The First Law of Thermodynamics

The photograph shows water boiling in a tea kettle kept on a stove. The water vapor is shown to emerge out of the nozzle of the kettle.

If we are interested in how heat transfer is converted into doing work, then the conservation of energy principle is important. The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system. In equation form, the first law of thermodynamics is

ΔU=Q−W. size 12{ΔU=Q - W} {}

Here ΔU size 12{ΔU} {}

is the change in internal energy U size 12{U} {}

of the system. Q size 12{Q} {}

is the net heat transferred into the system—that is, Q size 12{Q} {}

is the sum of all heat transfer into and out of the system. W size 12{W} {}

is the net work done by the system—that is, W size 12{W} {}

is the sum of all work done on or by the system. We use the following sign conventions: if Q size 12{Q} {}

is positive, then there is a net heat transfer into the system; if W size 12{W} {}

is positive, then there is net work done by the system. So positive Q size 12{Q} {}

adds energy to the system and positive W size 12{W} {}

takes energy from the system. Thus ΔU=Q−W size 12{ΔU=Q - W} {}

. Note also that if more heat transfer into the system occurs than work done, the difference is stored as internal energy. Heat engines are a good example of this—heat transfer into them takes place so that they can do work. (See [link].) We will now examine Q size 12{Q} {}

, W size 12{W} {}

, and ΔU size 12{ΔU} {}

further.

The figure shows a schematic diagram of a system shown by an ellipse. Heat Q is shown to enter the system as shown by a bold arrow toward the ellipse. The work done is shown pointing away from the system. The internal energy of the system is marked as delta U equals Q minus W. The second part of the figure shows two arrow diagrams for the heat change Q and work W. Q is shown as Q in minus Q out. W is shown as W out minus W in.

Making Connections: Law of Thermodynamics and Law of Conservation of Energy

The first law of thermodynamics is actually the law of conservation of energy stated in a form most useful in thermodynamics. The first law gives the relationship between heat transfer, work done, and the change in internal energy of a system.

Heat Q and Work W

Heat transfer (Q size 12{Q} {}

) and doing work (W size 12{W} {}

) are the two everyday means of bringing energy into or taking energy out of a system. The processes are quite different. Heat transfer, a less organized process, is driven by temperature differences. Work, a quite organized process, involves a macroscopic force exerted through a distance. Nevertheless, heat and work can produce identical results.**For example, both can cause a temperature increase. Heat transfer into a system, such as when the Sun warms the air in a bicycle tire, can increase its temperature, and so can work done on the system, as when the bicyclist pumps air into the tire. Once the temperature increase has occurred, it is impossible to tell whether it was caused by heat transfer or by doing work. This uncertainty is an important point. Heat transfer and work are both energy in transit—neither is stored as such in a system. However, both can change the internal energy U size 12{U} {}

of a system. Internal energy is a form of energy completely different from either heat or work.

Internal Energy U

We can think about the internal energy of a system in two different but consistent ways. The first is the atomic and molecular view, which examines the system on the atomic and molecular scale. The internal energy U size 12{U} {}

of a system is the sum of the kinetic and potential energies of its atoms and molecules. Recall that kinetic plus potential energy is called mechanical energy. Thus internal energy is the sum of atomic and molecular mechanical energy. Because it is impossible to keep track of all individual atoms and molecules, we must deal with averages and distributions. A second way to view the internal energy of a system is in terms of its macroscopic characteristics, which are very similar to atomic and molecular average values.

Macroscopically, we define the change in internal energy ΔU size 12{ΔU} {}

to be that given by the first law of thermodynamics:

ΔU=Q−W. size 12{ΔU=Q - W} {}

Many detailed experiments have verified that ΔU=Q−W size 12{ΔU=Q - W} {}

, where ΔU size 12{ΔU} {}

is the change in total kinetic and potential energy of all atoms and molecules in a system. It has also been determined experimentally that the internal energy U size 12{U} {}

of a system depends only on the state of the system and *not how it reached that state.* More specifically, U size 12{U} {}

is found to be a function of a few macroscopic quantities (pressure, volume, and temperature, for example), independent of past history such as whether there has been heat transfer or work done. This independence means that if we know the state of a system, we can calculate changes in its internal energy U size 12{U} {}

from a few macroscopic variables.

