| 
>>
View the other Key Equations and Concepts in this
chapter
Types of Energy
>> Parts of this equation/concept include:
There are two useful types of energy, heat (q) and work
(w). E = w + q. Energy itself is neither
positive nor negative; however, positive and negative values are
used to indicate direction.
positive values = energy added to system = endothermic
negative values = energy evolved by system = exothermic
It is possible to determine whether a value is positive or negative
by carefully keeping track of the position of initial and final
values for  V or T
calculations. However, it is normally easier to do
calculations as absolute values and then to assign the positive
or negative sign as a final step. That is the way examples in this
chapter are worked.
The usual type of work is PV
work, which occurs under constant pressure. If the pressure is constant,
then
w = PV
where V is the change in volume.
However, this formula does not give energy in its SI unit of joules.
The conversion factor between pressure • volume units and joules
can be derived from two versions of the gas constant (which appear
on most constant tables). R = 8.314 J/mol•K and R
= 0.08206 L•atm/mol•K. If one is divided by the other
(R/R = 1, so it is a fair conversion factor), then the conversion
factor is
|
|
= |
| 8.314 J/mol•K |
|
| 0.08206 L•atm/mol•K |
|
= |
101.3 J/L•atm |
If this conversion factor is used, then in the calculation of PV
must use pressure units of atm and volume units of liters.
>> Example 1
What is the work (in joules) required to expand a gas from 100.0
mL to 1.00 L at 750 torr?
Solution:
To use our conversion factor, volume must be in liters and pressure
in atmospheres, so convert units first.
| 750 torr |
 |
|
= |
0.987 atm |
| 100.0 mL |
|
|
= |
0.1000 L |
V = 1.00 L – 0.1000
L = 0.90 L
| w = PV = (0.987 atm)(0.90 L) |
= |
0.89 L•atm |
|
|
= |
90 J |
>> back
to the Top of the Page
The methods for determining H
values are discussed next. The H
= heat (q) in joules/mol reaction, with whatever sign is
appropriate for direction.
>> Using  H
to Find Energy
Mole of reaction refers to the reaction as it is written
(changing the reaction changes the value of H).
In this case, the stoichiometric coefficient refers to the number
of moles of the reactant or product in a mole of reaction.
>> Example 2
How much energy is involved if 0.263 g K reacts in the following
manner? (Is the energy produced or used up?)
4 K + O2 
2 K2O H
= –722 kJ/mol
Solution:
The stoichiometric coefficient of potassium metal is 4, so there
are 4 moles of potassium for each mole of reaction. H
gives the relationship between energy and moles of reaction. Dimensional
analysis can be used to determine energy.
Notice that the negative sign is dropped in the dimensional analysis.
The sign indicates direction; it does not indicate negative energy.
The negative sign indicates that energy is produced rather than
used up.
>> Example 3
How much energy is involved in the production of 4.00 g NO from
its elements? Is energy used up or produced in this process?
N2 + O2
2 NO H
= +180.4 kJ/mol
Solution:
of energy used up. (The positive sign indicates energy is used
up.)
>>> Phase Changes
A change in physical state involves a change in energy. Because
the energy is used in changing physical state, it is not used
to change the temperature.
The energy involved in changing from solid to liquid is called
the heat of fusion (Hfus).
The value is positive (energy is used up) in changing from solid
to liquid. The value is the same magnitude, but negative in changing
the same substance from liquid to solid.
The energy involved in changing from liquid to gas is called
the heat of vaporization (Hvap).
Hvap is positive
in the change from liquid to gas and negative in the change from
gas to liquid.
There is also a heat of sublimation (Hsub)
for changes from gas to solid.
>> Example 4
How much energy is required to melt 50.00 g water at its melting
point?
Solution:
The Hfus for
water is 6.01 kJ/mol (value from text).
| 50.00 g water |
|
|
|
|
= |
16.7 kJ |
must be added (since melting). |
>>> Fuel Values
The fuel value is the amount of energy produced per gram of fuel.
Since a combustion reaction is normally used to produce the energy
from a fuel, this reaction is assumed for fuel values. The Hcomb
can be changed to a fuel value by converting the moles part of
the unit to grams.
>> Example 5
The H for the combustion
of CH2O is –518 kJ/mol. What is the fuel value
of CH2O?
