Physics for ME · Module 12 of 16 · Thermo bridge
First Law of Thermodynamics and Energy Balance
Module 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 Modules 6 and 11, Math Module 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 Module 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.
The skills, taught in order
The first law is Module 6's energy audit with one more account and one more kind of transfer. Seven skills take you from drawing the boundary to walking a gas around an engine cycle.
12.1 System, boundary, and surroundings
Choose what you are analysing and draw a boundary around it; everything else is the surroundings. A closed system exchanges energy but not matter across that line. Drawing the boundary first is the habit every later term depends on.
12.2 Internal energy
Internal energy U is the molecular energy stored inside the system, the thermodynamic storage account. For an ideal gas it depends on temperature alone, so a temperature rise means a rise in U and nothing else.
12.3 Heat and work as transfers
Heat (Q) and work (W) are the two ways energy crosses the boundary. Neither is stored: a system does not "contain heat", it contains internal energy, and Q and W exist only while energy is in transit.
12.4 Boundary work
When the boundary moves against pressure, the system does work W = ∫P dV, the area under the path on a pressure-volume diagram. At constant pressure this collapses to the rectangle W = PΔV.
Micro-example. A piston pushed out by gas at a steady 200 kPa through 0.02 m³ does W = 200 000 × 0.02 = 4 kJ.
12.5 The first law
Energy is conserved across the boundary: ΔU = Q − W, with the convention that Q is positive flowing in and W is positive when the system does work on the surroundings. Most first-law errors are sign errors, so declare the convention and hold it.
12.6 Simple processes
Holding one quantity fixed simplifies the law. The four basic processes below each fix or kill one term, and every real cycle is built from them.
| Process | Held constant | What it gives |
|---|---|---|
| Isochoric | volume | W = 0, so ΔU = Q |
| Isobaric | pressure | W = PΔV |
| Isothermal | temperature | ΔU = 0, so Q = W |
| Adiabatic | no heat flow (Q = 0) | ΔU = −W |
12.7 Where the second law will enter
The first law says energy is never lost, but not which way a process runs. Heat flows hot to cold, and friction makes heat but never the reverse, even though energy would balance either way. That direction-and-quality story is the second law, formalised in the Thermodynamics course.
Engineering connection: the physics foundation of the whole Thermodynamics course: the energy-balance habit behind engines, compressors, and HVAC.
Worked example 1: 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: the next step is control-volume energy accounting.
Worked example 2: work around a P-V cycle
A fixed amount of gas is carried clockwise around the rectangular cycle in Figure 2, between 100 and 300 kPa and between 2 L and 6 L. Find the net work it does per cycle and the net heat it must absorb.
- ProblemFind the net work the gas does per cycle and the net heat it must absorb, for the clockwise cycle in Figure 2.
- Given / findPlow = 100 kPa, Phigh = 300 kPa, Vlow = 2 L = 0.002 m³, Vhigh = 6 L = 0.006 m³, clockwise. Find net W and net Q.
- AssumptionsClosed system of fixed gas, quasi-static legs, returning exactly to its starting state each cycle.
- ModelBoundary work is the area under each leg, W = ∫P dV; over a closed loop the net work is the enclosed area. Because the gas returns to its initial state, ΔU = 0 over the cycle.
- EquationsWcycle = enclosed area = (Phigh − Plow)(Vhigh − Vlow) ΔUcycle = 0 ⇒ Qnet = Wcycle
- SolveThe two constant-volume legs do no work; the top expansion does +PhighΔV = 300 000 × 0.004 = +1200 J; the bottom compression does Plow(−ΔV) = −400 J. Net = 1200 − 400 = 800 J, equal to the enclosed area (200 kPa)(0.004 m³). Since ΔU = 0, the gas absorbs Qnet = 800 J.
- CheckA clockwise loop encloses net work done by the gas (an engine); the same loop run counter-clockwise would consume work (a refrigerator). The enclosed area, not the path's fine detail, sets the number.
- ConclusionAn engine is a gas walked around a loop on the P-V plane, harvesting the enclosed area as net work every cycle. The Thermodynamics course is largely the study of which loops give the most area for the least heat.
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 Module 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
Use AI as an examiner, not a solver
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.