Energy
•
Law of conservation of energy is often
called the first law of thermodynamics.
• Internal energy E of a system is the sum of
the kinetic and potential energies of all the
“particles” in the system.
• To change the internal energy of a system:
ΔE = q + w
–
q represents heat
–
w represents work
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Work vs. Energy Flow
Click here to watch visualization.
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6–
Section 6.1
The Nature of Energy
Internal Energy
•
•
Sign reflects the system’s point of view.
Endothermic Process:
 q is positive (heat enters the system)
• Exothermic Process:
 q is negative (heat exits the system)
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Section 6.1
The Nature of Energy
Internal Energy
•
•
Sign reflects the system’s point of view.
Surroundings do work on the system:
 w is positive
• System does work on surroundings:
 w is negative
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Section 6.1
The Nature of Energy
Work = F x d (from 9th grade) - applying to a gas
•
Work = P × A × Δh = PΔV




P is pressure.
A is area.
Δh is the piston moving a
distance.
ΔV is the change in
volume.
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28
Section 6.1
The Nature of Energy
Work
•
•
•
For an expanding gas, ΔV is a positive
quantity because the volume is increasing.
Thus ΔV and w must have opposite signs:
w = –PΔV
To convert between L·atm and Joules, use
1 L·atm = 101.3 J.
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Work, Pressure, and Volume
Expansion
+V (increase)
-w results
Esystem decreases
Work has been done
by the system on the
surroundings
Compression
-V (decrease)
+w results
Esystem increases
Work has been done
on the system by the
surroundings
Figure 6.4
The Piston,
Moving a
Distance
Against a
Pressure P,
Does Work
On the
Surrounding
s
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6–
Section 6.1
The Nature of Energy
Exercise
w = –PΔV
•Which of the following performs more
work?
•a) A gas expanding against a pressure
of 2 atm from 1.0 L to 4.0 L.
b) A gas expanding against a pressure
of 3 atm from 1.0 L to 3.0 L.
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Section 6.1
The Nature of Energy
Exercise
w = –PΔV
•Which of the following performs more
work?
•a) A gas expanding against a pressure
of 2 atm from 1.0 L to 4.0 L.
b) A gas expanding against a pressure
of 3 atm from 1.0 L to 3.0 L.
• They perform the same amount of work.
(-6 AllL*atm)
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rights reserved
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33
E = q + w
E = change in internal energy of a system
q = heat flowing into or out of the system
-q if energy is leaving to the surroundings
+q if energy is entering from the surroundings
w = work done by, or on, the system
-w if work is done by the system on the
surroundings
+w if work is done on the system by the
surroundings
Section 6.1
The Nature of Energy
Concept Check •
ΔE = q + w
w = –PΔV
•Determine the sign of E for each of the following with
the listed conditions:
•a) An endothermic process that performs work on surrounding.

|work| > |heat| Δ E = negative wk (-) heat +

|work| < |heat| Δ E = positive
•b) Work is done on a gas and the process is exothermic.

|work| > |heat| Δ E = positive wk + and heat 
|work| < |heat| Δ E = negative
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Section 6.1
The Nature of Energy
Thermodynamic Quantities
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Energy Change in Chemical Processes
Endothermic:
Reactions in which energy flows into the
system as the reaction proceeds.
+ qsystem
- qsurroundings
Energy Change in Chemical Processes
Exothermic:
Reactions in which energy flows out of the
system as the reaction proceeds.
- qsystem
+ qsurroundings
Section 6.1
The Nature of Energy
Exercise 1
• Exercise 1 Internal Energy
• Calculate ΔE for a system undergoing an endothermic process
in which 15.6 kJ of heat flows and where 1.4 kJ of work is
done on the system.
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Section 6.1
The Nature of Energy
Exercise 1
• Exercise 1 Internal Energy
• Calculate ΔE for a system undergoing an endothermic process
in which 15.6 kJ of heat flows and where 1.4 kJ of work is
done on the system.
