1 Energy: The Ability to Do Work
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7.1 | Energy: The Ability to Do Work
Figure 7.1 | Liquid
hydrogen and oxygen
serve as fuel for the
space shuttle. The three
almost invisible points of
blue flame come from the
main engines of the space
shuttle, which consume
hydrogen and oxygen in
the formation of water.
(Corbis Images)
the book, which changes its position, increases the potential
energy. This energy is supplied by the person doing the lifting.
Letting the book fall allows the potential energy to decrease. The
lost potential energy is changed to kinetic energy as the book
gains speed during its descent.
How potential energy varies with position for objects that
attract or repel each other can be illustrated by magnets depicted
in Figure 7.2. When the magnets are held in a certain way, the
end of one magnet repels the end of another; we say the “north
pole” of one repels the “north pole” of the other. Because the
repulsions push the magnets apart, we have to do some work to
move them closer together. This causes their potential energy to
increase; the energy we expended pushing them together is now
stored by the magnets. If we release them, they will push each
other apart, and this “push” could also be made to do work for
us. Likewise, if we turn one magnet around, the ends of the two
magnets will attract each other—that is, the “south pole” of one
magnet attracts the “north pole” of the other. Now, to pull the
magnets apart, we have to do work, and this will increase their
potential energy.
Factors That Affect Potential Energy
• Potential energy increases when objects that
attract each other move apart, and decreases
when they move toward each other.
• Potential energy increases when objects that
repel each other move toward each other, and
decreases when they move apart.
255
+
Potential
energy
changes
If you ever have trouble remembering this, think back to the magnets
and how the poles attract or repel each other.
In chemical systems, the amount of chemical energy is determined by the type and
arrangement of the atoms. There aren’t north and south poles, but there are attractions
and repulsions between electrical charges. The attraction between opposite charges is
S
N
N
S
(a)
S
N
N
S
(b)
S
(c)
jespe_c07_253-302hr.indd 255
N
N
S
Figure 7.2 | The potential
energy of two magnets with the
same poles facing depends on
their distance apart. The red
arrows represent the amount of
repulsion between the two
magnets. (a) The north poles of
two magnets repel when placed
near each other. (b) When the
magnets are moved apart, their
potential energy decreases.
(c) When they are moved together,
their potential energy increases.
(Corbis Images)
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256 Chapter 7 | Energy and Chemical Change
similar to the attraction between two magnets in which the opposite poles are next to each
other. Electrons are attracted to protons because of their opposite electrical charges.
Electrons repel electrons and nuclei repel other nuclei because they have the same kind of
electrical charge. Changes in the relative positions of these particles, as atoms join to form
molecules or break apart when molecules decompose, leads to changes in potential energy.
These are the kinds of potential energy changes that lead to the release or absorption of
energy by chemical systems during chemical reactions.
Electron
–
Nucleus
d
+
Low potential energy
Atom A
Electron
–
Nucleus
d
+
High potential energy
Atom B
The positively charged nucleus
attracts the negatively charged
electron. As the electron is brought
closer to the nucleus, the potential
energy of the electron decreases. It
takes work to separate the electron
and the nucleus, so the electron in
atom B has more potential energy
than the electron in atom A.
Law of Conservation of Energy
One of the most important facts about energy is that it cannot be created or destroyed; it can
only be changed from one form to another. This fact was established by many experiments
and observations, and is known today as the law of conservation of energy. (Recall that when
scientists say that something is “conserved,” they mean that it is unchanged or remains
constant.) You observe this law whenever you toss something—a ball, for instance—into
the air. You give the ball some initial amount of kinetic energy when you throw it. As it
rises, its potential energy increases. Because energy cannot come from nothing, the rise in
potential energy comes at the expense of the ball’s kinetic energy. Therefore, the ball’s
1 mv 2
becomes smaller. Because the mass of the ball cannot change, the velocity (v)
2
becomes less and the ball slows down. When all of the kinetic energy has changed to
potential energy, the ball can go no higher; it has stopped moving and its potential energy
is at a maximum. The ball then begins to fall, and its potential energy is changed back into
kinetic energy.
