Energy, Heat, and Work

Chemistry runs on energy exchanges, not just matter exchanges. Every reaction that fizzes, warms a beaker, or drives a piston involves an accounting problem. The question is always: how much energy moved, and in which direction? This article builds the vocabulary and sign conventions chemists use. These tools track energy flow between a chosen "system" and everything else, the "surroundings." The two channels for that flow are heat and work. Both are ways energy crosses a boundary. They are not things a substance "contains." Separating this idea clearly is the single most important conceptual hurdle in introductory thermochemistry. Once the bookkeeping rules are fixed, they carry forward unchanged. They apply directly to the First Law of Thermodynamics. They also apply to calculating the enthalpy of reaction. Everything here uses only kinetic and potential energy and the joule as a unit. There is no new physics involved. It is simply a rigorous way of describing which way energy moves and how much.

Key takeaways

System = the part of the universe under study; surroundings = everything else; the boundary between them can be real or imaginary.
Energy is the capacity to do work or transfer heat; it exists as kinetic energy (motion) and potential energy (position/composition), and both convert to internal energy inside a system.
Heat (q) is energy transferred because of a temperature difference; work (w) is energy transferred by a force acting through a distance.
Sign convention (IUPAC-style, used throughout modern chemistry): $q>0$ and $w>0$ both mean energy flows into the system.
Heat and work are not state functions — they depend on the path taken, unlike internal energy.
Key idea: heat and work are the only two ways energy crosses the system–surroundings boundary, and consistent signs let you add them into one energy balance.

The core insight

Energy transferred into a system is positive; energy transferred out is positive when leaving. Track direction first, then magnitude.

The physical picture

Picture a chemical reaction happening inside a flask sitting on a lab bench. The chemist defines the system as the chemicals inside the flask — reactants and products. Everything else — the flask glass, the air, the bench, the chemist's hand, the rest of the room — is the surroundings. The boundary can be a physical wall (the flask itself) or a purely conceptual line drawn around the reacting substances. Energy can cross that boundary in exactly two ways. Heat ( $q$ ) is energy transfer driven by a temperature difference: if the reaction mixture is hotter than the flask wall, energy flows from mixture to wall until temperatures equalize. Work ( $w$ ) is energy transfer by a force acting through a distance — most commonly in chemistry, a gas expanding and pushing back the atmosphere, or the atmosphere compressing a gas. Both quantities describe energy in transit. Once heat or work has been transferred, it no longer exists as "heat" or "work" — it has become part of the internal energy of whichever side received it. This is the same distinction as a bank transfer: "deposit" and "withdrawal" describe a transaction, not a property of the resulting balance. The modern convention, adopted by IUPAC and standard in general-chemistry texts, treats the system as the reference point:

Heat absorbed by the system (flowing in): $q > 0$
Heat released by the system (flowing out): $q < 0$
Work done on the system (energy pushed in, e.g., surroundings compress the system): $w > 0$
Work done by the system (energy pushed out, e.g., system expands and pushes surroundings back): $w < 0$

A gas that expands against a piston is doing work on the surroundings, so from the system's perspective $w < 0$ . A gas compressed by an external force has work done on it, so $w > 0$ .

Figure

Figure 1 — System, surroundings, and the two energy channels

The core relations

The total change in a system's internal energy from heat and work combined is:

\Delta E = q + w

where:

$\Delta E$ — change in internal energy of the system (J)
$q$ — heat transferred (J); positive if absorbed by the system
$w$ — work done (J); positive if done on the system

For the common case of gas expansion or compression against a constant external pressure, work is:

w = -P_{\text{ext}}\,\Delta V

where:

$P_{\text{ext}}$ — constant external pressure opposing or driving the volume change (Pa, or atm with a unit-conversion factor)
$\Delta V = V_{\text{final}} - V_{\text{initial}}$ — change in system volume (m³, or L)
the negative sign encodes the convention: if the system expands ( $\Delta V > 0$ ), it does work on the surroundings, so $w$ (work done on the system) is negative

Heat itself, when no phase change or reaction occurs, connects to temperature change through the specific heat capacity:

q = m c_s \Delta T

where:

$m$ — mass of the substance (g)
$c_s$ — specific heat capacity (J g⁻¹ °C⁻¹), a property of the material
$\Delta T = T_{\text{final}} - T_{\text{initial}}$ — temperature change (°C or K, since the interval is the same size in both)

Derivation ABuilding the sign convention for work from a piston

Goal: show why $w = -P_{\text{ext}}\Delta V$ correctly gives $w<0$ for system expansion, starting only from the mechanical definition of work as force times distance.

