Work, Heat, and Internal Energy

Push a bicycle pump and the barrel warms up. No flame touched it. Squeezing the gas did the same job a Bunsen burner could have done. This puzzle sits at the foundation of thermodynamics. Chemists need a precise way to track energy as it moves between a system and its surroundings. That means separating two very different transfer modes, work and heat, from the energy actually stored inside the substance, called internal energy. This article builds the vocabulary and sign conventions used throughout thermochemistry. You will see how internal energy differs from heat and work. You will learn the modern IUPAC sign convention. Then you will apply it to two fully worked numerical problems. The next articles use these definitions to build the First Law and pressure-volume work.

Key takeaways

Internal energy $U$ is a state function: the total kinetic and potential energy of a system's molecules.
Heat $q$ and work $w$ are path functions: modes of energy transfer across a boundary, not stored quantities.
Modern convention: energy entering the system is positive, energy leaving is negative, for both $q$ and $w$ .
Work is transfer via organized (macroscopic) motion; heat is transfer via disordered (thermal) motion.
$U$ depends only on the current state; $q$ and $w$ depend on the path taken between states.
Key idea: Heat and work are two different bookkeeping categories for energy crossing a boundary, while internal energy is the running balance stored inside the system itself.

The core insight

Internal energy is the energy a system has; heat and work are the energy a system exchanges — and although the split between heat and work depends on the path chosen, their sum, which changes $U$ , does not.

The physical picture

Imagine a sealed cylinder of gas as the system, with a piston and a Bunsen burner nearby as the surroundings. The molecules inside possess kinetic energy from translation, rotation, and vibration, plus potential energy from intermolecular forces. The sum of all this microscopic energy is the internal energy, $U$ . It is impossible, and unnecessary, to know the absolute value of $U$ ; thermodynamics only ever needs the change in $U$ , written $\Delta U$ . Energy can enter or leave that gas in exactly two distinguishable ways. If the burner heats the cylinder, energy transfers because the surroundings are at a different temperature — the molecules in the wall jostle randomly, and that disordered thermal agitation is transmitted inward. This transfer mode is called heat, $q$ . Alternatively, if we push the piston down, we move it a macroscopic distance against a force, transferring energy through organized, directional motion. This transfer mode is called work, $w$ . Both processes raise $U$ , but they are experimentally and conceptually distinct. Heat requires a temperature difference; work requires a force acting through a displacement (or its electrical, magnetic, or other analog). Crucially, once the energy is inside the system, it no longer carries a label saying "this joule came from heat" or "this joule came from work" — it has simply become part of $U$ . This is why $U$ is a state function while $q$ and $w$ are path functions, a distinction introduced when defining systems and boundaries. The sign convention used throughout modern chemistry (IUPAC, matching Atkins) is system-centered: anything that increases the energy of the system is positive.

Figure

Energy crossing the system boundary: heat and work are transfer modes, U is the stored total

Figure

Same ΔU, different q/w split: path functions depend on route, U does not

The core relations

\Delta U = q + w

where $\Delta U$ is the change in internal energy, $q$ is heat absorbed by the system, and $w$ is work done on the system (all in joules).

w = -\int_{V_1}^{V_2} p_{\text{ext}}\, dV

where $p_{\text{ext}}$ is the external pressure opposing expansion; this is previewed here and derived fully under pressure-volume work.

dU = \delta q + \delta w

where $d$ denotes an exact differential (path-independent) and $\delta$ denotes an inexact differential (path-dependent), formalizing why $U$ is a state function while $q$ and $w$ are not.

Derivation AWhy ΔU is path-independent even though q and w are not

Goal. Show that although heat and work individually depend on the path taken between two states, their sum — the change in internal energy — does not.

Step 1. Internal energy is defined microscopically as the sum of kinetic and potential energies of all particles in the system. For a fixed amount of substance, this total depends only on the macroscopic state variables that fix the molecular distribution, principally temperature, volume, and composition. Two experiments that start and end at the same $T$ , $V$ , and composition therefore have the same $U_{\text{initial}}$ and the same $U_{\text{final}}$ , regardless of what happened in between.

