Thermodynamics and statistical reasoning
Learn the laws that govern heat, work, engines, fluctuations, and repeatable cycles.
Use system boundaries, state variables, entropy, efficiency, and statistics to evaluate ordinary engines and extraordinary energy claims on equal terms.
Before you begin
- • Mechanics and conservation laws
- • Basic probability and graph reading
By the end, you can
- • Apply the first law to a defined system.
- • Explain entropy statistically and operationally.
- • Calculate ideal efficiency bounds.
- • Separate fluctuations from sustained cyclic output.
Interactive model
Explore before calculating
Live laboratory
Heat-engine boundary studio
Set two reservoirs and an engine claim. The first law closes the energy flow; the Carnot bound asks whether the claimed conversion can be cyclic between those temperatures.
Carnot ceiling: 50.0%
Claimed efficiency: 40%
Work output: 400 J
Rejected heat: 600 J
The claim lies below the ideal ceiling. Real losses still require a measured full-cycle ledger.
Level 2 · Secondary physics teaching kit
Record the investigation. Teach the reasoning.
A learner-facing lab record and a course-specific instructor guide turn the live model into a repeatable classroom investigation.
Learner record
Heat-engine boundary and efficiency ledger
How do reservoir temperatures and uncounted transfers constrain the useful work of a cyclic engine?
Download learner recordInstructor guide
Teach for evidence, not button pushing
Learners close a cyclic energy ledger and distinguish measured efficiency from the reversible Carnot ceiling.
Download instructor guideLesson 1 of 3
The first law and system boundaries
Which energy crossed the boundary, and which energy remained stored inside?
The first law is energy conservation written for thermal systems: internal-energy change equals heat added plus work done on the system.
A boundary choice determines what counts as input, output, and storage. Hidden batteries, fuel, pressure, phase changes, and temperature drift all belong in the ledger.
Worked example
A gas absorbs 500 J of heat and does 180 J of work on its surroundings.
- 1. Use ΔU = Q − W_by.
- 2. Insert 500 − 180.
- 3. Track the sign convention explicitly.
The gas internal energy rises by 320 J.
Try it
Boundary redraw
Materials: A diagram of an electric kettle and paper.
- 1. Draw a boundary around water only.
- 2. Redraw around water plus kettle.
- 3. Redraw around kettle plus outlet and room.
- 4. List which transfers move from external to internal each time.
Notice: The physics is unchanged, but honest labels depend on the chosen boundary.
Check your understanding: Can a device show output with no simultaneous input and still conserve energy?
Answer: Yes, while stored energy decreases.
A full-duration test must measure initial and final storage states.
Lesson 2 of 3
Entropy and engine limits
Why can heat flow power an engine only when a temperature difference exists?
Entropy counts how broadly energy is distributed among accessible microscopic arrangements. In an isolated system, overwhelmingly probable evolution moves toward larger accessible multiplicity.
A heat engine converts part of a hot-to-cold energy flow into work. No cyclic engine can turn all heat from one reservoir into work.
Worked example
An ideal engine operates between 600 K and 300 K.
- 1. Use η_max = 1 − Tc/Th.
- 2. Compute 1 − 300/600.
- 3. Interpret as an upper bound.
Maximum ideal efficiency is 50%; real engines are lower.
Try it
Efficiency audit
Materials: Energy-flow data for any engine or appliance.
- 1. Identify hot and cold reservoirs.
- 2. Compute useful output divided by input.
- 3. Compare with a relevant ideal bound.
- 4. Account for rejected heat.
Notice: A result above the theoretical bound signals measurement, boundary, or model error before it signals new physics.
Check your understanding: Does the second law say local order can never increase?
Answer: No.
Local order can increase when greater entropy is exported to the surroundings.
Lesson 3 of 3
Fluctuations and complete cycles
When can random microscopic motion become useful work?
Small systems fluctuate. But equilibrium fluctuations do not provide a one-way resource to a passive cyclic device because the device itself fluctuates and detailed balance restores symmetry.
Nonequilibrium gradients, measurement, feedback, and switching can produce work, but their preparation and information costs belong in the complete cycle.
Worked example
A nanoscale paddle sometimes turns forward from random impacts. Is that an engine?
- 1. Measure reverse turns too.
- 2. Include the ratchet's own thermal motion.
- 3. Test for a temperature or chemical gradient.
- 4. Close the reset cycle.
At one equilibrium temperature there is no sustained preferred work output; a gradient or controlled feedback supplies the resource.
Try it
Random-walk power test
Materials: A coin, graph paper, and 100 trials.
- 1. Move +1 for heads and −1 for tails.
- 2. Plot position over time.
- 3. Find intervals with apparent trends.
- 4. Repeat and test whether the trend persists predictably.
Notice: Short random sequences can look directional; repeatable power requires a stable asymmetry and energy source.
Check your understanding: Why must a Casimir or vacuum-energy proposal specify its reset step?
Answer: Because net work is defined over a repeatable closed cycle.
A favorable one-way force does not establish positive total work after separation, switching, and control costs.
Formula-to-meaning deck
Read the equation in ordinary language.
ΔU = Q − W_by
Internal-energy change equals heat added minus work done by the system.
Units: J
η = W_out/Q_in
Efficiency is useful work divided by energy taken from the hot input.
Units: dimensionless
η_Carnot = 1 − T_c/T_h
Reservoir temperatures set the maximum reversible heat-engine efficiency.
Units: temperatures in K
Independent practice
Problem set
Work each problem before opening its hint and solution.
1. A system receives 900 J of heat and its internal energy rises 250 J. How much work did it do?
Reveal hint
Rearrange ΔU = Q − W.
Reveal solution
650 J.
2. An engine absorbs 2000 J and delivers 600 J work. Find efficiency and rejected heat.
Reveal hint
η = W/Q; energy conservation gives the remainder.
Reveal solution
30% efficiency; 1400 J rejected.
3. Find the Carnot bound between 500 K and 300 K.
Reveal hint
Use 1 − 300/500.
Reveal solution
40%.
Continue into the evidence