• ANSWER: 17.0 kJ
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Section 6.1
The Nature of Energy
Exercise 2
•Exercise 2 PV Work
•Calculate the work associated with the expansion of a gas from
46 L to 64 L at a constant external pressure of 15 atm.
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Section 6.1
The Nature of Energy
Exercise 2
•Exercise 2 PV Work
•Calculate the work associated with the expansion of a gas from
46 L to 64 L at a constant external pressure of 15 atm.
•ANSWER: ‒270 L•atm
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Section 6.1
The Nature of Energy
Exercise 3 Internal Energy, Heat, and Work
•Exercise 3 A balloon is being inflated to its full extent by heating
the air inside it. In the final stages of this process, the volume of
the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the
addition of 1.3 • 108 J of energy as heat. Assuming that the
balloon expands against a constant pressure of 1.0 atm, calculate
ΔE for the process. (To convert between L ⋅ atm and J, use 1 L ⋅
atm = 101.3 J.)
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Section 6.1
The Nature of Energy
Exercise 3 Internal Energy, Heat, and Work
•Exercise 3 A balloon is being inflated to its full extent by heating
the air inside it. In the final stages of this process, the volume of
the balloon changes from 4.00 • 106 L to 4.50 • 106 L by the
addition of 1.3 • 108 J of energy as heat. Assuming that the
balloon expands against a constant pressure of 1.0 atm, calculate
ΔE for the process. (To convert between L ⋅ atm and J, use 1 L ⋅
atm = 101.3 J.)
•ANSWER: 8.0 • 107 J
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Stopped here 2-5-13 - need to
make ch.3-4 test shorter.
Section 6.2
Temperature and Heat
Temperature - Check out this simulation regarding the
effect temperature on molecular motion.
• Click on the following link and
• http://phet.colorado.edu/en/simulation/states-of-matterbasics
• Click on Run Now. Open with Java.
• Adjust the temperatures, elements, and states of matter
to see what happens from the solid, liquid, gas tab.
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Section 6.2
Temperature and Heat
Temperature
• A measure of the random motions of the components
of a substance.
Temperature (T)—is
proportional to the average
kinetic energy of the
molecules, KEave .
“Heat ‘em up and speed
‘em up”.
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Section 6.2
Temperature and Heat
Heat
•
A flow of energy between two objects due to a
temperature difference between the objects.
 Heat is the way in which thermal energy is
transferred from a hot object to a colder object.
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Section 6.2
Temperature and Heat
Heat
• Two systems with different temperatures that are in
thermal contact will exchange thermal energy, the
quantity of which is call heat. This transfer of energy
in a process (flows from a warmer object to a cooler
one, transfers heat because of temperature difference
but, remember, temperature is not a measure of
energy—it just reflects the motion of particles)
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Section 6.2
Enthalpy
Enthalpy and Calorimetry
•Enthalpy (H)– flow of energy (heat exchange) at constant pressure
when two systems are in contact.
• Enthalpy of reaction (ΔHrxn) – amount of heat released (negative
values) or absorbed (positive values) by a chemical reaction at constant
pressure in kJ/molrxn
• Enthalpy of combustion (ΔHcomb)—heat absorbed or released by
burning (usually with O2) in kJ/molrxn; note that combustion reactions
yield oxides of that which is combusted
• Enthalpy of formation (ΔHf) – heat absorbed or released when
ONE mole of compound is formed from elements in their standard
states in kJ/molrxn
• Enthalpy of fusion (ΔHfus)—heat absorbed to melt (overcome
IMFs) 1 mole of solid to liquid @ MP expressed in kJ/molrxn
• Enthalpy of vaporization (ΔHvap)—heat absorbed to vaporize or
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boil ©(overcome
IMFs)
50
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Section 6.2
Enthalpy and Calorimetry
Change in Enthalpy (ΔH )
•
•
•
State function
ΔH = q at constant pressure (ex. atmospheric pressure)
ΔH = Hproducts – Hreactants
•
Measure only the change in enthalpy, ΔH ( the difference between
the potential energies of the products and the reactants)
•
A system that releases heat to the surroundings, an exothermic reaction, creates a negative
ΔH because the enthalpy of the products is lower than the enthalpy of the reactants of the
system (Figure 1).
•
A system that absorbs heat from the surroundings, an endothermic reaction, creates a
positive ΔH, because the enthalpy of the products is higher than the enthalpy of the
reactants of the system.