Step 2. Consider two different routes from state A to state B. Route 1 might add energy mostly as heat (slow thermal contact, little piston motion). Route 2 might add energy mostly as work (rapid adiabatic compression, then cooling). The intermediate values of $q$ and $w$ accumulated along each route generally differ, because they depend on the mechanism of transfer, not just the endpoints.

Step 3. Despite this, calorimetric experiments (first systematically confirmed by Joule's paddle-wheel and electrical heating experiments) show that whenever a system is taken from the same A to the same B by any adiabatic path, the total work required is the same. This establishes that $w_{\text{adiabatic}} = U_B - U_A$ is path-independent, which is the empirical basis for defining $U$ as a state function in the first place.

Step 4. For a general (non-adiabatic) path, some of the energy required arrives as heat instead of work, but the total energy delivered to reach state B from state A must still equal the same fixed difference $U_B - U_A$ , because $U$ only depends on the endpoints (Step 1). Therefore the heat and work contributions can trade off against each other along different routes, while their sum is locked to the same value.

Step 5. Writing this algebraically, for any path, the incremental energy change is $dU = \delta q + \delta w$ . Integrating along any path from A to B gives $\int dU = U_B - U_A$ , an integral that depends only on the endpoints because $dU$ is an exact differential. In contrast, $\int \delta q$ and $\int \delta w$ separately depend on the path, which is why they are written with $\delta$ , not $d$ .

\boxed{\Delta U = q + w \text{, independent of path, even though } q \text{ and } w \text{ separately are not}}

Worked examples

Example 1 — Foundational: electrical heating at constant volume. A rigid, sealed steel vessel (fixed volume, so no expansion work is possible) contains 50.0 g of water. An electrical immersion heater inside the vessel supplies a current of 2.00 A at a potential difference of 12.0 V for exactly 5.00 minutes. No other work is done, and the vessel is otherwise thermally isolated except for the deliberate electrical input. Step 1: Electrical work done on the system.

w = P \times t = (I \times \Delta V) \times t

w = (2.00\ \text{A} \times 12.0\ \text{V}) \times (5.00 \times 60\ \text{s}) = 24.0\ \text{W} \times 300\ \text{s} = 7200\ \text{J}

Since this energy is delivered to the system (the water), and no expansion work occurs (rigid container), this is the only work term, and by the sign convention $w = +7200\ \text{J} = +7.20\ \text{kJ}$ . Step 2: Heat. The vessel is sealed and thermally isolated apart from the electrical leads, so no heat crosses the boundary: $q = 0$ . Step 3: Apply the First Law relation.

\Delta U = q + w = 0 + 7200\ \text{J} = +7.20\ \text{kJ}

Sanity check: this matches the everyday fact that an immersion heater raises water's internal energy (and hence temperature) purely through electrical work, exactly the way stirring or Joule heating does — no flame required, consistent with the bicycle-pump intuition from the overview.

Example 2 — Applied: gas expansion doing work while absorbing heat. A gas sample expands against a constant external pressure of $1.00\ \text{atm}$ ( $101{,}325\ \text{Pa}$ ), and its volume increases from $2.00\ \text{L}$ to $5.50\ \text{L}$ . During this expansion, the surroundings supply $850\ \text{J}$ of heat to the gas to keep the temperature from dropping. Find $\Delta U$ . Step 1: Convert the volume change to SI units.

\Delta V = (5.50 - 2.00)\ \text{L} = 3.50\ \text{L} = 3.50 \times 10^{-3}\ \text{m}^3

Step 2: Compute the work using $w = -p_{\text{ext}}\Delta V$ (previewed here; justified fully when pressure–volume work is treated).

w = -(101{,}325\ \text{Pa})(3.50 \times 10^{-3}\ \text{m}^3) = -354.6\ \text{J}

The negative sign correctly indicates the system does work on the surroundings as it expands, so energy leaves via work. Step 3: Identify heat. The problem states $850\ \text{J}$ is absorbed by the gas, so by convention $q = +850\ \text{J}$ . Step 4: Apply the First Law.