CH2O + O2
CO2 + H2O
Solution:
The fuel value should have units of kJ/g CH2O. The
negative sign is not included. In H
it represents direction, but since a fuel must be exothermic,
that information is not needed.
| 518 kJ/mol rxn |
|
|
|
|
= |
17.3 kJ/g |
Note: mol of reaction is the unit being changed. Since
it is on the bottom, it must be on top in the dimensional analysis
to cancel.
Since energy of the reaction comes from bonds being formed and
broken, the ratio of bonds to molar mass is one estimate of fuel
value. Consequently, molecules with lower molar mass tend to be
molecules with a large fraction of hydrogens and fewer of the
higher molar mass atoms like oxygen.
>> Example 6
Rank the following by fuel value.
CH2O, H2, CH4, C2H6
Solution:
H2 has the highest fuel value, since it has the
lowest mass-to-bond ratio. CH2O has the lowest fuel
value because the oxygen has a relatively high mass. Methane
(CH4) has a higher ratio of bonds to molar mass than
ethane (C2H6). Consequently, the ranking
is
highest H2, CH4, C2H6,
CH2O lowest
>> Calorimetry
A calorimetry experiment is designed so that no work is done and
all energy is transferred in the form of heat. This experiment takes
advantage of the first law of thermodynamics in the form of heat
lost = heat gained. The heat lost or gained by the surroundings
indicates what happened in the system. The change in energy is determined
from the change in temperature. Energy and temperature are related
by
q = ncPT
(Equation 11.8)
where cP is the molar heat capacity (J/mol•°C),
n is the number of moles, and T
is the change in temperature. The molar heat capacity depends on
the type of substance. Values can be found in appropriate tables.
These values are determined experimentally and are therefore given
in tables.
Instead of molar heat capacity, some tables use specific heat (J/g•°C).
If specific heat is used, the equation becomes
q = (mass)(specific heat)(T)
Since this text uses molar heat capacity, all examples are worked
using Equation 11.8.
>>> Relating Heat and Temperature Change
Equation 11.8 is used to relate temperature change to heat energy.
>> Example 7
How much heat is lost when 36.0 g platinum (cP
= 25.94 J/mol•°C) at 35.0 °C is cooled to 27.4
°C?
Solution:
q = ncPT
The heat lost is q.
| n = moles Pt |
= |
36.0 g |
|
|
= |
0.184 mol |
| cP |
= |
heat capacity of platinum, given in problem |
| |
= |
25.94 J/mol•°C |
T = 35.0 – 27.4
= 7.6 °C
q = (0.184 mol)(25.94 J/mol•°C)(7.6 °C)
= 36 J
The use of two significant figures is due to the T
value.
>> Example 8
If 72.5 g of gold (cP = 25.41 J/mol•°C)
at 64.2 °C is added to 100.0 g of water (cP
= 75.3 J/mol•°C) at 23.0 °C, what is the final
temperature of the water, assuming no energy is lost to the
container?
Solution:
All the heat energy is lost by the gold and gained by the water.
The final temperature of the gold and water will be the same
(T) when all the heat is transferred. The temperature
of the water will increase, so its T
= T – 23.0. The temperature of the gold will decrease,
so its T = 64.2 – T.
By writing the T so that
the value is positive, the equation is worked in absolute value
and the signs will all work out correctly.
| n of gold |
= |
72.5 g |
|
|
= |
0.368 mol Au |
| n of water |
= |
100.0 g |
|
|
= |
5.55 mol water |
| q lost by gold |
= |
ncPT |
| |
= |
(0.368 mol)(25.41 J/mol•°C)(64.2 – T) |
| q gained by water |
= |
ncPT |
| |
= |
(5.55 mol)(75.3 J/mol•°C)(T – 23.0) |
According to the first law of thermodynamics
q lost by gold = q gained by water
(0.368)(25.41)(64.2 – T) = (5.55)(75.3)(T
– 23.0)
9.35(64.2 – T) = 417.915(T – 23.0)
600.27 – 9.35T = 417.915T – 9612.045
10212.315 = 427.265T
23.9 °C = T
>>> Heat Capacity
A calorimeter is a device designed so that all energy is transferred
in the form of heat to the calorimeter. The calorimeter acts as
the surroundings in a calorimetry experiment. The calorimeter
consists of both the container and its contents. The heat capacity
of the calorimeter (cP) refers to the entire
calorimeter, so its units are J/°C or kJ/°C. (No quantity
unit is needed because the quantity is one calorimeter.) The heat
gained or lost by the calorimeter is then
q = cPT
(Equation 11.11)
The heat capacity of a calorimeter must be determined experimentally,
since even instruments of the same type will not have the same
heat capacity. This requires that a known reaction be used. The
traditional reaction is the combustion of benzoic acid (C7H6O2).