Heat and Temperature
Figure 7.3 | Energy transfer
from a warmer to a cooler
object. (a) The longer trails on the
left denote higher kinetic energies
of a hot object like hot water just
before it’s poured into a cooler
coffee cup, indicated with shorter
arrows. (b) Collisions between the
hot water molecules and the cooler
molecules in the cup’s material
cause the water molecules to slow
down and the cup’s molecules to
speed up, as kinetic energy is
transferred from the water to the
cup. (c) Thermal equilibrium is
established: the temperatures of the
water and the cup wall are now
equal.
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If we think about the molecular level, atoms, molecules, or ions are constantly moving and
colliding with each other, which means that any object, even if we cannot see it moving,
will have some kinetic energy. To get an idea of the average kinetic energy of the particles,
we can measure its temperature, since the temperature of an object is proportional to the
average kinetic energy of its particles. The higher the average kinetic energy is, the higher the
temperature. What this means is that when the temperature of an object is raised, the molecules move faster, giving rise to more collisions with the thermometer. (Recall that KE =
1 mv 2 . Increasing the average kinetic energy doesn’t increase the masses of the atoms, so it
2
must increase their speeds.) Likewise, when the temperature is reduced, the particles move
slower and the average kinetic energy of the molecules is lower, so the number of collisions
and the force of the collisions of the particles is lower.
Heat is the kinetic energy (also called thermal energy) that
is transferred on the molecular level between objects caused
by differences in their temperatures. As you know, heat
always passes spontaneously from a warmer object to a cooler
one. This energy transfer continues until both objects come
(a)
to the same temperature. Whereas heat is the amount of
energy that is transferred, the temperature is proportional
to the average kinetic energy of the object.
On the molecular level, when a hot object is placed in
contact with a cold one, the faster atoms of the hot object
collide with and lose some kinetic energy to the slower
(b)
atoms of the cold object (Figure 7.3). This decreases the
average kinetic energy of the particles of the hot object,
causing its temperature to drop. At the same time, the average kinetic energy of the particles in the cold object is
raised, causing the temperature of the cold object to rise.
Eventually, the average kinetic energies of the atoms in
(c)
both objects become the same and the objects reach the
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7.2 | Internal Energy
257
same temperature. Thus, the transfer of heat is interpreted as a transfer of kinetic energy
between two objects.
The Joule
The SI unit of energy is a derived unit called the joule (symbol J) and corresponds to the
amount of kinetic energy possessed by a 2 kilogram object moving at a speed of
1 meter per second. Using the equation for kinetic energy, KE = 12 mv 2,
2
( )
1
1m
(2 kg)
2
1s
2 –2
1 J = 1 kg m s
1J =
The joule is actually a rather small amount of energy and in most cases we will use the
larger unit, the kilojoule (kJ); 1 kJ = 1000 J = 103 J.
Another energy unit you may be familiar with is called the calorie (cal). Originally, it was
defined as the energy needed to raise the temperature of 1 gram of water by 1 degree Celsius.
With the introduction of the SI, the calorie has been redefined as follows:
1 cal = 4.184 J (exactly)
(7.2)
The larger unit kilocalorie (kcal), which equals 1000 calories, can also be related to the
kilojoule:
1 kcal = 4.184 kJ
The nutritional or dietary Calorie (note the capital), Cal, is actually one kilocalorie.
1 Cal = 1 kcal = 4.184 kJ
While joules and kilojoules are the standard units of energy, calories and kilocalories are
still in common use, so you will need to be able to convert joules into calories and vice versa.
7.2 | Internal Energy
In Section 7.1 we introduced heat as a transfer of energy that occurs between objects with
different temperatures. For example, heat will flow from a hot cup of coffee into the cooler
surroundings. Eventually the coffee and surroundings come to the same temperature and
we say they are in thermal equilibrium with each other. The temperature of the coffee has
dropped and the temperature of the surroundings has increased a bit.