•
A positive ΔH is endothermic rxn
•
A negative ΔH is exothermic rxn
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51
Section 6.2
Enthalpy and Calorimetry
Enthalpy Calculations
• Enthalpy can be calculated from several
• sources including:
•
•
•
•
•





Stoichiometry
Calorimetry
From tables of standard values
Hess’s Law
Bond energies
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Section 6.2
Stoichiometrically:
Enthalpy and Calorimetry
• Exercise 4
• Upon adding solid potassium hydroxide pellets to water the
following reaction takes place:
•
KOH(s) → KOH(aq) + 43 kJ/mol
• Answer the following questions regarding the addition of 14.0 g of
KOH to water:
• a) Does the beaker get warmer or colder? How do you know?
• b) Is the reaction endothermic or exothermic?
• c) What is the enthalpy change for the dissolution of the
•
14.0 grams of KOH?
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Section 6.2
Stoichiometrically:
Enthalpy and Calorimetry
• Exercise 4
• Upon adding solid potassium hydroxide pellets to water the
following reaction takes place:
•
KOH(s) → KOH(aq) + 43 kJ/mol
• Answer the following questions regarding the addition of 14.0 g of
KOH to water:
• a) Does the beaker get warmer or colder?
• b) Is the reaction endothermic or exothermic?
• c) What is the enthalpy change for the dissolution of the
•
14.0 grams of KOH?
• ANSWER: (a) warmer (the +43 kJ/mole is heat on the product side) (b)
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exothermic (c) −10.7 kJ/molrxn (take 14 g divide by the molar mass of KOH 54
- to get #
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Section 6.2
Enthalpy and Calorimetry
Exercise 5
•
Consider the combustion of
propane:
•
C3H8(g) + 5O2(g) → 3CO2(g) +
4H2O(l)
•
ΔH = –
2221 kJ
•Assume that all of the heat comes from the
combustion of propane. Calculate ΔH in which
5.00 g of propane is burned in excess oxygen at
constant pressure.
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Section 6.2
Enthalpy and Calorimetry
Exercise 5
•
Consider the combustion of
propane:
•
C3H8(g) + 5O2(g) → 3CO2(g) +
4H2O(l)
•
ΔH = –
2221 kJ
•Assume that all of the heat comes from the
combustion of propane. Calculate ΔH in which
5.00 g of propane is burned in excess oxygen at
constant pressure.
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Section 6.2
Calorimetry
Enthalpy and Calorimetry
•
Science of measuring heat
•
Heat capacity – energy required to raise temp. by 1
degree (Joules/ °C)
– Specific heat capacity: (s)
• The energy required to raise the temperature
of one gram of a substance by one degree
Celsius.
– Molar heat capacity: (Cp)
• The energy required to raise the temperature
of one mole of substance by one degree
Celsius.
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Section 6.2
Enthalpy and Calorimetry
Calorimetry
•
If two reactants at the same temperature
are mixed and the resulting solution gets
warmer, this means the reaction taking
place is exothermic.
•
An endothermic reaction cools the
solution.
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58
Calorimetry
The amount of heat absorbed or released during a
physical or chemical change can be measured,
usually by the change in temperature of a known
quantity of water in a calorimeter.
Section 6.2
Enthalpy and Calorimetry
A Coffee–Cup
Calorimeter Made of
Two Styrofoam Cups
Coffee-cup calorimetry – in the lab
this is how we experiment to find
energy of a particular system. We use
a Styrofoam® cup, reactants that
begin at the same temperature and
look for change in temperature. After
all data is collected (mass or volume;
initial and final temperatures) we can
use the specific formula to find the
energy released or absorbed. We
refer to this process as constant
pressure calorimetry. ** q = ΔH @
these conditions**
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60
Section 6.2
Enthalpy and Calorimetry
Calorimetry
•
•
•
•
q = s × m × ΔT
• q = Energy released (heat) (or
gained)
s = specific heat capacity (J/°C·g)
m = mass (g)
ΔT = change in temperature (°C)
•
(Q = Cp x m × ΔT) - some use Cp as specific heat
variable.
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61
Units for Measuring Heat
The Joule is the SI system unit for measuring
heat:
The calorie is the heat required to raise the
temperature of 1 gram of water by 1 Celsius
degree
1 BTU (British Thermal Unit) is the heat
required to raise the temperature of 1 pound
of water by 1 F
Section 6.2
Measuring Energy Changes
•
The common energy units for heat are
the calorie and the joule.
 calorie – the amount of energy (heat)
required to raise the temperature of
one gram of water 1oC.
