Physics for ME · Module 11 of 16
Thermal Physics: Temperature, Heat, and Material Response
Materials expand, store heat, and change phase. This module is the bridge to Thermodynamics, Heat Transfer, and Materials Science.
Readiness check
From Modules 1 and 6. Tick only what you can do closed-notes.
- Keep joules, watts, and degrees distinct.
- Work with 10⁻⁶-scale coefficients without slips.
- Run an energy audit with named accounts (Module 6).
- Convert Celsius and kelvin, knowing when each matters.
- Read a simple material-property table.
The core idea
Temperature measures molecular agitation; heat is energy in transit; materials respond by expanding, storing, or transforming.
ΔL = αLΔTQ = mcΔTQ = mLfThree material responses cover most engineering: expansion (α), sensible heating (c), and phase change (latent heat). Heat moves by conduction, convection, and radiation: named here, quantified in the Heat Transfer course.
The skills, taught in order
Heat is energy accounting continued into materials. Six skills cover how a part responds to it: how hot it gets, how much it grows, when it transforms, and how the heat arrives and leaves.
11.1 Temperature and thermometers
Temperature measures how vigorously molecules jostle, and two bodies in contact drift to the same value (thermal equilibrium). Use Celsius for differences, but switch to the absolute Kelvin scale (TK = TC + 273.15) for any ratio or gas law, because only Kelvin starts at true zero motion.
11.2 Thermal expansion
Heat a solid and it grows: ΔL = αL₀ΔT for length, with volume expanding about three times as fast (β ≈ 3α). When expansion is blocked, the suppressed strain becomes stress instead, σ = EαΔT, which is the sting in the rail example.
| Material | Linear α (×10⁻⁶ per K) |
|---|---|
| Steel | 12 |
| Copper | 17 |
| Brass | 20 |
| Aluminium | 24 |
| Glass | 4 to 9 |
Micro-example. Aluminium expands twice as much as steel for the same heating (24 versus 12), the mismatch a bimetallic thermostat strip turns into motion.
11.3 Heat and heat capacity
Adding heat without a phase change raises temperature by Q = mcΔT, where the specific heat c is the energy each kilogram needs per degree. Water's c is unusually large, which is why it dominates the temperature of any system it is part of.
| Material | Specific heat c (J per kg·K) |
|---|---|
| Water | 4190 |
| Ice | 2100 |
| Aluminium | 910 |
| Steel (iron) | 470 |
| Copper | 390 |
11.4 Phase changes
While a material melts or boils its temperature holds steady and the heat goes into breaking molecular bonds: Q = mL, with L the latent heat. For water, melting costs 334 kJ/kg and boiling 2256 kJ/kg, so vaporising is the expensive step by far.
Micro-example. Melting 1 kg of ice takes 334 kJ, the same energy that would warm that 1 kg of liquid water by 80 K.
11.5 A glimpse of kinetic theory
Temperature is, at root, the average kinetic energy of the molecules: in an ideal gas that average is proportional to the absolute temperature. This is why Kelvin is the physically meaningful scale, and it is the doorway to the gas laws of Module 12.
11.6 Heat transfer modes
Heat travels three ways: conduction through a material (rate H = kA·ΔT/L), convection carried by a moving fluid, and radiation across empty space (climbing sharply with absolute temperature). This module names them; the Heat Transfer course quantifies each in full.
Engineering connection: the bridge to Thermodynamics, Heat Transfer, and Materials Science; the blocked-expansion stress previews Mechanics of Materials.
Worked example 1: the rail gap
A 30 m steel rail (α = 12 × 10⁻⁶ /K) is laid at 15 °C and can reach 55 °C in summer. What expansion gap must the track design allow, and what happens if the rail is fully clamped instead?
- ProblemSize the expansion gap in Figure 1 and assess the clamped case.
- Given / findL = 30 m, α = 12 × 10⁻⁶ /K, ΔT = 40 K. Find ΔL; then the clamped-rail stress.
- AssumptionsUniform temperature, constant α over the range, free expansion in the first case.
- ModelLinear expansion for the gap; for the clamped rail, the suppressed strain αΔT becomes elastic stress σ = EαΔT (preview of Mechanics of Materials, E = 200 GPa for steel).
- EquationsΔL = αLΔT σ = EαΔT (if blocked)
- SolveΔL = 12 × 10⁻⁶ × 30 × 40 = 1.44 × 10⁻² m = 14.4 mm. Clamped: σ = 200 × 10⁹ × 12 × 10⁻⁶ × 40 = 96 MPa of compression.
- CheckUnits: /K × m × K = m. Scale: millimetres on a 30 m rail (about 0.05%) is the right order for metals. The 96 MPa is near half of mild steel's yield: severe, as the buckled-track photos in any railway handbook confirm.
- ConclusionGive the rail its 14.4 mm or it will take 96 MPa instead: expansion either moves or loads, never disappears. Every bridge roller, pipeline loop, and engine clearance traces back to this trade.