Energy that is transferred as heat comes from an object’s internal energy. Internal energy
is the sum of energies for all of the individual particles in a sample of matter. All of the
particles within any object are in constant motion. For example, in a sample of air at room
temperature, oxygen and nitrogen molecules travel faster than rifle bullets, constantly colliding with each other and with the walls of their container. The molecules spin as they
move, the atoms within the molecules jiggle and vibrate, and the electrons move around
the atoms; these internal molecular motions also contribute to the kinetic energy of the
molecule and, thus, to the internal energy of the sample. We’ll use the term molecular
kinetic energy for the energy associated with such motions. Each particle has a certain value
of molecular kinetic energy at any given moment. Molecules are continually exchanging
energy with each other during collisions, but as long as the sample is isolated, the total
kinetic energy of all the molecules remains constant.
Internal energy is often given the symbol E.1 In studying both chemical and physical
changes, we will be interested in the change in internal energy that accompanies the process. This is defined as DE, where the symbol D (Greek letter delta) signifies a change.
n Recall that kinetic energy is the
energy of motion and is given by
KE = 1/2 mv 2, where m is the mass
of an object and v is its velocity.
DE = Efinal - Einitial
1
Sometimes the symbol U is used for internal energy.
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258 Chapter 7 | Energy and Chemical Change
For a chemical reaction, Efinal corresponds to the internal energy of the products, so we’ll
write it as Eproducts. Similarly, we’ll use the symbol Ereactants for Einitial. So for a chemical reaction the change in internal energy is given by
DE = Eproducts - Ereactants
(7.3)
Notice an important convention illustrated by this equation. Namely, changes in something like temperature (D t) or in internal energy (DE ) are always figured by taking “final
minus initial” or “products minus reactants.” This means that if a system absorbs energy
from its surroundings during a change, its final energy is greater than its initial energy and
DE is positive. This is what happens, for example, when photosynthesis occurs or when a
battery is charged. As the system (the battery) absorbs the energy, its internal energy
increases and is then available for later use elsewhere.
Temperature and Average Molecular Kinetic Energy
n Temperature is related to the
average molecular kinetic energy, but
it is not equal to it. Temperature is not
an energy.
The idea, introduced in Section 7.1, that atoms and molecules are in constant random
motion forms the basis for the kinetic molecular theory. It is this theory that tells us in part
that temperature is related to the average kinetic energy of the atoms and molecules of an
object. Be sure to keep in mind the distinction between temperature and internal energy;
the two are quite different. Temperature is related to the average molecular kinetic energy,
whereas internal energy is related to the total molecular kinetic energy.2
The concept of an average kinetic energy implies that there is a distribution of kinetic
energies among the molecules in an object. Let’s examine this further. At any particular
moment, the individual particles in an object are moving at different velocities, which
gives them a broad range of kinetic energies. An extremely small number of particles will
be standing still, so their kinetic energies will be zero. Another small number will have very
large velocities; these will have very large kinetic energies. Between these extremes there
will be many molecules with intermediate amounts of kinetic energy.
The graphs in Figure 7.4 show the kinetic energy distributions among molecules in a
sample at two different temperatures. The vertical axis represents the fraction of molecules
with a given kinetic energy (i.e., the number of molecules with a given KE divided by the total
number of molecules in the sample). Each curve in Figure 7.4 describes how the fraction of
molecules with a given KE varies with KE.
Each curve starts out with a fraction equal to zero when KE equals zero. This is because
the fraction of molecules with zero KE (corresponding to molecules that are motionless)
is essentially zero, regardless of the temperature. As we move along the KE axis we find that
the fraction of the molecules increases, reaches a peak, and then declines toward zero
again. For very large values of KE, the fraction drops off again because very few molecules
have very large velocities.
Each curve in Figure 7.4 has a characteristic peak or maximum corresponding to the
most frequently experienced values for molecular KE. Because the curves are not symmetrical, the average values of molecular KE lie slightly to the right of the maxima. Notice
that when the temperature increases (going from curve 1 to 2), the curve flattens and the
maximum shifts to a higher value, as does the average molecular KE. In fact, if we double
the Kelvin temperature, the average KE also doubles, so the Kelvin temperature is directly proportional to the average KE. The reason the curve flattens is because the area under each
curve corresponds to the sum of all the fractions, and when all the fractions are added, they
must equal one. (No matter how you cut up a pie, if you add all the fractions, you must
end up with one pie!)