The combustion of benzoic acid will produce 26.38 kJ for each
gram of benzoic acid.
>> Example 9
What is the heat capacity of the calorimeter if 1.395 g benzoic
acid changes the temperature of the calorimeter from 21.9 °C
to 39.9 °C?
Solution:
| Heat evolved by the benzoic acid |
= |
heat absorbed by the calorimeter |
| Heat evolved by the benzoic acid |
= |
1.395 g |
|
|
| |
|
|
| |
= |
36.80 kJ |
| |
|
|
| |
= |
q |
Change in temperature = 39.9 – 21.9 = 18.0 °C
q = CPT
36.80 kJ = CP(18.0 °C)
2.04 kJ/°C = CP = heat capacity of
calorimeter
>> Example 10
Using the preceding calorimeter, how much heat changes the
temperature from 23.4 °C to 20.3 °C?
Solution:
The question asked for heat energy, q.
q = CPT
The heat capacity was calculated in the previous example =
2.04 kJ/mol
The T = 3.1 °C
Therefore
q = (2.04 kJ/°C)(3.1 °C) = 6.3 kJ
>>> Determining
H
of Reaction
The Hrxn can be
determined from q/mol reaction. The sign is determined
from the direction of the temperature change. If the temperature
increases, the reaction must have given off energy; therefore
the sign on H is negative.
If the temperature decreases, the sign on H
is positive.
The value of q can be calculated from either Equation
11.8 or Equation 11.11, depending on the data given.
Moles of reaction are determined from the chemical reaction and
the amount of reactant used up or the amount of product produced.
>> Example 11
If 1.902 g of hydrogen gas reacting with excess oxygen raises
the temperature of 150.0 g of water from 21.2 °C to 57.5
°C, what is the Hrxn
(in kJ/mol)? Assume no energy is lost to the surroundings.
2 H2 + O2
2 H2O
Solution:
H = –q/(mol
rxn). Because the temperature increased, the H
value must be negative.
| mol rxn |
= |
0.1902 g H2 |
|
|
|
|
= |
0.0471 mol rxn |
q lost by rxn = q gained by water =
nwatercP,waterT
| nwater |
= |
150.0 g |
|
|
= |
8.329 mol water |
T = 57.5 – 21.2
= 36.3 (remember to use the absolute value)
| q |
= |
(8.329)(75.3)(36.3) |
| |
|
|
| |
= |
2.28 x 104 J |
|
|
| |
|
|
| |
= |
22.8 kJ |
| H |
= |
|
| |
|
|
| |
= |
|
| |
|
|
| |
= |
–484 kJ/mol |
>> Example 12
What is the Hrxn
if the dissolution of 1.645 g of ammonium chloride in 200.0
mL water decreases the temperature from 23.4 °C to 18.3
°C? (Assuming no energy is lost to the surroundings.)
Solution:
H = (+q NH4Cl)/(mol
reaction). Because the temperature of the surroundings decreased,
the reaction must have taken energy from the surroundings and
H is positive. The dissolution
reaction is
NH4Cl
NH4+ + Cl–
| mol rxn |
= |
1.645 g NH4Cl |
|
|
|
|
= |
0.03075 mol |
| mol water |
= |
200.0 g |
|
|
= |
11.10 mol |
cP = 75.3 J/mol•°C for water
T = 23.4 – 18.3
= 5.1 °C
| q NH4Cl |
= |
qwater |
| |
= |
nwatercPT |
| |
= |
(11.10)(75.3)(5.1) |
| |
= |
4262.73 J (only two digits are significant) |
| H |
= |
|
| |
|
|
| |
= |
+138625 J/mol |
|
|
| |
|
|
| |
= |
+138.6 kJ/mol |
H = +1.4 x 105
J/mol = +1.4 x 102 kJ/mol
>> Example 13
What is the Hrxn
if the reaction of 0.556 g Na raises the temperature of a calorimeter
(CP = 3.145 kJ/°C) from 19.23 °C to
20.65 °C?