Making Connections: Macroscopic and Microscopic

In thermodynamics, we often use the macroscopic picture when making calculations of how a system behaves, while the atomic and molecular picture gives underlying explanations in terms of averages and distributions. We shall see this again in later sections of this chapter. For example, in the topic of entropy, calculations will be made using the atomic and molecular view.

To get a better idea of how to think about the internal energy of a system, let us examine a system going from State 1 to State 2. The system has internal energy U1 size 12{U rSub { size 8{1} } } {}

in State 1, and it has internal energy U2 size 12{U rSub { size 8{2} } } {}

in State 2, no matter how it got to either state. So the change in internal energy ΔU=U2−U1 size 12{ΔU=U rSub { size 8{2} } - U rSub { size 8{1} } } {}

is independent of what caused the change. In other words, ΔU size 12{ΔU} {}

is independent of path. By path, we mean the method of getting from the starting point to the ending point. Why is this independence important? Note that ΔU=Q−W size 12{ΔU=Q - W} {}

. Both Q size 12{Q} {}

and W size 12{W} {}

depend on path, but ΔU size 12{ΔU} {}

does not. This path independence means that internal energy U size 12{U} {}

is easier to consider than either heat transfer or work done.

Calculating Change in Internal Energy: The Same Change in U size 12{U} {} is Produced by Two Different Processes

(a) Suppose there is heat transfer of 40.00 J to a system, while the system does 10.00 J of work. Later, there is heat transfer of 25.00 J out of the system while 4.00 J of work is done on the system. What is the net change in internal energy of the system?

(b) What is the change in internal energy of a system when a total of 150.00 J of heat transfer occurs out of (from) the system and 159.00 J of work is done on the system? (See [link]).

Strategy

In part (a), we must first find the net heat transfer and net work done from the given information. Then the first law of thermodynamics (ΔU=Q−W size 12{ΔU=Q - W} {})

can be used to find the change in internal energy. In part (b), the net heat transfer and work done are given, so the equation can be used directly.

Solution for (a)

The net heat transfer is the heat transfer into the system minus the heat transfer out of the system, or

Q=40.00J−25.00 J=15.00J. size 12{Q="40" "." "00"" J"-"25" "." "00"" J"="15" "." "00"" J"} {}

Similarly, the total work is the work done by the system minus the work done on the system, or

W=10.00J−4.00J=6.00 J. size 12{W="10" "." "00"" J"-4 "." "00"" J"=6 "." "00"" J"} {}

Thus the change in internal energy is given by the first law of thermodynamics:

ΔU=Q−W=15.00J−6.00J=9.00 J. size 12{DU=Q-W="15" "." "00"" J"-6 "." "00"" J"=9 "." "00"" J"} {}

We can also find the change in internal energy for each of the two steps. First, consider 40.00 J of heat transfer in and 10.00 J of work out, or

ΔU1=Q1−W1=40.00J−10.00J=30.00 J. size 12{DU rSub { size 8{1} } =Q rSub { size 8{1} } -W rSub { size 8{1} } ="40" "." "00"" J"-"10" "." "00"" J"="30" "." "00"" J"} {}

Now consider 25.00 J of heat transfer out and 4.00 J of work in, or

ΔU2=Q2−W2=-25.00J−(−4.00J)=–21.00 J. size 12{DU rSub { size 8{2} } =Q rSub { size 8{2} } -W rSub { size 8{2} } "=-""25" "." "00"" J"- \( -4 "." "00"" J" \) "=-""21" "." "00"" J"} {}

The total change is the sum of these two steps, or

ΔU=ΔU1+ΔU2=30.00J+−21.00J=9.00 J. size 12{DU=DU rSub { size 8{1} } +DU rSub { size 8{2} } ="30" "." "00"" J"+ left (-"21" "." "00"" J" right )=9 "." "00"" J"} {}

Discussion on (a)

No matter whether you look at the overall process or break it into steps, the change in internal energy is the same.