2
When we’re looking at heat flow, the total molecular kinetic energy is the part of the internal energy we’re
most interested in. However, internal energy is the total energy of all the particles in the object, so molecular
potential energies (from forces of attraction and repulsion that operate between and within molecules) can
and do make a large contribution to the internal energy. This is especially true in liquids and solids.
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Fraction of the total
number of particles having
a particular kinetic energy
7.3 | Measuring Heat
0.6
1
0.4
Average kinetic energy
2
0.2
0
0
5
10
15
20
Kinetic energy in arbitrary units
25
259
Figure 7.4 | The distribution of kinetic energies among gas
particles. The distribution of individual kinetic energies changes
in going from a lower temperature, curve 1, to a higher temperature, curve 2. The highest point on each curve is the most
probable value of the kinetic energy for that temperature—that
is, the one we would most frequently find if we could observe
and measure the kinetic energy of each molecule. At the lower
temperature, this peak is at lower values for the kinetic energy. At
the higher temperature, more molecules have high speeds and
fewer molecules have low speeds, so the maximum shifts to the
right and the curve flattens.
We will find the graphs illustrated in Figure 7.4 very useful later when we discuss the
effect of temperature on such properties as the rates of evaporation of liquids and the rates
of chemical reactions.
State Functions
Equation 7.3 defines what we mean by a change in the internal energy of a chemical system during a reaction. This energy change can be made to appear entirely as heat, and as
you will learn soon, we can measure heat. However, there is no way to actually measure
either Eproducts or Ereactants, since this would require measuring all of the molecular motions,
attractions, and repulsions. Fortunately, we are more interested in DE than we are in the
absolute amounts of energy that the reactants or products have.
Even though we cannot measure internal energy, it is important to discuss it. As it turns
out, the energy of an object depends only on the object’s current condition. It doesn’t matter
how the object acquired that energy, or how it will lose it. This fact allows us to be concerned with only the change in energy, DE.
The complete list of properties that specify an object’s current condition is known as the state
of the object. In chemistry, it is usually enough to specify the object’s pressure, temperature, volume, and chemical composition (numbers of moles of all substances present) to
give the state of the object.
Any property, like energy, that depends only on an object’s current state is called a state
function. Pressure, temperature, and volume are state functions. A system’s current temperature, for example, does not depend on what it was yesterday. Nor does it matter how
the system acquired it—that is, the path to its current value. If it’s now 25 °C, we know
all we can or need to know about its temperature. Also, if the temperature were to
increase—say, to 35 °C—the change in temperature, D t, is simply the difference between
the final and the initial temperatures:
D t = tfinal - tinitial
(7.4)
n “State,” as used in
thermochemistry, doesn’t have the
same meaning as when it’s used in
terms such as “solid state” or “liquid
state.”
n The symbol D denotes a change
between some initial and final state.
To make the calculation of D t, we do not have to know what caused the temperature
change; all we need are the initial and final values. This independence from the method or
mechanism by which a change occurs is the important feature of all state functions. As we’ll see
later, the advantage of recognizing that some property is a state function is that many
calculations are then much easier.
7.3 | Measuring Heat
By measuring the amount of heat that is absorbed or released by an object, we are able to
quantitatively study nearly any type of energy transfer. For example, if we want to measure
the energy transferred by an electrical current, we can force the electric current through
something with high electrical resistance, like the heating element in a toaster, and the
energy transferred by the current becomes heat.
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260 Chapter 7 | Energy and Chemical Change
In our study of energy transfer, it is very important to specify the
boundary across which the heat flows. The boundary might be visible
(like the walls of a beaker) or invisible (like the boundary that separates warm air from cold air along a weather front). The boundary
encloses the system, which is the object we are interested in studying. Everything outside the system is called the surroundings. The
system and the surroundings together comprise the universe.
Three types of systems are possible, depending on whether matter or energy can cross the boundary as shown in Figure 7.5.
(a)
(b)
Figure 7.5 | Open, closed, and
isolated systems. The open
system (a) allows the exchange of
both energy and mass; the closed
system (b) only allows the
exchange of energy; and the
isolated system (c) is sealed and
insulated so no heat or mass can be
exchanged with the surroundings.