Step 1. Consider a gas confined in a cylinder by a frictionless, massless piston of cross-sectional area $A$ , with constant external pressure $P_{\text{ext}}$ pushing on the piston from outside.

Step 2. The external force pressing inward on the piston has magnitude

F_{\text{ext}} = P_{\text{ext}} A

This is the force the surroundings exert on the system boundary.

Step 3. Suppose the gas expands, pushing the piston outward through a distance $d$ . The gas (the system) exerts a force on the surroundings and moves the piston outward, so the system does positive mechanical work on the surroundings, equal in magnitude to

w_{\text{on surroundings}} = F_{\text{ext}} \cdot d = P_{\text{ext}} A d

Step 4. The volume swept out by the piston is $\Delta V = A d$ (positive, since the gas expanded), so

w_{\text{on surroundings}} = P_{\text{ext}}\,\Delta V

Step 5. By the sign convention adopted for the system, work done on the system is the negative of work done by the system on the surroundings:

w = -w_{\text{on surroundings}} = -P_{\text{ext}}\,\Delta V

Step 6. Check the direction: if the gas expands, $\Delta V>0$ , so $w<0$ — the system loses energy by pushing the surroundings back, matching physical intuition that expansion costs the system energy. If instead the surroundings compress the gas, $\Delta V<0$ , giving $w>0$ — the system gains energy, consistent with being pushed on. This confirms the formula reproduces the correct sign in both directions without any additional assumption beyond force times distance and the system-centered sign convention.

Worked examples

Example 1 — Combining heat and work with asymmetric numbers A gas inside a cylinder absorbs $q = 245\ \text{J}$ of heat from a burner while expanding against a constant external pressure. During the expansion the gas does $312\ \text{J}$ of work on the surroundings (a value obtained separately from $P_{\text{ext}}\Delta V$ ). Find $\Delta E$ for the gas. Step 1 — assign signs. Heat is absorbed by the system, so

q = +245\ \text{J}

The system does work on the surroundings (expansion), so work done on the system is negative:

w = -312\ \text{J}

Step 2 — apply the first-law-style energy balance.

\Delta E = q + w = 245\ \text{J} + (-312\ \text{J})

\Delta E = -67\ \text{J}

Interpretation. Even though heat flowed in, more energy left as work than entered as heat, so the internal energy of the gas decreased by 67 J. This illustrates why $q$ and $w$ must be tracked with independent signs — one cannot assume they reinforce each other. Limiting-case check. If instead the gas were compressed (surroundings do work on it) with the same heat absorbed, say $w=+312\ \text{J}$ , then $\Delta E = 245+312 = 557\ \text{J}$ , a large gain — confirming the formula responds correctly when the work term flips sign.

Example 2 — Heat from a temperature change, then combining with expansion work A $37.5\ \text{g}$ sample of liquid water ( $c_s = 4.184\ \text{J g}^{-1}\,^{\circ}\text{C}^{-1}$ ) is heated from $19.2\,^{\circ}\text{C}$ to $46.8\,^{\circ}\text{C}$ in an open flask, while the container walls allow the system to expand slightly against constant atmospheric pressure, performing $8.5\ \text{J}$ of work on the surroundings. Find $q$ , $w$ (system convention), and $\Delta E$ . Step 1 — compute $\Delta T$ .

\Delta T = 46.8\,^{\circ}\text{C} - 19.2\,^{\circ}\text{C} = 27.6\,^{\circ}\text{C}

Step 2 — compute heat absorbed.

q = m c_s \Delta T = (37.5\ \text{g})(4.184\ \text{J g}^{-1}\,^{\circ}\text{C}^{-1})(27.6\,^{\circ}\text{C})

q = 37.5 \times 4.184 \times 27.6\ \text{J}

q = 4331.6\ \text{J} \approx 4.33 \times 10^{3}\ \text{J}

Since the water is heated (absorbs energy), this is consistent with the system convention: $q = +4.33 \times 10^{3}\ \text{J}$ . Step 3 — assign the sign to work. The system expands and does work on the surroundings, so

w = -8.5\ \text{J}

Step 4 — combine.