\Delta U = q + w = 850\ \text{J} + (-354.6\ \text{J}) = +495\ \text{J}

Sanity check: the gas both expands (losing energy as work) and absorbs heat, and since the heat input exceeds the work output, $\Delta U$ is positive — internal energy rose overall, consistent with the numbers ( $850 > 354.6$ ).

Limiting cases & common traps

Situation	What happens	Common error
Adiabatic process	$q = 0$ , so $\Delta U = w$ exactly	Forgetting adiabatic means no heat, not no temperature change
Rigid, sealed container	No expansion work possible, $w_{\text{exp}} = 0$	Assuming all work vanishes even if electrical/stirring work is present
Cyclic process	$\Delta U = 0$ over the full cycle	Assuming $q = 0$ and $w = 0$ individually; only their sum is zero
Free expansion into vacuum	$p_{\text{ext}} = 0$ , so $w = 0$	Confusing this with $\Delta V = 0$ ; volume still changes, but no work is done

The chemist behind it

James Prescott Joule's mechanical-equivalent-of-heat experiments in the 1840s established that a fixed amount of mechanical work always produces the same temperature rise in water as a fixed amount of heat, regardless of the mechanism — paddle wheels, electrical resistors, or compression. This was the decisive empirical evidence that heat and work are interconvertible transfer modes of a single conserved quantity, internal energy, rather than separate indestructible fluids (the old "caloric" theory). His careful calorimetry gave the numerical link between mechanical and thermal units that underlies the modern joule.

Origins · 1 minJames Prescott Joule and how this was discovered

James Prescott Joule

1818–1889 · English brewer-physicist whose calorimetry experiments established the mechanical equivalent of heat, unifying work and heat as interconvertible energy transfers

Ask the tutor: How does this sign convention extend when we specifically isolate pressure-volume work in the next article?

Why does internal energy include kinetic and potential energy but not nuclear or rest-mass energy?

Thermodynamics is only interested in energy changes accessible under normal chemical conditions: translation, rotation, vibration, and intermolecular or intramolecular potential energy. Nuclear energy and rest mass do not change during chemical reactions or physical processes like heating and compression, so they are simply excluded from the working definition of $U$ used in chemistry. Only the accessible, changeable part of the total energy matters for $\Delta U$ , which is all that is ever measured.

Is it true that heat always flows from hot to cold?

Yes, spontaneously heat always flows from a higher-temperature body to a lower-temperature one, a statement closely related to the Second Law. However, the sign convention for $q$ in $\Delta U = q + w$ is about direction relative to the system, not about which body is hotter. A system can absorb heat ( $q>0$ ) from a hotter surrounding, or release heat ( $q<0$ ) to a colder one; both obey the hot-to-cold rule while having opposite signs in the bookkeeping.

Why do older textbooks write ΔU = q - w instead of q + w?

Older conventions defined $w$ as work done by the system on the surroundings, which is the opposite sign to the modern IUPAC convention where $w$ is work done on the system. Both conventions are self-consistent internally, but mixing them causes sign errors. Atkins and most current curricula use $\Delta U = q + w$ with $w$ defined as work done on the system, which is the convention used throughout this article.

Can work and heat be negative at the same time?

Yes. If a system expands against external pressure (releasing energy as work, $w<0$ ) while simultaneously losing heat to cooler surroundings ( $q<0$ ), both terms are negative and $\Delta U$ decreases sharply. This is common in rapid gas expansions, such as when a compressed gas cylinder valve is opened and the escaping gas cools while also pushing back the atmosphere.

Does ΔU = 0 mean nothing happened to the system?

No. $\Delta U = 0$ only means the internal energy is unchanged; it says nothing about the individual heat and work terms. In a full thermodynamic cycle, a system can absorb heat, perform work, release heat, and have work done on it, tracing out a rich path in state space, yet return exactly to its starting internal energy. Engines rely on exactly this behavior every cycle.