• Open systems can gain or lose mass and energy across their
boundaries. The human body is an example of an open system.
• Closed systems can absorb or release energy, but not mass, across the boundary. The
mass of a closed system is constant, no matter what happens inside. A lightbulb is
an example of a closed system.
• Isolated systems cannot exchange matter or energy with their surroundings. Because
energy cannot be created or destroyed, the energy of an isolated system is constant,
no matter what happens inside. Processes that occur within an isolated system are
called adiabatic, from the Greek a + diabatos, meaning not passable. A stoppered
Thermos bottle is a good approximation of an isolated system.
(c)
The Heat and Temperature Change
If the only type of energy that is transferred between two objects is heat energy, then all of
the heat lost by one object must be gained by the second object since we have to obey the
law of conservation of energy. Unfortunately, there is no instrument available that directly
measures heat. Instead, we measure temperature changes, and then use them to calculate
changes in heat energy. Experience tells you that the more heat you add to an object, the
more its temperature will rise. In fact, experiments show that the temperature change, D t,
is directly proportional to the amount of heat absorbed, which we will identify by the symbol q. This can be expressed in the form of an equation as
q = C D t
Heat capacity
(7.5)
where C is a proportionality constant called the heat capacity of the object. The units of
heat capacity are usually J °C-1 (or J/°C), which expresses the amount of energy needed to
raise the temperature of an object by 1 °C.
Heat capacity depends on two factors. One is the size of the sample; if we double the
size of the sample, we need twice as much heat to cause the same temperature increase.
Heat capacity also depends on the sample’s composition. For example, it’s found that it
takes more heat to raise the temperature of one gram of water by 1 °C than to cause the
same temperature change in one gram of iron.
Specific Heat
From the preceding discussion, we see that different samples of a given substance will have
different heat capacities, and this is what is observed. For example, if we have 10.0 g of
water, it must absorb 41.8 J of heat energy to raise its temperature by 1.00 °C. The heat
capacity of the sample is found by solving Equation 7.5 for C :
C =
q
41.8 J
=
= 41.8 J/°C
∆t
1.00 °C
On the other hand, a 100 g sample of water must absorb 418 J of heat to have its temperature raised by 1.00 °C. The heat capacity of this sample is
C =
jespe_c07_253-302hr.indd 260
q
418 J
=
= 418 J/°C
∆t
1.00 °C
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7.3 | Measuring Heat
261
Thus, a 10-fold increase in sample size has produced a 10-fold increase in the heat capacity.
This indicates that the heat capacity is directly proportional to the mass of the sample.
C=m×s
(7.6)
where m is the mass of the sample and s is a constant called the specific heat capacity (or simply
the specific heat). If the heat capacity has units of J °C-1 and if the mass is in grams, the specific heat capacity has units of J g-1 °C-1. The units tell us that the specific heat capacity is
the amount of heat required to raise the temperature of 1 gram of a substance by 1 °C.
Because C depends on the size of the sample, heat capacity is an extensive property. When
we discussed density in Section 2.5, you learned that if we take the ratio of two extensive
properties, we can obtain an intensive property—one that is independent of the size of the
sample. Because heat capacity and mass both depend on sample size, their ratio, obtained
by solving Equation 7.6 for s, yields an intensive property that is the same for any sample
of a substance.
If we substitute Equation 7.6 into Equation 7.5, we can obtain an equation for q in
terms of specific heat.
q = ms D t
(7.7)
Specific heat capacity
We can use this equation to calculate the specific heat of water from the information given
above, which states that 10.0 g of water requires 41.8 J to have its temperature raised by
1.00 °C. Solving for s and substituting gives
s=
q
41.8 J
=
m ∆t
10.0 g × 1.00 °C
= 4.18 J g-1 °C-1
Recall that the older energy unit calorie (cal) is currently defined as 1 cal = 4.184 J.
Therefore, the specific heat of water expressed in calories is
s = 1.000 cal g-1 °C-1
(for H2O)
In measuring heat, we often use an apparatus containing water, so the specific heat of
water is a quantity we will use in working many problems. Every substance has its own
characteristic specific heat, and some are listed in Table 7.1.