 Joule – 1 calorie = 4.184 joules
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63
Section 6.2
Measuring Energy Changes
Exercise 6
•
Convert 60.1 cal to joules and kJ
1000 J = 1 kiloJoule (kJ)
251 J = 0.251 kJ
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64
Section 6.2
Measuring Energy Changes
•
•
•
•
•
•
 Energy (q) released or gained at constant pressure:
q = m x s x ΔT
q = quantity of heat (Joules or calories)
m = mass in grams
ΔT = Tf −Ti (final – initial)
s = specific heat capacity ( J/g°C) (or Cp with some)
•  Specific heat of water (liquid state) = 4.184 J/g°C (or 1.00
cal/g°C)
• *Water has one of the highest specific heats known! This property
makes life on earth possible and regulates earth’s temperature year
round!
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65
Section 6.2
Measuring Energy Changes
Exercise 7
•
Calculate the amount of heat energy (in
joules) needed to raise the temperature of
6.25 g of water from 21.0°C to 39.0°C.
•
•
Where are we going?
We want to determine the amount of energy
needed to increase the temperature of 6.25 g of
water from 21.0°C to 39.0°C.
What do we know?
The mass of water and the temperature increase.
•
•
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66
Section 6.2
Measuring Energy Changes
q = s × m × ΔT
Example 7
•
Calculate the amount of heat energy (in
joules) needed to raise the temperature of
6.25 g of water from 21.0°C to 39.0°C.
•
•
What information do we need?
We need the specific heat capacity of water.
 4.184 J/g°C
How do we get there?
•
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67
Section 6.2
Measuring Energy Changes
q = s × m × ΔT
Exercise 8
•A sample of pure iron requires 142 cal of energy to raise its
temperature from 23ºC to 92ºC. What is the mass of the sample?
(The specific heat capacity of iron is 0.45 J/gºC.)
–
–
–
–
a)
b)
c)
d)
0.052 g
4.6 g
19 g
590 g
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68
Section 6.2
Measuring Energy Changes
q = s × m × ΔT
Exercise 8
•A sample of pure iron requires 142 cal of energy to raise its
temperature from 23ºC to 92ºC. What is the mass of the sample?
(The specific heat capacity of iron is 0.45 J/gºC.)
–
–
–
–
a)
b)
c)
d)
0.052 g
4.6 g
19 g
590 g
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69
Section 6.2
q = s × m × ΔT
Enthalpy and Calorimetry
Concept Check
Example 9
•A 100.0 g sample of water at 90°C is
added to a 100.0 g sample of water at
10°C.
•The final temperature of the water is:
– a) Between 50°C and 90°C
– b) 50°C
– c) Between 10°C and 50°C
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Section 6.2
q = s × m × ΔT
Enthalpy and Calorimetry
Concept Check
Example 9
•A 100.0 g sample of water at 90°C is
added to a 100.0 g sample of water at
10°C.
•The final temperature of the water is:
– a) Between 50°C and 90°C
– b) 50°C - work on next screens
– c) Between 10°C and 50°C
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71
Section 6.2
Enthalpy and Calorimetry
Problem
•
We start by calling the final, ending temperature 'x.' Keep in mind that BOTH
water samples will wind up at the temperature we are calling 'x.' Also, make sure
you understand that the 'x' we are using IS NOT the Δt, but the FINAL
temperature. This is what we are solving for.
•
The warmer water goes down from to 90 to x, so this means its Δt equals 90 minus
x. The colder water goes up in temperature, so its Δt equals x minus 10.
•
That last paragraph may be a bit confusing, so let's compare it to a number line:
•
---10-------------------------x-----------------------------------90----
•
To compute the absolute distance, it's the larger value minus the smaller value, so
90 to x is 90 minus x and the distance from x to 10 is x - 10.
•
These two distances on the number line represent our two Δt values:
•
a) the Δt of the warmer water is 90 minus x
•
b) the Δt of the cooler water is x minus 10
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72
Section 6.2
Problem
Enthalpy and Calorimetry
• q lost = q gain
• q = (mass) (Δt) (s)
• So:
• (mass) (Δt) (s) = (mass) (Δt) (s)
• With qlost on the left side and qgain on the right side. Since these
are both liquid water the specific heat capacity (s) are equal and
can be eliminated from the equation. Also, since the amount of
water is the same, it too could be eliminated. Then we let x be
final temperature
• Substituting values into the above, we then have
•
•
•
90 - x = x - 10 Then solve for x (the final temperature)
100 = 2x so x = 50
The final temperature is at 50 degrees C.