2 Na + 2 H2O
2 NaOH + H2
Solution:
H = –q/mol.
Since the temperature increases, the reaction is exothermic
and H is negative.
| mol rxn |
= |
0.556 g |
|
|
|
|
= |
0.0121 mol rxn |
cP = 3.145 kJ/°C from the problem
T = 20.65 – 19.23
q = cPT
= (3.145 kJ/°C)(1.42) = 4.46 kJ
| H |
= |
|
= |
–369 kJ/mol |
>> Determining H
from Bond Energies
Average bond energies (BE) are listed in Table 11.2. To determine
what bonds exist in the molecules of the products and reactants,
you may have to draw the Lewis structure. The Hrxn
is determined from the formula
 BEbroken – BEmade
= Hrxn
Since it takes energy to break bonds, those values are positive.
Energy is released when bonds are made; therefore those values are
negative. The formula is set up to introduce the positive and negative
values, so use the absolute values in the table when using the formulas.
One method is to break all bonds of the reactants and form all
bonds of the products. This will always work. However, sometimes
the bonds that break and form are obvious. In this case, it is unnecessary
to break and form every bond. Instead, look only at the bonds breaking
and forming. This method uses fewer numbers, so math errors are
less likely.
>> Example 14
What is the Hrxn
of the following reactions, based on bond energies?
- 2 HCl + F2
Cl2 + 2 HF
- C2H4 + H2
C2H6
- CH4 + 2 O2
CO2 + 2 H2O
Solution:
-
Written in semi-Lewis structures (just showing bonds, not
lone pairs of electrons), this reaction is
H–Cl + H–Cl + F–F
Cl–Cl + H–F + H–F
Therefore two H–Cl bonds and one F–F bond are broken;
two H–F and one Cl–Cl bonds are formed. The bond
energies are H–Cl = 431 kJ/mol, F–F = 155 kJ/mol,
Cl–Cl = 242 kJ/mol and H–F = 565 kJ/mol. Using the
preceding formula
H = [2(431) + 155]
– [242 + 2(565)] = 1017 – 1372 = –355 kJ/mol
-
Written as semi-Lewis structures, this reaction is
The carbon–hydrogen bonds on the first structure are
not broken, so one way to do this is to say the bonds broken
are C=C (612 kJ/mol) and H–H of H2 (436 kJ/mol)
and bonds formed are one C–C (348 kJ/mol) and two C–H
(413 kJ/mol), so
H = [612 + 436] –
[348 + 2(413)] = 1048 – 1174 = –126 kJ/mol
If that is not obvious, all the bonds in the reactants (C=C
is 612 kJ/mol, C–H is 413 kJ/mol and H–H is 436
kJ/mol) can be broken and all the bonds in the products formed
(C–C is 348 kJ/mol and C–H is 413 kJ/mol).
H = [612 + 4(413) +
436] – [348 + 6(413)] = –126 kJ/mol
-
CH4 consists of 4 C–H bonds, O2
has a O=O bond, and the product carbon dioxide is O=C=O and
water is H–O–H, so bonds broken are 4 C–H bonds
(413 kJ/mol) and one O=O bond (497 kJ/mol) and bonds formed
are two C=O bonds (743 kJ/mol) and two O–H bonds (463
kJ/mol). So
H = [4(413) + 497]
– [2(743) + 2(463)] = 2149 – 2412 = –263
kJ/mol
>> Determining H
from Formation Reactions
>>> Writing Formation Reactions
A formation reaction forms 1 mole of product from its elements
in standard state. This requires knowing the standard state of
the elements. The standard states of the elements are the following:
metals and metalloids are monoatomic solids, except Hg,which is
a monoatomic liquid; the noble gases are all monoatomic gases;
halogens, hydrogen, oxygen, and nitrogen are all diatomic gases,
except iodine, which is a diatomic solid and bromine, which is
a diatomic liquid; carbon's standard state is graphite; and phosphorus
is solid P4. There is some dispute as to whether the
standard state of sulfur is S(s) or S8(s).
(Check with your professor for his or her preference.)
The other important part of formation reactions is that only
1 mole of product is formed. Because it is 1 mole, the
stoichiometric coefficients can be and often are fractions.