When comparing or working with mole-sized quantities of substances, we can use the
molar heat capacity, which is the amount of heat needed to raise the temperature of 1 mol
of a substance by 1 °C. Molar heat capacity equals the specific heat times the molar mass
and has units of J mol-1 °C-1.
Table 7.1
water:
s = 1.000 cal g-1 °C-1
s = 4.184 J g-1 °C-1
Specific Heats
Substance
jespe_c07_253-302hr.indd 261
n You should learn these values for
Specific Heat, J g-1 °C-1 (25 °C)
Carbon (graphite)
0.711
Copper
0.387
Ethyl alcohol
2.45
Gold
0.129
Granite
0.803
Iron
0.4498
Lead
0.128
Olive oil
2.0
Silver
0.235
Water (liquid)
4.184
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262 Chapter 7 | Energy and Chemical Change
Chemistry Outside the ClassrOOm
7.1
Water, Climate, and the Body’s “Thermal Cushion”
Compared with most substances, water has a very high specific
heat. A body of water can therefore gain or lose a substantial
amount of heat without undergoing a large change in temperature.
Because of this, the oceans of the world have a very significant
moderating effect on climate. This is particularly apparent when
we compare temperature extremes of locales near the oceans with
those inland away from the sea and large lakes (such as the Great
Lakes in the upper United States). Places near the sea tend to
have cooler summers and milder winters than places located inland
because the ocean serves as a thermal “cushion,” absorbing heat
in the summer and giving some of it back during the winter.
Warm and cool ocean currents also have global effects on climate. For e xample, the warm waters of the Gulf of Mexico are
c arried by the Gulf Stream across the Atlantic Ocean and keep
winter relatively mild for Ireland, England, and Scotland. By comparison, northeastern Canada, which is at the same latitude as the
British Isles, has much colder winters.
Water also serves as a thermal cushion for the human body. The
adult body is about 60% water by mass, so it has a high heat capacity. In other words, the body can exchange considerable energy with
the environment but experience only a small change in temperature. This makes it relatively easy for the body to maintain a steady
temperature of 37 °C, which is vital to survival. With a substantial
thermal cushion, the body adjusts to large and sudden changes in
outside temperature while experiencing very small fluctuations of
its core temperature.
Direction of Heat Flow
Heat is energy that is transferred from one object to another. This means that the heat lost
by one object is the same as the heat gained by the other. To indicate the direction of heat
flow, we assign a positive sign to q if the heat is gained and a negative sign if the heat is lost.
For example, if a piece of warm iron is placed into a beaker of cool water and the iron loses
10.0 J of heat, the water gains 10.0 J of heat. For the iron, q = -10.0 J and for the water,
q = +10.0 J.
The relationship between the algebraic signs of q in a transfer of heat can be stated in a
general way by the equation
q1 = -q2
Heat transfer
(7.8)
where 1 and 2 refer to the objects between which the heat is transferred.
Example 7.1
Determining the Heat Capacity of an Object
If you dry your hair with a blow drier and wear earrings at the same time, sometimes you
feel the earrings get very warm—almost enough to burn your ear. Suppose that a set of
earrings at 85.4 °C is dropped into an insulated coffee cup containing 25.0 g of water at
25.00 °C. The temperature of the water rises to 25.67 °C. What is the heat capacity of the
earrings in J/°C?
n Analysis:
In this question, heat is being transferred from a warm object, the earrings, to
the cooler water. We are being asked for the heat capacity, C, of the warm object. We can
calculate the amount of heat gained by the water. This amount of heat is the same as the
amount of heat lost by the earring. Using this amount of heat, we can then calculate the
heat capacity of the earring.
n Assembling the Tools: First, for the amount of heat gained by the water, qH O, we will
apply Equation 7.7 as our tool, which uses the specific heat of water (4.184 J g-1 °C-1),
the mass of water, and the temperature change for water to calculate specific heat.