Solution for (b)

Here the net heat transfer and total work are given directly to be Q=–150.00 J size 12{Q"=-""150" "." "00"" J"} {}

and W=–159.00 J size 12{W"=-""159" "." "00"" J"} {}

, so that

ΔU=Q–W=–150.00 J–(−159.00 J)=9.00 J. size 12{DU=Q-W"=-""150" "." "00"" J"- \( -"159" "." "00"" J" \) =9 "." "00"" J"} {}

Discussion on (b)

A very different process in part (b) produces the same 9.00-J change in internal energy as in part (a). Note that the change in the system in both parts is related to ΔU size 12{ΔU} {}

and not to the individual Q size 12{Q} {}

s or W size 12{W} {}

s involved. The system ends up in the same state in both (a) and (b). Parts (a) and (b) present two different paths for the system to follow between the same starting and ending points, and the change in internal energy for each is the same—it is independent of path.

The first part of the picture shows a system in the form of a circle for explanation purposes. The heat entering and work done are represented by bold arrows. A quantity of heat Q in equals forty joules, is shown to enter the system and Q out equals negative twenty five joules is shown to leave the system. The energy of the system in is marked as fifteen joules. At the right-hand side of the circle, a work W in equals negative four joules is shown to be applied on the system and a work W out equals ten joules is shown to leave the system. The energy of the system out is marked as six joules. The second part of the picture shows a system in the form of a circle for explanation purposes. The heat entering and work done are represented by bold arrows. A work of negative one hundred fifty nine is shown to enter the system. The energy in the system is shown as one hundred fifty nine joules. The out energy of the system is one hundred fifty joules. A heat Q out of negative one hundred fifty joules is shown to leave the system as an outward arrow.

Human Metabolism and the First Law of Thermodynamics

Human metabolism is the conversion of food into heat transfer, work, and stored fat. Metabolism is an interesting example of the first law of thermodynamics in action. We now take another look at these topics via the first law of thermodynamics. Considering the body as the system of interest, we can use the first law to examine heat transfer, doing work, and internal energy in activities ranging from sleep to heavy exercise. What are some of the major characteristics of heat transfer, doing work, and energy in the body? For one, body temperature is normally kept constant by heat transfer to the surroundings. This means Q size 12{Q} {}

is negative. Another fact is that the body usually does work on the outside world. This means W size 12{W} {}

is positive. In such situations, then, the body loses internal energy, since ΔU=Q−W size 12{ΔU=Q - W} {}

is negative.

Now consider the effects of eating. Eating increases the internal energy of the body by adding chemical potential energy (this is an unromantic view of a good steak). The body metabolizes all the food we consume. Basically, metabolism is an oxidation process in which the chemical potential energy of food is released. This implies that food input is in the form of work. Food energy is reported in a special unit, known as the Calorie. This energy is measured by burning food in a calorimeter, which is how the units are determined.

In chemistry and biochemistry, one calorie (spelled with a lowercase c) is defined as the energy (or heat transfer) required to raise the temperature of one gram of pure water by one degree Celsius. Nutritionists and weight-watchers tend to use the dietary calorie, which is frequently called a Calorie (spelled with a capital C). One food Calorie is the energy needed to raise the temperature of one kilogram of water by one degree Celsius. This means that one dietary Calorie is equal to one kilocalorie for the chemist, and one must be careful to avoid confusion between the two.

Again, consider the internal energy the body has lost. There are three places this internal energy can go—to heat transfer, to doing work, and to stored fat (a tiny fraction also goes to cell repair and growth). Heat transfer and doing work take internal energy out of the body, and food puts it back. If you eat just the right amount of food, then your average internal energy remains constant. Whatever you lose to heat transfer and doing work is replaced by food, so that, in the long run, ΔU=0 size 12{ΔU=0} {}

. If you overeat repeatedly, then ΔU size 12{ΔU} {}

is always positive, and your body stores this extra internal energy as fat. The reverse is true if you eat too little. If ΔU size 12{ΔU} {}

is negative for a few days, then the body metabolizes its own fat to maintain body temperature and do work that takes energy from the body. This process is how dieting produces weight loss.

