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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.



jespe_c07_253-302hr.indd 256



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



11/11/10 5:12 PM



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.



jespe_c07_253-302hr.indd 257



11/11/10 5:12 PM



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.



jespe_c07_253-302hr.indd 258



11/11/10 5:12 PM



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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11/11/10 5:12 PM



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



11/11/10 5:12 PM



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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11/11/10 5:12 PM



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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7.4 | Energy of Chemical Reactions



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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