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73
Section 6.2
Concept Check
Enthalpy and Calorimetry
Exercise 10
•A 100.0 g sample of water at 90.°C is added
• to a 500.0 g sample of water at 10.°C.
q = s × m × ΔT
•The final temperature of the water is:
– a) Between 50°C and 90°C
– b) 50°C
– c) Between 10°C and 50°C
•Calculate the final temperature of the water.
--------•100 x 4.19 x (90-t) = 500 x 4.19 x (t - 10) =
•9000 -100t = 500t - 5000 so 14000 = 600t
•t = 23.333 = 23°C
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74
Section 6.2
Enthalpy and Calorimetry
Concept Check
Exercise 11
q = s × m × ΔT
•You have a Styrofoam cup with 50.0 g of water at
10.C. You add a 50.0 g iron ball at 90.C to the
water. (sH2O = 4.18 J/°C·g and sFe = 0.45 J/°C·g)
•The final temperature of the water is:
– a) Between 50°C and 90°C
– b) 50°C
– c) Between 10°C and 50°C
•Calculate the final temperature of the water.
•(4.18) (t-10) = (.45) (90-t) t = 17.77 = 18°C
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75
Table 6.1
The
Specific
Heat
Capacities
of Some
Common
Substances
Copyright © Houghton Mifflin Company. All rights reserved.
6–
Section 6.2
Enthalpy and Calorimetry
Type 2 Problem Exercise 12
A 110.0 g sample of Cu (specific heat capacity = .20 J/°C·g is heated to
82.4 °C and then placed in a container of water at 22.3°C. The
final temperature of the water and the copper is 24.9°C. What
was the mass of the water in the original container, assuming that
all of the heat lost by the copper is gained by the water?
The heat lost by copper is equal to:
110.0g Cu x 0.20 J/°C·g x (82.4°C - 24.9°C)= 1265 J
The heat lost (-value) by Cu is equal to the heat gained by water.
mH2O = q/(s x Δt) so 1265 J / (4.184 J x (24.9-22.3)) = 116.3 g water.
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77
120
The Heat Curve for Water, going from -20 to 120 oC,
78
The liquid is boiling at
100o
120
C; no temperature change
(use q = mol x ΔHvaporization
)
vap.)
The gas temperature is rising
from 100 to 120 oC
(use
(use q = s x m x ΔT)
The liquid temperature is rising
from 0 to 100 oC
(q
= s x m x ΔT)
The solid is melting at 0o C; no temperature change
(use q = mol x ΔHfusion)
The solid temperature is rising from -20 to 0 oC
(use q = s x m x ΔT)
79
Section 6.2
Exercise 13
Enthalpy and Calorimetry
• 20.0 mL of pure water at 285 K is mixed with 48.0 mL of water
at 315 K. What is the final temperature of the mix? Requires 5
steps. Assume all heat from warm water transferred to cold
water. q gain = q lost
(Density pure water is 1.0 g/mL so 1 gm/L = m/20 mL
•
•
•
•
•
a)
b)
c)
d)
e)
m × ΔT x s = m × ΔT x s
(20 g) (tf-285) (s) = (48 g) (315 - tf) (s)
20tf - 5700 = 15120 - 48tf
68tf = 20820
tf = 306.17 = 306 K
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80
Section 6.2
Exercise 14
Enthalpy and Calorimetry
• 20.0 g of water at 28.5 C is mixed with 32.5 g of water at
C. What is the final temperature after complete mixing?
(Assume no energy is lost to the surroundings.)
•
q cold = q warm
•
•
•
•
•
a)
b)
c)
d)
e)
77.0
m × ΔT x s = m × ΔT x s
(20 g) (tf-28.5) (s) = (32.5 g) (77 - tf) (s) (liquids - s cancels)
20tf - 570 = 2502.5 - 32.5tf
52.5tf = 3072.5
tf = 58.5 C
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Section 6.2
Exercise 15
Enthalpy and Calorimetry
• Exercise 15 In a coffee cup calorimeter, 100.0 mL of 1.0 M
NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions
were originally at 24.6°C. After the reaction, the final
temperature is 31.3°C. Assuming that all solutions have a
density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C,
calculate the enthalpy change for the neutralization of HCl by
NaOH. Assume that no heat is lost to the surroundings or the
calorimeter.
