Energy is of no use without work.

A spring can be compressed to store potential energy, but it's of little use to us if we can't use that stored energy to do something. That stored energy can be used to do useful work. The spring is a storehouse of energy: We have to do work to compress it, and at a later time it can unleash that energy as the work of pushing or pulling something.

Work transfers energy from one place or time to another, and possibly from one form to another.

Work is caused by a force. Without any force, there is no work. When a force works over a distance, work has been done, and its value, in units of energy, can be calculated.

Work (w) is force (F, see Newton's laws) multiplied by the distance (d) over which the force was exerted:

$$w = F \cdot d$$

The units of work are the units of energy, Joules (J). Joules are the SI unit of energy. Here's a unit analysis of F·d :

$$\left( \frac{Kg \cdot m}{s^2} \right) \cdot m = \frac{Kg\cdot m^2}{s^2} = \bf J$$

Notice also that 1J = 1 N·m, or 1 Newton-meter (N·m). Once in a while you'll see N·m used instead of Joules.


SI units

SI stands for Système international (of units). In 1960, the SI system of units was published as a guide to the preferred units to use for a variety of quantities. Here are some common SI units


Definition of work

Work is force exerted over a distance.

w = F · d

The SI unit of work is the unit of energy, the Joule (J). 1 J = 1 Kg · m2 · s -2.

Work can be interchanged with potential energy (PE) and kinetic energy (KE).

Equivalence of work and energy

Work is equivalent to (or can be "traded for" either kinetic energy (KE) or potential energy (PE). Two examples might help.

Gravitational PE

We know that the potential energy of an object of mass m lifted to a height h is PE = mgh, where m is the mass (in Kg), g is the acceleration of gravity (g = 9.8 m/s2 near the surface of Earth), and h is the height in meters. The work that it takes to lift an object to such a height, giving it that PE, is exactly equal to the PE gained. In this case we express the equivalence of work and PE as

$$w = -\Delta PE$$

The negative is because the work done to "buy" PE is always in the opposite direction of the motion that would result if the PE were translated into KE.

The work of brakes in a car

A car moving in a straight line at a velocity v has kinetic energy, KE = ½mv2. If the brakes in that car are applied to bring it to a stop, the amount of work done by the brakes in counteracting the KE is exactly equal to the amount of KE that the car initially had.

There are other kinds of work, such as the pressure-volume (PV) work we encounter in chemistry, and we will discuss some others below.

$$w = -\Delta KE$$

The sign of the work in this case is dependent on what we get an object to do by doing work on it. We can, for example, speed up or slow down a moving object by doing work – exerting a force over a distance. Each results on a change in KE equal in size to the work done.

Work against gravity

Equivalence of work and gravitational PE

The work done in lifting an object a vertical distance is equivalent to the potential energy gained in doing so. The gravitational PE is

$$PE = mgh$$

where m is the mass, g is the acceleration of gravity near the surface of Earth, and h is the height.

A nice example of this is a roller coaster. Consider the roller coaster diagram below:

The first step in any roller coaster's journey is to be lifted to the top of the tallest hill (A). The amount of work done is w = mgh, which is exactly equal to the amount of potential energy gained: PE = mgh.

As the coaster begins to descend hill A, it loses potential energy (PE) and gains kinetic energy (KE), the energy of motion. In a sense, the coaster is "trading in" PE for KE.

At point B, which is a little higher than the origin of the coaster, most of the PE has been converted to KE, though some remains.

As the coaster passes B and begins to climb hill C, it loses KE and gains PE. All of this is due to the original work done in lifting the coaster to the top of hill A.

Upon descending hill C, the coaster returns to its initial height, thus converting all of its PE (therefore all of the work that was done) to KE.

A roller coaster could keep going like that forever if it wasn't for friction (between the wheels and the track and due to air resistance). The friction force does work on the coaster, much as the brakes on a car do work to stop it, reducing its KE. Because of friction, such a coaster could never climb another hill as high as hill A after the initial lift.

Work in chemistry: PV work

The kind of work we encounter most often in chemistry is pressure-volume (PV) work.

The product of pressure and volume has units of energy. Here's how it works: The units of pressure are force divided by area, which we can write as Newtons (N) divided by m2,

$$P = \frac{N}{m^2}$$

Multiplying by matching units of volume, m3, gives

$$PV = \frac{n}{m^2} m^3 = N\cdot m = J$$

Recall that 1 N = 1 Kg·m·s-2, so multiplying by meters gives 1 J = 1 Kg·m2·s-2.

PV work is apparent when we heat a container of gas under constant pressure. This can be done using a cylinder fitted with a movable piston, atop of which sits a weight. The weight produces a constant force on the piston (which has a fixed area), and thus a constant pressure.

If we then heat the container, the volume must increase (recall from the ideal gas law that V = nRT/P). Play the animation to see how it works.

When the volume expands, we say (by convention) that the surroundings have done work on the system, and we call this positive work (+w). When the volume decreases, we say that the surroundings have done work on the system, and we call that negative work (-w).

Play the animation (arrow) a few times to get the idea.

The signs of heat and work (+ / -)

When the surroundings do work on the system (e.g. to compress the system), positive work has been done from the point of view of the system.

When the system does work on the surroundings (e.g. to expand its volume), negative work has been done.

Think of the sign conventions for heat and work as system-centric. Heat added to or work done on a system is positive. Otherwise, both are negative.

Equivalence of heat and work

Doing work on a chemical system, such as a liquid, is completely equivalent to adding an equal amount of heat (remember, the units are the same). The total energy input into a system is the heat it receives plus the work it does; it's part of the principle of conservation of energy.

Prescott Joule, one of the early developers of thermodynamics, performed an experiment that showed that mechanical work done on a container of water produces a temperature rise in the water equivalent to the amount of heat needed to cause the temperature change.

In Joule's experiment, a weight of known mass was dropped a known distance, thus producing a known amount of kinetic energy which produced mechanical stirring of a quantity of water. The friction of the stirring process produced a temperature rise equivalent to that kinetic energy.

Heat and work are equivalent in chemical systems.


Suppose that the weight in Joules experiment is 1 Kg and drops a distance of 10 m (which could be accomplished by ten 1m drops if the weight was cranked back to the top of its 1m range slowly), and suppose our jar contains 1 L of water (1 Kg because the density of water is 1 g/mL).

Now PE = mgh = (1 Kg)(9.8 m·s-2)(10 m) = 98 J. So that 98 J is "released" into the water through friction between the turning paddle blades and the water. The KE of motion of the paddles is translated to the motion of the water molecules, which is what heat is.

Now to calculate the temperature rise, we need a formula with which you might not be familiar (it's here): The heat (q) required to change the temperature of m grams of a substance by ΔT ˚C is

$$q = mC\Delta T$$

where C is a property (which Joule measured) called the specific heat capacity. The specific heat capacity of water is C = 4.184 J/g˚C, so the temperature rise would be

$$ \begin{align} \Delta T &= \frac{q}{mC} \\ &= \frac{98 \, J}{(1000 \, g)(4.184 \, J/g˚C)} \\ \\ &= 0.023 \, ˚C \end{align}$$

which is a small, but measurable temperature rise.

More work to do ...

As you continue to study thermodynamics, you'll find that calculus will become more important. We've just scratched the surface of work and thermo here. A planned section will cover work using calculus.

Creative Commons License   optimized for firefox by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012-2019, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to