Worked example 2: quenching a hot part
A 2.0 kg steel part at 500 °C (c = 470 J/kg·K) is quenched into 8.0 kg of water at 20 °C (c = 4190 J/kg·K). Find the final temperature once everything settles, assuming the bath does not boil and loses no heat to its surroundings.
- ProblemFind the equilibrium temperature when the steel part is quenched in the water bath of Figure 2.
- Given / findms = 2.0 kg, cs = 470 J/kg·K at 500 °C; mw = 8.0 kg, cw = 4190 J/kg·K at 20 °C. Find the final T.
- AssumptionsInsulated bath (no loss to surroundings), no water boils away, specific heats constant over the range.
- ModelEnergy conservation as a heat balance: the heat the steel gives up equals the heat the water takes in, at one shared final temperature.
- Equationsmscs(Ts − T) = mwcw(T − Tw)
- Solve2.0(470)(500 − T) = 8.0(4190)(T − 20). That is 940(500 − T) = 33 520(T − 20), so 1 140 400 = 34 460 T and T = 33.1 °C.
- CheckThe steel releases Q = 2.0(470)(500 − 33) = 439 kJ; that raises 8 kg of water by 439 000/(8 × 4190) = 13 K, from 20 to 33 °C, matching the answer and staying well below boiling.
- ConclusionA glowing 500 °C part barely warms the bath, because water stores about nine times the heat per kilogram that steel does and there is four times the mass of it. That is why water is the workhorse quenchant and coolant, and the same heat balance sizes every quench tank and cooling loop.
Misconceptions and diagnostics
| Mistake | Symptom | Diagnostic question | Correction |
|---|---|---|---|
| Heat and temperature interchanged | "The oven has lots of temperature" | "Amount of energy, or level of agitation?" | Heat is energy in transit (J); temperature is the level (K). A sparkler is hot, a bathtub holds more heat. |
| Celsius in ratio formulas | Gas-law and radiation answers nonsensical | "Does this formula compare absolute levels?" | Differences may use °C; ratios and gas laws demand kelvin. |
| Phase change expected to warm | Melting ice "should" rise above 0 °C | "Where is the energy going during the change?" | Into breaking bonds: temperature holds while Q = mL is paid. |
| Holes expected to shrink on heating | Shrink-fit logic inverted | "Does the hole behave like the missing disc?" | Holes expand like the material that would fill them: that is why heated bearings slip onto shafts. |
Practice ladder
How much energy heats 2.0 kg of aluminium (c = 910 J/kg·K) from 20 °C to 70 °C?
Show answer
Q = 2 × 910 × 50 = 91 kJ.
A 3 kW kettle holds 1.5 kg of water at 20 °C (c = 4190 J/kg·K). How long to reach 100 °C, and how much longer to boil 0.2 kg away (Lv = 2.26 MJ/kg)?
Show answer
Heating: Q = 1.5 × 4190 × 80 = 503 kJ, t = 503/3 = 168 s. Boiling 0.2 kg: Q = 452 kJ, another 151 s. The latent step nearly matches the entire 80-degree climb: phase change is expensive.
A steel shaft of 80.000 mm must enter a bearing bored to 79.940 mm by cooling the shaft (α = 12 × 10⁻⁶ /K). What temperature drop is needed for 0.02 mm of working clearance?
Show answer
Required shrink = 0.060 + 0.020 = 0.080 mm. ΔT = ΔL/(αL) = 0.08/(12 × 10⁻⁶ × 80) = 83 K: cool to about −63 °C, a dry-ice or nitrogen job. The same arithmetic runs every shrink-fit drawing.
Find one real expansion provision (bridge joint, pipeline loop, slotted screw hole on a fence rail) and one real shrink/interference fit. Estimate the ΔL each must handle from sizes and a plausible ΔT.
What good work looks like
Photos or sketches, the αLΔT estimate for each with a stated α, and a comparison of the provision's travel against the computed need.
Working with AI, and proving it yourself
Use AI as an examiner, not a solver
Portfolio task
Measure water's heat capacity with a kettle: known power, known mass, timed temperature rise. Compare your c with 4190 J/kg·K and account for the losses that explain your (inevitable) overshoot of the true value.
Retrieval and spaced review
Closed notes. Answer out loud, then reveal.
1. Distinguish heat, temperature, and internal energy.
Temperature: the agitation level (K). Internal energy: the stored molecular energy (J). Heat: energy moving between bodies because of a temperature difference (J).
2. Write the three material-response equations.
ΔL = αLΔT; Q = mcΔT; Q = mL (latent, at constant temperature).
3. What happens when thermal expansion is prevented?
The suppressed strain becomes stress: σ = EαΔT, often structurally serious (96 MPa in the rail example).
4. Name the three heat-transfer modes with one example each.
Conduction (handle of a pan), convection (radiator warming a room), radiation (sun, glowing heater). Quantified in Heat Transfer.
5. Why must gas laws use kelvin?
They relate ratios to absolute molecular energy; Celsius has an arbitrary zero and breaks the proportionality.