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82
Section 6.2
Exercise 15
Enthalpy and Calorimetry
• Exercise 15 In a coffee cup calorimeter, 100.0 mL of 1.0 M
NaOH and 100.0 mL of 1.0 M HCl are mixed. Both solutions
were originally at 24.6°C. After the reaction, the final
temperature is 31.3°C. Assuming that all solutions have a
density of 1.0 g/cm3 and a specific heat capacity of 4.184 J/g°C,
calculate the enthalpy change for the neutralization of HCl by
NaOH. Assume that no heat is lost to the surroundings or the
calorimeter.
• ANSWER: ‒5.6 kJ/mol (m = mass = 200 g (because we have 100 mL NaOH plus 100 mL HCl each
with a density of 1 g / mL) Then using the equation q = m * s * ΔT = (200 g) (4.184 J/g*C) * 6.7 C) = 5606.56 J = 5.6 kJ
•
Since this is an exothermic reaction, q is negative.
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83
Section 6.2
Exercise 16
Enthalpy and Calorimetry
• When 1 mole of methane (CH4) is burned
at constant pressure, 890 kJ/mol of
energy is released as heat. Calculate ΔH
for a process in which a 5.8 gram sample
of methane is burned at constant
pressure.
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84
Section 6.2
Exercise 16
Enthalpy and Calorimetry
• When 1 mole of methane (CH4) is burned
at constant pressure, 890 kJ/mol of
energy is released as heat. Calculate ΔH
for a process in which a 5.8 gram sample
of methane is burned at constant
• pressure.
ANSWER: ΔH = heat flow = ‒320 kJ/mol
5.8 g / 16.0 g = .3625 moles so set up proportion
1 mole/890 kJ = 0.36 mol/x
so x = (890)(.36)/(1) = 320.4 = 320 kJ/mol and is -320 kJ/mol since
exothermic reaction
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85
Section 6.2
Exercise 17
Enthalpy and Calorimetry
• Exercise 17 Constant-Pressure Calorimetry
• When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed
with 1.00 L of 1.00 M Na2SO4 solution at 25C in a
calorimeter, the white solid BaSO4 forms and the temperature
of the mixture increases to 28.1C. Assuming that the
calorimeter absorbs only a negligible quantity of heat, and that
the specific heat capacity of the solution is 4.18 J/C ⋅ g, and
that the density of the final solution is 1.0 g/mL, calculate the
enthalpy change per mole of BaSO4 formed.
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86
Section 6.2
Exercise 17
Enthalpy and Calorimetry
• Exercise 17 Constant-Pressure Calorimetry
• When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0°C is mixed
with 1.00 L of 1.00 M Na2SO4 solution at 25°C in a
calorimeter, the white solid BaSO4 forms and the temperature
of the mixture increases to 28.1C. Assuming that the
calorimeter absorbs only a negligible quantity of heat, and that
the specific heat capacity of the solution is 4.18 J/C ⋅ g, and
that the density of the final solution is 1.0 g/mL, calculate the
enthalpy change per mole of BaSO4 formed.
• ANSWER = ‒26 kJ/mol
• 2 L total volume = 2000 mL x (1.0 g/mL) = 2000 grams of sol.
•
ΔH = q at constant pressure = q = s × m × ΔT
•
= (4.18 J/C*g) (2000 g) (28.1-25.0 C) = 25,916 Joules or 25.9 = 26 kJ per mole
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87
Specific Heat
The amount of heat required to raise the temperature of
one gram of substance by one degree Celsius.
Substance
Specific Heat (J/g·C)
Water (liquid)
4.18
Ethanol (liquid)
2.44
Water (solid)
2.06
Water (vapor)
1.87
Aluminum (solid)
0.897
Carbon (graphite,solid)
0.709
Iron (solid)
0.449
Copper (solid)
0.385
Mercury (liquid)
0.140
Lead (solid)
0.129
Gold (solid)
0.129
Section 6.2
Enthalpy and Calorimetry
END OF NOTES 6.1-6.2
• HW: Read Section 6.1-6.3 by next Tuesday.
• HW: Pre-lab Calorimetry due on Friday for lab
• Turn in assignments including take home test ch. 5 if not
done already no later than Friday if not turned in today.
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