Physics for ME · Chapter 12 of 16 · Intermediate · Thermo bridge
First Law of Thermodynamics and Energy Balance
Chapter 6's energy audit, extended to heat: energy crossing a boundary as heat or work changes what is stored inside. The full Thermodynamics course builds on this page.
Readiness check
From Chapters 6 and 11, Math Chapter 6. Tick only what you can do closed-notes.
- Run a mechanical energy audit with named accounts.
- Use Q = mcΔT for sensible heating.
- Distinguish heat, temperature, and internal energy.
- Use the ideal gas law P = mRT/V (Math Chapter 6's example).
- Hold a sign convention through a whole problem.
The core idea
Energy is conserved across the system boundary: what enters as heat, leaves as work, or stays as internal energy.
ΔU = Q − WDraw the boundary first (the thermodynamics habit). Q counts positive entering; W positive when the system does work on the surroundings (the piston convention). Internal energy U is the storage account, and for an ideal gas it tracks temperature alone.
What this chapter covers
- 12.1 System, boundary, surroundings: the thermodynamic worldview.
- 12.2 Internal energy: the storage account; temperature-linked for ideal gases.
- 12.3 Heat and work as transfers: two ways across the line.
- 12.4 Boundary work: W = ∫P dV, the piston integral (Math Chapter 5).
- 12.5 The first law: ΔU = Q − W and its sign discipline.
- 12.6 Simple processes: constant volume, constant pressure, by behavior.
- 12.7 Where the second law will enter: Chapter 6's friction leak, formalized later.
Engineering connection: the physics foundation for the full Thermodynamics course (Çengel and Boles as its text).
Worked example: the heated piston
Gas in a piston-cylinder receives 65 kJ of heat. While expanding, it pushes the piston with 25 kJ of work. Find the internal energy change, and the gas temperature trend.
- ProblemFind ΔU for the gas in Figure 1 and state what happens to its temperature.
- Given / findQ = +65 kJ (in), W = +25 kJ (done by the gas). Find ΔU.
- AssumptionsThe gas is the system; no leaks; kinetic and potential energy of the gas bulk negligible.
- ModelFirst law with the piston sign convention: ΔU = Q − W.
- EquationsΔU = Q − W
- SolveΔU = 65 − 25 = +40 kJ. For a (near-ideal) gas, U tracks temperature, so the gas ends warmer: of the 65 kJ supplied, 25 left as piston work and 40 stayed as molecular agitation.
- CheckLimiting cases discipline: locked piston (W = 0) would give ΔU = 65 kJ, the hottest outcome; a perfectly insulated expansion (Q = 0) would give ΔU = −25 kJ and cooling. Our answer sits between, as it must.
- ConclusionOne dashed boundary and one sign convention settled the whole audit. Engines, compressors, and turbines are this picture with flow added: exactly where Çengel and Boles begin.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Sign conventions mixed | ΔU = Q + W in one line, Q − W in the next | "Which convention did I declare, and where?" | Write the convention beside the boundary sketch and never renegotiate mid-problem. |
| Heat treated as a substance stored inside | "The gas contains 65 kJ of heat" | "Is this energy crossing the boundary right now?" | Heat exists only in transit. Inside, it is internal energy. |
| Work forgotten because nothing "mechanical" is visible | Expansion problems with ΔU = Q | "Did the boundary move against pressure?" | A moving piston does work W = ∫P dV whether or not machinery is attached. |
| Boundary never drawn | Energy terms double-counted or lost | "Inside or outside: which is this device?" | Draw the dashed line first. Every term is then classifiable. |
Practice ladder
A rigid (locked-volume) tank of gas receives 12 kJ of heat. Find W, ΔU, and the temperature trend.
Show answer
No boundary motion: W = 0, so ΔU = +12 kJ and the gas warms. Constant-volume heating is pure storage.
An air compressor's gas gives up 30 kJ of heat to its cooling fins while 50 kJ of work is done on it. Find ΔU with careful signs.
Show answer
Q = −30 kJ; work done by the gas W = −50 kJ. ΔU = −30 − (−50) = +20 kJ: the gas warms even while losing heat, because compression work outpaces the cooling. Every bicycle pump confirms it.
Gas expands at constant pressure P = 200 kPa from 0.10 m³ to 0.16 m³ while absorbing 45 kJ of heat. Find the boundary work and ΔU.
Show answer
W = PΔV = 200 000 × 0.06 = 12 kJ. ΔU = 45 − 12 = +33 kJ. The ∫P dV integral collapsed to a rectangle because P held constant: Math Chapter 5 at work.
Audit one real device as a first-law system (kettle, hair dryer, bike pump, car engine at idle): draw its boundary, list every Q and W crossing with directions, and estimate the dominant term.
What good work looks like
A boundary sketch, all crossings signed, magnitudes estimated with stated assumptions, and the storage term ΔU judged (steady devices: near zero, and saying why).
Working with AI, and proving it yourself
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Portfolio task
Produce a one-page "Boundary Album": four devices, each with a dashed boundary, signed arrows for every crossing, and a one-line first-law statement. Reuse the bike-pump warm-barrel observation as your experimental anchor.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. State the first law with the piston convention and define each term.
ΔU = Q − W: storage change equals heat in minus work done by the system.
2. What is boundary work, and its integral?
Work done by a moving boundary against pressure: W = ∫P dV; a rectangle PΔV when P is constant.
3. Why does a pumped tire's barrel warm with no flame near it?
Work is done on the gas faster than heat leaves: ΔU rises, and U tracks temperature.
4. For an ideal gas, what does U depend on?
Temperature only: ΔU = 0 in an isothermal process even with heat and work flowing.
5. What question does the first law not answer?
Direction and quality: why heat flows hot-to-cold and why some energy is unrecoverable: the second law's territory, in the Thermodynamics course.
Textbook mapping
| Item | Mapping |
|---|---|
| Main source | OpenStax University Physics Vol. 2, The First Law of Thermodynamics |
| Bridge reference | Çengel and Boles, Thermodynamics: An Engineering Approach (the full course's text, later) |
| Core topics | 12.1 System and boundary · 12.2 Internal energy · 12.3 Heat and work transfers · 12.4 Boundary work · 12.5 The first law · 12.6 Simple processes · 12.7 Second-law preview |
| Engineering connection | The foundation of the entire Thermodynamics course; energy-balance habit for engines, compressors, HVAC. |
| Read next | Chapter 13: Fluids: Pressure, Buoyancy, and Flow Intuition. |