>> Example 15
Write formation reactions for the following compounds.
- H3PO4(l)
- C2H6O(l)
- Zn(NO3)2(s)
- XeF4(g)
Solution:
-
The standard state of hydrogen is H2(g).
The standard state of oxygen is O2(g).
The standard state of phosphorous is P4(s).
Therefore the formation reaction is
H2(g) + P4(s) + O2(g)
H3PO4(l)
The reaction is balanced so that there is 1 mole of product.
Therefore, the balanced reaction is
3/2 H2(g) + 1/4 P4(s)
+ 2 (O2(g)
H3PO4(l)
-
The standard state of carbon is C(graphite). The
standard state of hydrogen is H2(g). The
standard state of oxygen is O2(g). Therefore
the formation reaction is
C(graphite) + H2(g) + O2(g)
C2H6O(l)
Balanced, the reaction is
2 C(graphite) + 3 H2(g) + 1–2
O2(g)
C2H6O(l)
-
The standard state of zinc is Zn(s). The standard
state of nitrogen is N2(g). The standard
state of oxygen is O2(g). There the reaction,
balanced, is
Zn(s) + N2(g) + 3 O2(g)
Zn(NO3)2(s)
-
The standard state of xenon is Xe(g). The standard
state of fluorine is F2(g). Therefore
the balanced formation reaction is
Xe(g) + 2 F2(g)
XeF4(g)
>>> DH
from Hf°
If the H is determined solely
from enthalpies of formation, the following formula can be used.
Hrxn = nHf,products
– mHf,reactants
where n is the stoichiometric coefficient of each product
and m is the stoichiometric coefficient of each reactant.
Values for Hf are
listed in the appendix.
>> Example 16
Using heats of formation, what is the Hrxn
for the following? (All reactants and products are solids.)
Al + Fe2O3
Al2O3 + Fe
Solution:
Since Al and Fe are elements in their standard states, their
Hf is zero. The
Hrxn for ferric
oxide is –824.2 kJ/mol and that for aluminum oxide is –1676
kJ/mol. Therefore
Hrxn = [–1676
+ 0] – [–824.2 + 0] = –851.8 kJ/mol
>> Example 17
What is the Hrxn
of the following reactions based on heats of formation?
H2SO4(aq) + 2 NaOH(aq)
2 Na+(aq) + SO42–(aq)
+ 2 H2O(l)
Solution:
From the appendix
| Compound |
H (kJ/mol) |
| H2SO4 |
–909.3 |
| NaOH |
–470.1 |
| Na+ |
–240.1 |
| SO42– |
–909.3 |
| H2O |
–285.8 |
Be sure—particularly with hydrochloric acid, sodium hydroxide,
and water—that you use H
appropriate to the physical state. There is often more than
one listing for these compounds.
| H |
= |
[2(–240.1) + (–909.3)
+ 2(–285.8)] – [–909.3 + 2(–470.1)] |
| |
= |
[–1961.1] – [–1849.5] |
| |
= |
–111.6 kJ/mol |
>> Determining H
from Hess's Law
To determine the enthalpy of a reaction from other reactions, the
other reactions must be manipulated so that the sum of the reactions
is the reaction of interest. Reactions may be reversed or the stoichiometric
coefficients may be multiplied by a factor to achieve this.
The first step in doing this is to make sure the products and reactants
are in the appropriate positions. Therefore reverse the reactions
as necessary to accomplish this. If you reverse a reaction, change
the sign on the H value.
The next step is to make sure the stoichiometric coefficients match.
Determine the multiplication factor necessary to match the stoichiometric
coefficients, and multiply every stoichiometric coefficient of the
reaction and the H value by
that factor.
Add the reactions together, canceling any substances that are the
same on either side of the equation. Remember that the stoichiometric
coefficients refer to how many of each substance, so these may be
added or partially canceled.
If any substances remain that are not in the desired reaction,
more reactions may be needed to cancel those compounds. Reverse
or multiply as above to accomplish this.
>> Example 18
What is the Hrxn
for C2H4 + H2O
C2H6O based on the following reactions?
H2 + C2H4
C2H6 H
= –137 kJ/mol
2 C2H6 + O2
2 C2H6O H
= –300 kJ/mol
2 H2 + O2
2 H2O H = –572
kJ/mol
Solution:
For clarity, let's label the reactions.