\Delta E = q + w = 4331.6\ \text{J} + (-8.5\ \text{J})

\Delta E \approx 4.32 \times 10^{3}\ \text{J}

Interpretation. The expansion work is a tiny correction compared to the heat term — typical for liquids in open vessels, where volume changes are small. This foreshadows why chemists often approximate $\Delta E \approx q$ for reactions at constant volume, and why a separate quantity, enthalpy, is defined to absorb the pressure–volume work term automatically at constant pressure. Limiting-case check. As the expansion work term approaches zero (a rigid, sealed container, $\Delta V = 0$ ), $w \to 0$ and $\Delta E \to q$ exactly — matching the constant-volume limit used later when defining internal energy changes calorimetrically.

Limiting cases & common traps

Situation	What happens to signs	Common mistake
Gas expands freely into vacuum ( $P_{\text{ext}}=0$ )	$w = -P_{\text{ext}}\Delta V = 0$ regardless of $\Delta V$	Assuming expansion always does nonzero work
System and surroundings reach thermal equilibrium	$q \to 0$ as $\Delta T \to 0$	Forgetting $q=0$ does not mean no energy was ever transferred, only that net flow has stopped
Reaction run in a sealed, rigid container	$\Delta V = 0 \Rightarrow w=0$ , so $\Delta E = q_v$	Confusing this constant-volume heat with the constant-pressure heat used for enthalpy
Exothermic reaction releasing heat	$q<0$ by system convention	Older textbooks/thermochemical conventions reverse this sign; mixing conventions mid-problem causes errors

The chemist behind it

The modern language of heat, work, and internal energy owes its precision largely to the work of Julius Robert von Mayer, James Prescott Joule, and Rudolf Clausius, who in the mid-1800s established that heat and mechanical work are interconvertible forms of energy transfer rather than distinct fluids. Joule's mechanical equivalent of heat experiments — using falling weights to turn a paddle wheel in insulated water and measuring the resulting temperature rise — provided the quantitative link between mechanical work and thermal energy that makes the equation $\Delta E = q + w$ possible at all.

Origins · 1 minJames Prescott Joule and how this was discovered

James Prescott Joule

1818–1889 · Manchester, England, during the early industrial rise of thermodynamics

Joule, trained in his family's brewing business rather than a university, carried out exquisitely careful measurements showing that a fixed amount of mechanical work always produced the same amount of heat, regardless of the mechanism used to generate it. This experimentally established the equivalence of heat and work as energy in transit — the physical basis for treating $q$ and $w$ as two contributions to the same energy balance, later formalized as the First Law of Thermodynamics. The SI unit of energy bears his name.

Ask the tutor: If a reaction is run in an open flask rather than a sealed rigid container, how does the extra expansion work term change the way chemists measure and define the heat released — and why does this motivate defining enthalpy?

Why does chemistry use "internal energy" instead of just calling it "heat energy"?

Because a system's energy content is not made of heat — heat is only one way energy can be added or removed. Internal energy $E$ (sometimes written $U$ ) is the total kinetic and potential energy of all particles in the system, a property of the state itself. Heat and work are the two mechanisms by which that internal energy changes; neither exists "inside" the system waiting to be measured.

Is it possible for both q and w to be negative at once?

Yes. If a system releases heat to the surroundings ( $q<0$ ) while also doing work on the surroundings, such as an exothermic reaction that also produces expanding gas ( $w<0$ ), both terms subtract from the internal energy. The system loses energy through both channels simultaneously, and $\Delta E = q+w$ becomes more negative than either term alone.

Why is the negative sign in

w = -P_{\text{ext}}\Delta V

sometimes missing in other textbooks?

Some older or engineering-oriented texts define work from the surroundings' perspective, writing $w = +P\Delta V$ as "work done by the system." Chemistry generally adopts the IUPAC convention where all energy changes are reported from the system's point of view, so expansion (energy leaving the system) is negative. Always check which convention a given source uses before combining formulas.

Does q = mcΔT apply to every heating process in chemistry?

Only when no phase change and no chemical reaction occurs, and only over a temperature range where the specific heat capacity is roughly constant. Melting, boiling, or reacting substances require additional terms (latent heats, or reaction enthalpies) since energy goes into breaking or forming interactions rather than only raising kinetic energy of motion.