Then, we can use Equation 7.8, the tool for calculating the transfer of heat from the
earrings to the water, and write:
2
qearrings = -qH2O
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7.3 | Measuring Heat
263
We have to be very careful about the algebraic signs. Because the water is gaining heat,
qH2O will be a positive quantity, and because the earrings lose heat, qearrings will be a negative
quantity. The negative sign on the right assures us that when we substitute the positive
value for qH2O, the sign of qearrings will be negative.
Once we’ve found qearrings, we will utilize Equation 7.5,
q = C Dt
the tool that relates heat capacity to temperature change and amount of heat exchanged.
To find C, we divide both sides by the temperature change, D t :
C = q/D t
n Solution:
The temperature of the water rises from 25.00 °C to 25.67 °C, so for the
water,
D tH O = tfinal - tinitial = 25.67 °C - 25.00 °C = 0.67 °C
2
The specific heat of water is 4.184 J g-1 °C-1 and its mass is 25.0 g. Therefore, for water,
the heat absorbed is
qH O = ms D t = 25.0 g × 4.184 J g-1 °C-1 × 0.67 °C
2
= +7.0 × 101 J
Therefore, qearrings = -7.0 × 101 J.
The temperature of the earring decreases from 85.40 °C to 25.66 °C, so
D tearrings = tfinal - t initial = 25.66 °C - 85.40 °C = -59.74 °C
The heat capacity of the earring is then
C =
q earrings
−7.0 × 101 J
=
= 1.2 J/ °C
∆t earrings
−59.74 °C
n Is
the Answer Reasonable? In any calculation involving energy transfer, we first check
to see that all quantities have the correct signs. Heat capacities are positive for common
objects, so the fact that we’ve obtained a positive value for C tells us we’ve handled the
signs correctly.
For the transfer of a given amount of heat, the larger the heat capacity, the smaller the
temperature change. The heat capacity of the water is C = ms = 25.0 g × 4.18 J g-1 °C-1 =
105 J °C-1 and water changes temperature by 0.66 °C. The size of the temperature change for
the earrings, 59.74 °C, is almost 100 times as large as that for the water, so the heat capacity
should be about 1/100 that of the water. Dividing 105 J °C-1 by 100 gives 1.05 J °C-1, which
is not too far from our answer, so our calculations seem to be reasonable.
Example 7.2
Calculating Heat from a Temperature Change, Mass, and Specific Heat
If a piece of copper wire with a mass of 20.9 g changes in temperature from 25.00 to
28.00 °C, how much heat has it absorbed?
n Analysis:
The question asks us to connect the heat absorbed by the wire with its temperature change, D t. We need to know either the heat capacity of the wire or the specific
heat of copper and its mass. We don’t know the heat capacity of the wire. However, we do
know the mass of the wire and the fact that it is made of copper, so we can look up the
specific heat and calculate the amount of heat absorbed.
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264 Chapter 7 | Energy and Chemical Change
n Assembling
the Tools: We can start with Equation 7.7, the equation that relates mass,
specific heat, and temperature change to the amount of heat absorbed, as our tool to solve
the problem. To find the specific heat of copper, we will look at Table 7.1, which gives the
specific heat of copper as 0.387 J g-1 °C-1.
n Solution:
The mass m of the wire is 20.9 g, the specific heat s is 0.387 J g-1 °C-1, and
the temperature increases from 25.00 to 28.00 °C, so D t is 3.00 °C. Using these values
in Equation 7.7 gives
n The sign of D t will
determine the sign of the heat
exchanged. D t is positive
because (tfinal - tinitial) is
positive.
q = ms D t
= (20.9 g) × (0.387 J g-1 °C-1) × (3.00 °C)
= 24.3 J
Thus, only 24.3 J raises the temperature of 20.9 g of copper by 3.00 °C. Because D t is
positive, so is q, 24.3 J. Thus, the sign of the energy change is in agreement with the fact
that the wire absorbs heat.
n Is
the Answer Reasonable? If the wire had a mass of only 1 g and its temperature increased by 1 °C, we’d know from the specific heat of copper (let’s round it to
0.39 J g-1 °C-1) that the ring would absorb 0.39 J. For a 3 °C increase, the answer would
be three times as much, or 1.2 J. For a piece of copper a little heavier than 20 g, the heat
absorbed would be 20 times as much, or about 24 J. So our answer (24.3 J) is reasonable.
