Heat and Work Sign Conventions

Introduction

The signs of heat and work are bookkeeping directions, not labels for “good” and “bad” energy. In the common chemistry convention, energy entering the system as heat or work is positive; energy leaving is negative. Choosing the system and applying one convention consistently makes the first law straightforward.

Core explanation

Heat q is energy transferred across a boundary because of a temperature difference. If the surroundings warm the system, q sys > 0; if the system releases heat to the surroundings, q sys < 0. Work w is energy transferred by a force acting through a displacement or another organized mode. If surroundings compress a gas, they do work on the gas and w sys > 0. If a gas expands against external pressure, it does work on the surroundings and w sys < 0.

The first law for a closed system in this convention is ΔU = q + w. A positive ΔU means the system's internal energy increases. The signs of q and w may oppose one another. For example, a gas can absorb 500 J as heat and do 200 J of expansion work, giving ΔU = +300 J. The word “absorbs” identifies q as positive; “does work on surroundings” identifies w as negative.

Other disciplines sometimes define work done by the system as positive and write ΔU = q − W by. This is equivalent if W by = −w in the chemistry convention. Problems become wrong when one copies a formula from one convention but assigns signs from another. Write the equation and label whether work is on or by the system before calculating.

Exothermic and endothermic refer to heat flow in a defined context. An exothermic reaction releases heat to surroundings, so reaction-system q is negative along a suitable constant-pressure route. Its reaction enthalpy is often negative. An endothermic reaction absorbs heat and has positive q under analogous conditions. Yet in a process with substantial work or unusual constraints, q and ΔU can have different signs. Heat transfer is not identical to total internal-energy change.

For pressure-volume work at constant external pressure, w = −P extΔV. Expansion has ΔV > 0 and w < 0; compression has ΔV < 0 and w > 0. The pressure is the external pressure opposing the boundary, not automatically the gas's instantaneous internal pressure. A freely expanding gas into a vacuum has P ext = 0 and does no P–V work in this simple model, even though its volume increases.

Heat and work are process quantities, so neither is “contained” in the system. The system contains internal energy and has a state; q and w describe how energy crossed its boundary. This language guards against the idea that a thermometer directly reads the heat stored in a sample.

Step-by-step reasoning

1. Name and sketch the system boundary. 2. Draw arrows for heat and work crossing it. 3. Assign positive signs to transfers entering the system. 4. Assign negative signs to transfers leaving the system. 5. Add q + w and check the result against the physical situation.

Visual explanation

Draw a gas cylinder as the system. A red arrow entering through a warm wall is +q; a blue arrow leaving with piston expansion is −w. Below it write ΔU = (+q) + (− w ). Reverse both arrows for cooling and compression examples.

Real-world analogy

A bank account gains money from deposits and loses it through withdrawals. Heat and work are two transaction channels; internal energy is the account balance. The analogy captures signs, though energy can also be transported with matter in open systems.

Real-world example

During a piston expansion, a heated gas can both absorb heat from a flame and lift a weight. The flame supplies positive q to the gas, while lifting the weight means negative w for the gas. The internal-energy change depends on the difference between those magnitudes.

Why?

Why is expansion work negative in chemistry notation? The system transfers energy outward by pushing the surroundings. Work is signed from the system's viewpoint, so outward energy flow reduces its balance unless offset by heat input.

Common misconception

“If the system absorbs heat, ΔU must increase.” It may also do more work than the heat it receives, leaving ΔU negative. Both transfer modes must be included.

Worked example

A closed gas sample receives 750 J of heat and performs 300 J of work on a piston. Write q = +750 J and w = −300 J. Then ΔU = q + w = +450 J. If a second sample receives 200 J but does 350 J of work, its ΔU is −150 J despite positive heat input. These calculations show why words must be translated to signs before arithmetic.

Quick check

1. What sign does work have when surroundings compress the gas system? Answer: Positive, because work energy enters the gas system.

Exam focus

State the convention ΔU = q + w, then translate verbs into signs. Do not use “work done by system” as a positive number in that equation without converting it to negative w.

Advanced insight

Heat transfer is associated with thermal interaction, while work can include electrical and surface work as well as P–V work. In a battery, electrical work may cross the system boundary even when volume barely changes. The same sign rule applies to any work mode under the chosen convention.

Summary

In chemistry convention, q and w are positive when energy enters the system and negative when it leaves. Expansion work is negative, compression work positive. A consistent boundary and first-law equation prevent sign errors.

Practice questions

1. A system releases 100 J of heat and receives 40 J of work. Find ΔU. Answer: q = −100 J, w = +40 J, so ΔU = −60 J. 2. A gas absorbs 300 J and performs 500 J of work. Find ΔU. Answer: q = +300 J, w = −500 J, so ΔU = −200 J. 3. Is “work done by the system” positive w in the chemistry convention? Answer: No. Work done by the system transfers energy out, so w is negative.