Life is not always this simple, as any dieter knows. The body stores fat or metabolizes it only if energy intake changes for a period of several days. Once you have been on a major diet, the next one is less successful because your body alters the way it responds to low energy intake. Your basal metabolic rate (BMR) is the rate at which food is converted into heat transfer and work done while the body is at complete rest. The body adjusts its basal metabolic rate to partially compensate for over-eating or under-eating. The body will decrease the metabolic rate rather than eliminate its own fat to replace lost food intake. You will chill more easily and feel less energetic as a result of the lower metabolic rate, and you will not lose weight as fast as before. Exercise helps to lose weight, because it produces both heat transfer from your body and work, and raises your metabolic rate even when you are at rest. Weight loss is also aided by the quite low efficiency of the body in converting internal energy to work, so that the loss of internal energy resulting from doing work is much greater than the work done.**It should be noted, however, that living systems are not in thermalequilibrium.

The body provides us with an excellent indication that many thermodynamic processes are irreversible. An irreversible process can go in one direction but not the reverse, under a given set of conditions. For example, although body fat can be converted to do work and produce heat transfer, work done on the body and heat transfer into it cannot be converted to body fat. Otherwise, we could skip lunch by sunning ourselves or by walking down stairs. Another example of an irreversible thermodynamic process is photosynthesis. This process is the intake of one form of energy—light—by plants and its conversion to chemical potential energy. Both applications of the first law of thermodynamics are illustrated in [link]. One great advantage of conservation laws such as the first law of thermodynamics is that they accurately describe the beginning and ending points of complex processes, such as metabolism and photosynthesis, without regard to the complications in between. [link] presents a summary of terms relevant to the first law of thermodynamics.

Part a of the figure is a pictorial representation of metabolism in a human body. The food is shown to enter the body as shown by a bold arrow toward the body. Work W and heat Q leave the body as shown by bold arrows pointing outward from the body. Delta U is shown as the stored food energy. Part b of the figure shows the metabolism in plants .The heat from the sunlight is shown to fall on a plant represented as Q in. The heat given out by the plant is shown as Q out by an arrow pointing away from the plant.

Summary of Terms for the First Law of Thermodynamics, ΔU=Q−W
Term Definition
U size 12{U} {} Internal energy—the sum of the kinetic and potential energies of a system’s atoms and molecules. Can be divided into many subcategories, such as thermal and chemical energy. Depends only on the state of a system (such as its P size 12{P} {}, V size 12{V} {}, and T size 12{T} {}), not on how the energy entered the system. Change in internal energy is path independent.
Q size 12{Q} {} Heat—energy transferred because of a temperature difference. Characterized by random molecular motion. Highly dependent on path. Q size 12{Q} {} entering a system is positive.
W size 12{W} {} Work—energy transferred by a force moving through a distance. An organized, orderly process. Path dependent. W size 12{W} {} done by a system (either against an external force or to increase the volume of the system) is positive.

Section Summary

Conceptual Questions

Describe the photo of the tea kettle at the beginning of this section in terms of heat transfer, work done, and internal energy. How is heat being transferred? What is the work done and what is doing it? How does the kettle maintain its internal energy?

The first law of thermodynamics and the conservation of energy, as discussed in Conservation of Energy, are clearly related. How do they differ in the types of energy considered?

Heat transfer Q size 12{Q} {}

and work done W size 12{W} {}

are always energy in transit, whereas internal energy U size 12{U} {}

is energy stored in a system. Give an example of each type of energy, and state specifically how it is either in transit or resides in a system.

How do heat transfer and internal energy differ? In particular, which can be stored as such in a system and which cannot?

If you run down some stairs and stop, what happens to your kinetic energy and your initial gravitational potential energy?

Give an explanation of how food energy (calories) can be viewed as molecular potential energy (consistent with the atomic and molecular definition of internal energy).

Identify the type of energy transferred to your body in each of the following as either internal energy, heat transfer, or doing work: (a) basking in sunlight; (b) eating food; (c) riding an elevator to a higher floor.

