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Thermodynamic systems, heat and work

Distinguish thermodynamic systems, surroundings, heat and work, then compare state and path variables using a clear sign convention.

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  • University Physics Year 1
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Why this matters + quick links

This page gives the UY1 working model/result for Definition and concepts of thermodynamics. You reuse it when you turn thermal intuition into calculable models (energy balances, process reasoning, and kinetic/statistical links), with careful sign/units checks.

Thermodynamics studies how energy is transferred and transformed, especially as heat and work.

Thermodynamics is mostly modelling discipline: before you compute anything, you must (1) choose the system and boundary, (2) say what counts as heat vs work, and (3) lock a sign convention so your energy balance has meaning.

1) At a glance

  • Prerequisites: basic mechanics work-energy ideas; ideal gas equation PV = nRT
  • Outcomes: define system/boundary/surroundings; classify energy transfer as heat or work; apply one sign convention consistently
  • Key result: for this module, energy crossing the boundary is tracked by Δ E = Q + W where Q > 0 is heat into the system and W > 0 is work done on the system
Sign convention (used throughout UY1 Thermodynamics here)
  • Q > 0: heat enters the system.
  • W > 0: work is done on the system.
  • Work done by the system is Wₒᵤₜ = -W.
  • Every term in the first law has units of energy (J). If your answer has units like Pa or K, something went wrong.
  • Limiting case: if the system is isolated (Q = 0 and W = 0), then Δ E = 0.
Common traps (sign + definition mistakes)
  • Mixing conventions between sources (some books write Δ E = Q-W with W defined as work by the system). - Treating Q or W like stored energy (writing Δ Q or Δ W as if they were state changes). - Misclassifying energy transfer: stirring/shaft/electrical input is work, not heat.

2) Setup (model, assumptions, sign conventions)

Take a system separated from surroundings by a boundary.

  • System: the matter/region we analyze (for example, gas in a piston)
  • Boundary: real or imaginary; fixed or moving
  • Heat Q: energy transfer caused by temperature difference
  • Work W: any other boundary energy transfer (mechanical, electrical, shaft work, etc.)

3) Core derivation/explanation

Two important categories:

  • State variables: P,V,T,E depend only on state, not on how the state was reached
  • Path variables: Q,W depend on the process path

So, between the same initial and final states:

  • Δ E is fixed
  • Q and W can change individually, but their sum must satisfy Δ E = Q + W

Mathematically, we write exact differential dE for a state variable, and inexact differentials δ Q,δ W for path variables: dE = δ Q + δ W

Assigning signs before adding energy transfers

Worked example 1

Heating and expansion

Problem

A gas receives 120 J of heat and does 45 J of work on its surroundings. Find its internal-energy change using positive work done on the gas.
Show full solution

  1. Assign the transfer signs

    Method

    Take incoming heat as positive and outgoing work as negative.

    Reason

    Both signs refer to the gas as the system; expansion transfers energy away through work.

    Working

    Answer

    Q = +120 J and W = -45 J.

  2. Calculate the energy gain

    Working

    Answer

    Δ E = Q + W = 120 + (-45) = 75 J

    The gas gains 75 J: incoming heat exceeds the energy delivered as work.

Practice set

Check your understanding 1

A system loses 60 J as heat and has 30 J of work done on it. Find Δ E.

Show hint
Assign signs for transfers into the chosen system.
Show answer
Δ E = Q + W = -60 + 30 = -30 J. Work input replaces only part of the heat loss, so the energy decreases.

Check your understanding 2

A system has Δ E = 0 and receives 90 J of heat. Find the work W done on it.

Show hint
Rearrange the first law for W.
Show answer
W = Δ E-Q = 0-90 = -90 J. The system must do 90 J of work on its surroundings to leave its energy unchanged.

Check your understanding 3

One process has Q = 40 J and W = 10 J. A second process between the same initial and final states has Q = 100 J. Find W for the second process; W is work done on the system.

Show hint
Find the state change from the first path and reuse it for the second.
Show answer
Both paths have Δ E = 40 + 10 = 50 J. Hence W = 50-100 = -50 J for the second path. State changes agree even though the heat and work transfers differ.

6) Summary + next steps

  • Thermodynamics tracks energy crossing a boundary as heat or work.
  • In this module, use Δ E = Q + W with Q > 0 in and W > 0 on system.
  • E is a state variable; Q and W are path variables.

Next: work output by gas pistons Previous: heat capacity of solids Back to thermodynamics

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Syllabus and review details

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