- H2 + C2H4
C2H6 H
= –137 kJ/mol
- 2 C2H6 + O2
2 C2H6O H
= –300 kJ/mol
- 2 H2 + O2
2 H2O H
= –572 kJ/mol
- C2H4 + H2O
C2H6O
Reaction 1 contains a reactant (C2H4) from
the desired equation (#4), which is also in the reactant position.
Since the stoichiometric coefficient of C2H4
reaction 1 is the same as reaction 4, it can remain unchanged.
- H2 + C2H4
C2H6 H
= –137 kJ/mol
Reaction 2 contains the product (C2H6O)
of the reaction 4. The product is in the appropriate position,
but the stoichiometric coefficient is incorrect. To make that
correct, the entire equation needs to be multiplied by 1/2. If
the equation is multiplied by 1/2, the H
must also be multiplied by 1/2.
- C2H6 + 1/2 O2
C2H6O H
= –150 kJ/mol
Reaction 3 contains the other reactant (H2O), but
it is in the product position, so the reactions should be reversed.
If the reaction is reversed, the sign is changed.
- 2 H2O
2 H2 + O2 H
= +572 kJ/mol
In addition, the stoichiometric coefficient must be changed,
so the equation should be multiplied by 1/2.
- H2O
H2 + 1/2 O2 H
= +286 kJ/mol
Next, equations, 1, 2, and 3 are added together.
H2O + C2H6 + 1/2 O2
+ H2 + C2H4
C2H6 + C2H6O + H2
+ 1/2 O2
Substances that are the same on each side can be canceled, giving
the desired equation:
C2H4 + H2O
C2H6O
As the equations are added, so are the H
values, so
H = –137 – 150
+ 286 = –1 kJ/mol
>> Example 19
What is the Hrxn
for Pb2+ + 2 Br–
PbBr2 based on the following reactions?
Pb + Br2
PbBr2 H
= –279 kJ/mol
Pb
Pb+ + e– H
= +716 kJ/mol
Pb+
Pb2+ + e– H
= +844 kJ/mol
Br + e–
Br– H
= –324 kJ/mol
Br2
2 Br H
= +111 kJ/mol
Solution:
Labeling the reactions for convenience,
- Pb + Br2
PbBr2 H
= –279 kJ/mol
- Pb
Pb+ + e– H
= +716 kJ/mol
- Pb+
Pb2+ + e– H
= +844 kJ/mol
- Br + e–
Br– H
= –324 kJ/mol
- Br2
2 Br H
= +111 kJ/mol
Reaction 1 contains the product with the appropriate stoichiometric
coefficient, so it remains unchanged.
- Pb + Br2
PbBr2 H
= –279 kJ/mol
Reaction 3 contains a reactant (Pb2+), but in the
product position, so it should be reversed.
- Pb2+ + e–
Pb+ H
= –844 kJ/mol
Reaction 4 contains a reactant (Br–), but in the product position, so it should be reversed.
- Br–
Br + e– H
= +324 kJ/mol
The stoichiometric coefficient should also be changed by multiplying
reaction 4 by 2.
- 2 Br–
2 Br + 2e– H
= +648 kJ/mol
Adding reactions 1, 3, and 4, you get
2 Br– + Pb2+ + e–
+ Pb + Br2
PbBr2 + Pb+ + 2Br + 2e–
H = –279 – 844
+ 648 = –475 kJ/mol
Only one electron cancels from this reaction.
- 2 Br– + Pb2+ + Pb + Br2
PbBr2 + Pb+ + 2 Br + e– H
= –475 kJ/mol
To remove the Pb, the reverse of reaction 2 is needed:
- Pb+ + e–
Pb H
= –716 kJ/mol
To remove Br2, the reverse of equation 5 is needed:
- 2 Br
Br2 H
= –111 kJ/mol
Adding these to equation 6:
Pb+ + e– + 2 Br + 2Br–
+ Pb2+ + Pb + Br2
PbBr2 + Pb+ + 2Br + e–
+ Br2 + Pb
The extra substances cancel to get the correct equation:
Pb2+ + 2 Br–
PbBr2
H = –475 – 716
– 111 = –1302 kJ/mol
>> View
the other Key Equations and Concepts in this chapter
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H to Find Energy