Problems & Exercises

What is the change in internal energy of a car if you put 12.0 gal of gasoline into its tank? The energy content of gasoline is 1 . 3 × 10 8 J/gal size 12{1 "." 3 times "10" rSup { size 8{8} } " J/gal"} {}

. All other factors, such as the car’s temperature, are constant.

1 . 6 × 10 9 J size 12{1 "." 6 times "10" rSup { size 8{9} } " J"} {}

How much heat transfer occurs from a system, if its internal energy decreased by 150 J while it was doing 30.0 J of work?

A system does 1.80×108 J size 12{1 "." "80"´"10" rSup { size 8{8} } " J"} {}

of work while 7.50×108 J size 12{7 "." "50"´"10" rSup { size 8{8} } " J"} {}

of heat transfer occurs to the environment. What is the change in internal energy of the system assuming no other changes (such as in temperature or by the addition of fuel)?

- 9 . 30 × 10 8 J size 12{ +- 9 "." "30"´"10" rSup { size 8{8} } " J"} {}

What is the change in internal energy of a system which does 4.50×105 J size 12{4 "." "50"´"10" rSup { size 8{5} } " J"} {}

of work while 3.00×106 J size 12{3 "." "00"´"10" rSup { size 8{6} } " J"} {}

of heat transfer occurs into the system, and 8.00×106 J size 12{8 "." "00"´"10" rSup { size 8{6} } " J"} {}

of heat transfer occurs to the environment?

Suppose a woman does 500 J of work and 9500 J of heat transfer occurs into the environment in the process. (a) What is the decrease in her internal energy, assuming no change in temperature or consumption of food? (That is, there is no other energy transfer.) (b) What is her efficiency?

(a) − 1 . 0 × 10 4 J size 12{ - 1 "." 0 times "10" rSup { size 8{4} } " J"} {}

, or − 2 . 39 kcal

(b) 5.00%

(a) How much food energy will a man metabolize in the process of doing 35.0 kJ of work with an efficiency of 5.00%? (b) How much heat transfer occurs to the environment to keep his temperature constant? Explicitly show how you follow the steps in the Problem-Solving Strategy for thermodynamics found in Problem-Solving Strategies for Thermodynamics.

(a) What is the average metabolic rate in watts of a man who metabolizes 10,500 kJ of food energy in one day? (b) What is the maximum amount of work in joules he can do without breaking down fat, assuming a maximum efficiency of 20.0%? (c) Compare his work output with the daily output of a 187-W (0.250-horsepower) motor.

(a) 122 W

(b) 2.10×106 J size 12{2 "." "09" times "10" rSup { size 8{6} } " J"} {}

(c) Work done by the motor is 1.61×107 J size 12{1 "." "61" times "10" rSup { size 8{7} } " J"} {}

;thus the motor produces 7.67 times the work done by the man

(a) How long will the energy in a 1470-kJ (350-kcal) cup of yogurt last in a woman doing work at the rate of 150 W with an efficiency of 20.0% (such as in leisurely climbing stairs)? (b) Does the time found in part (a) imply that it is easy to consume more food energy than you can reasonably expect to work off with exercise?

(a) A woman climbing the Washington Monument metabolizes 6 . 00 × 10 2 kJ size 12{6 "." "00" times "10" rSup { size 8{2} } " kJ"} {}

of food energy. If her efficiency is 18.0%, how much heat transfer occurs to the environment to keep her temperature constant? (b) Discuss the amount of heat transfer found in (a). Is it consistent with the fact that you quickly warm up when exercising?

(a) 492 kJ

(b) This amount of heat is consistent with the fact that you warm quickly when exercising. Since the body is inefficient, the excess heat produced must be dissipated through sweating, breathing, etc.

Glossary

first law of thermodynamics
states that the change in internal energy of a system equals the net heat transfer into the system minus the net work done by the system
internal energy
the sum of the kinetic and potential energies of a system’s atoms and molecules
human metabolism
conversion of food into heat transfer, work, and stored fat
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