Example 7.3
Calculating a Final Temperature from the Specific Heat,
Heat Lost or Gained, and the Mass
If a 25.2 g piece of silver absorbs 365 J of heat, what will the final temperature of the silver
be if the initial temperature is 22.2 °C? The specific heat of silver is 0.235 J g-1 °C-1.
n Analysis:
We are being asked to find the final temperature of the silver after it absorbs
365 J of heat. This question is easier to solve in two steps. First, we will manipulate Equation 7.7 to determine the change in temperature, and then we will use the initial temperature and the change in temperature to find the final temperature.
n Assembling
the Tools: Just as in Example 7.2, we will use Equation 7.7, q = ms D t, as
our tool to solve the problem. Then, we will rearrange Equation 7.4, D t = tfinal - tinitial,
and use it to calculate the final temperature.
n Solution:
We are looking for the final temperature, so let’s start by rearranging
Equation 7.7 and then using the numbers.
q = ms D t
Divide both sides by m and s:
q
ms ∆t
=
ms
ms
Cancel out what is the same:
q
= ∆t
ms
The amount of heat gained is 365 J, the mass of the silver is 25.2 g, and the specific heat
for silver is 0.235 J g-1 °C-1. Use these numbers in the equation and solve:
365 J
25 . 2 g × 0.235 J g −1 °C−1
= ∆t
D t = 61.6 °C
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265
This just gives us the change in temperature. We are really interested in the final temperature, so we will use this change of temperature and the initial temperature to determine
the final temperature.
D t = tfinal - tinitial
61.6 °C = tfinal - 22.2 °C
tfinal = 61.6 °C + 22.2 °C
tfinal = 83.8 °C
n Is
the Answer Reasonable? Heat is being added to the system, so we would expect the
temperature of the silver to increase, and it does. How much it increases depends on the
amount of silver and the specific heat of silver. If we round off the numbers, we can see
whether we are on the right track: the amount of heat is about 350 J, the mass is about
25 g, and the specific heat is about 0.2 J g-1 °C-1.
350 J
= 70 °C
25 g × 0.2 J g −1 °C −1
The change in temperature of 70 °C is very close to 61.6 °C, so it is reasonable to think
that we are correct.
7.1 | A ball bearing at 220.0 °C is dropped into a cup containing 250.0 g of water at
20.0 °C. The water and ball bearing come to a temperature of 30.0 °C. What is the heat
capacity of the ball bearing in J/°C? (Hint: How are the algebraic signs of q related for the
ball bearing and the water?)
Practice Exercises
7.2 | The temperature of 255 g of water is changed from 25.0 to 30.0 °C. How much
energy was transferred into the water? Calculate your answer in joules, kilojoules, calories,
and kilocalories.
7.3 | Silicon, used in computer chips, has a heat capacity of 0.712 J g-1 °C-1. If 549 J of
heat is absorbed by 7.54 g of silicon, what will the final temperature be if the initial temperature is 25.0 °C?
7.4 | Energy of Chemical Reactions
Almost every chemical reaction involves the absorption or release of energy. As the reaction
occurs, the potential energy (also called chemical energy) changes. To understand the origin
of this energy change we need to explore the origin of potential energy in chemical systems.
In Chapter 3 we introduced you to the concept of chemical bonds, which are the attractive
forces that bind atoms to each other in molecules, or ions to each other in ionic compounds.
In this chapter you learned that since particles experience attractions or repulsions, potential
energy changes occur when the particles come together or move apart. We can now bring
these concepts together to understand the origin of energy changes in reactions.
Exothermic and Endothermic Reactions
Chemical reactions generally involve both the breaking and making of chemical bonds. In
most reactions, when bonds form, things that attract each other move closer together,
which tends to decrease the potential energy of the reacting system. When bonds break,
on the other hand, things that are normally attracted to each other are forced apart, which
increases the potential energy of the reacting system. Every reaction, therefore, has a certain net overall potential energy change, the difference between the “costs” of breaking
bonds and the “profits” from making them.
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