
Thermal unit · institution-dependent · Interactive lesson
Temperature, Equilibrium & Thermometry
Before T can appear in an equation you need a number a thermometer can deliver and a reason two thermometers should agree on it. This lesson builds that number from thermal equilibrium upward: a property that changes with hotness, fixed points that pin its scale, and a dilute gas whose low-density limit erases the choice of substance.
UPIII-06.01 · Lesson 1 of 9 in this unit. Position is not a completion score.Thermal unit · institution-dependent. Many departments teach thermal physics separately or within Physics II. Check whether your university includes this numbered unit in Physics III.
Build the model
Bodies in mutual equilibrium share one number. A thermometer's job is to print it.
The key idea
Temperature is the number shared by every body in mutual thermal equilibrium, made measurable by choosing a property that changes with hotness and fixing its scale at reproducible states; the kelvin takes that property to be the pressure of a dilute gas at constant volume.
Example. A constant-volume gas thermometer at 51.20 kPa at the triple point of water reads 68.83 kPa in a bath: T ≈ 273.16 K × 68.83/51.20 = 367.2 K, or 94.1 °C, pending the small correction for the gas's own non-ideality.
The complete physical model and assumptions
Temperature begins as a relation, not a quantity. Put two bodies in thermal contact and their measurable properties drift until nothing changes further: they are in thermal equilibrium. The zeroth law says the relation is transitive — if A and B are each in equilibrium with C, they are in equilibrium with each other — and transitivity is exactly what lets every body be given one number such that equal numbers mean no drift on contact. That number is defined, not discovered: pick a thermometric property X that changes with hotness, assign values at reproducible fixed points, and interpolate. Mercury length, platinum resistance and thermocouple emf each yield a scale, and the scales agree only where the fixed points force them to, because none of those properties is exactly linear in any other. The dilute gas breaks the tie. At constant volume its pressure ratio p/pₜₚ, extrapolated to zero filling, is the same for every gas, and the line of p against Celsius temperature crosses zero at −273.15 °C whatever the filling. Setting T = 273.16 K × lim(p/pₜₚ) makes the kelvin substance-independent; since 2019 the kelvin is fixed through the Boltzmann constant instead, with the gas thermometer as one way to realise it. The model stops where thermometry does: at a body too small or too fast-changing to reach equilibrium with the probe, and at temperatures where no gas stays dilute.
1Thermal contact, equilibrium and the zeroth law
Two bodies are in thermal contact when heat can pass between them without either doing work on the other — a copper wall, not a piston. Bring them together and their measurable properties change, then stop changing: that final state is thermal equilibrium, and reaching it can take a second for a thermocouple bead or an hour for a stirred bath. The zeroth law adds transitivity: if A and B are each in equilibrium with C, they are in equilibrium with each other. That is a physical claim, not a triviality — nothing in mechanics forces it — and it is what makes a thermometer possible. C can be a small probe carried from A to B; if its reading is unchanged, A and B need never touch. It also makes equilibrium an equivalence relation, so the states of any system split into classes and each class can be labelled with one number. Everything else in thermometry is the choice of that labelling.
2Build an empirical scale from a thermometric property
A thermometric property is anything that changes monotonically with hotness and can be read: the length of a mercury column, the resistance of a platinum wire, the emf of a thermocouple, the pressure of a gas held at fixed volume. Choosing one and decreeing how its value maps to temperature defines an empirical scale. The historical Celsius recipe uses two fixed points and a straight line: t = 100 °C × (X − X₀)/(X₁₀₀ − X₀). A Pt100 element with R₀ = 100.00 Ω and R₁₀₀ = 138.51 Ω that reads 111.67 Ω is assigned t = 100 °C × 11.67/38.51 = 30.30 °C. Notice what has and has not been done. The reading is exactly right at the two fixed points because it was made so; between them it is a definition, and a mercury thermometer built by the same recipe will not agree with it, because mercury length is not a linear function of platinum resistance. Each empirical scale is only as linear as its working substance.
3Let the dilute gas settle the argument
One property behaves better than the rest. Hold a gas at constant volume — a bulb joined to a manometer whose mercury is brought back to a fiducial mark after every change — and record its pressure p in the bath and pₜₚ at the triple point of water. The ratio p/pₜₚ depends slightly on which gas and how much of it was used, but reduce the filling and every gas converges on one value. The reason is that the correction to p ∝ T is proportional to the gas density, so a plot of p/pₜₚ against pₜₚ is a straight line whose intercept at zero filling is gas-independent. Defining T = 273.16 K × lim(p/pₜₚ) therefore erases the working substance. For the steam point the intercept is 1.365954, giving 373.124 K, or 99.974 °C: once 273.16 K was fixed at the triple point in 1954, the boiling point of water became a measured quantity, and it is no longer exactly 100 °C.
4Extrapolate to absolute zero and set the kelvin
Plot the pressure of a dilute gas at fixed volume against Celsius temperature and the points fall on a line, p = p₀(1 + t/273.15 °C), where p₀ is the ice-point pressure. Change the filling and the slope changes with p₀, but the intercept does not: every line reaches p = 0 at t = −273.15 °C. No gas gets there — it condenses first — so the intercept is a limit of dilute-gas behaviour, not a state anything reaches. Moving the origin to that limit gives T/K = t/°C + 273.15 and turns the line into a proportionality, p ∝ T, the form the kinetic model will later explain. Two nearby numbers do different jobs. 273.15 K is the ice point, used for conversion; 273.16 K is the triple point, chosen as the calibration state because one pressure and temperature fix it, reproducible to 0.1 mK, whereas the ice point drifts with air pressure and dissolved gas. Since 2019 the kelvin is defined by fixing k = 1.380649 × 10⁻²³ J K⁻¹ exactly, so the triple point is now a measured 273.16 K, and the gas thermometer is one primary way to realise the unit rather than its definition.
5Quantify the error of a linear scale
How far is a two-point scale wrong between its fixed points? Take the platinum relation R = R₀(1 + At + Bt²) with A = 3.9083 × 10⁻³ °C⁻¹ and B = −5.775 × 10⁻⁷ °C⁻². The linear scale reads tₗᵢₙ = 100 °C × (R − R₀)/(R₁₀₀ − R₀) = (At + Bt²)/(A + 100 °C × B), so the error is δ = tₗᵢₙ − t = B t (t − 100 °C)/(A + 100 °C × B). It vanishes at both fixed points, as it must, and dδ/dt = B (2t − 100 °C)/(A + 100 °C × B) = 0 puts the extreme at t = 50 °C, where δ = −2500 °C² × B/(A + 100 °C × B) = +0.37 °C. A Pt100 read linearly therefore shows 50.37 °C in a bath the gas scale calls 50.00 °C. The cure is not a purer wire but a better interpolation: add a third fixed point and fit the quadratic. That is how practical scales such as ITS-90 work — many fixed points, a named interpolating instrument, and a specified formula between them.
6Demand a monotonic property and a small probe
A property can be sensitive and still fail as a thermometer. A thermocouple's emf against a 0 °C reference junction is close to E = αt + βt² over a wide range; for α = 14.0 μV °C⁻¹ and β = −0.020 μV °C⁻² the emf peaks at the neutral temperature tₙ = −α/(2β) = 350 °C, and E(100 °C) = E(600 °C) = 1200 μV. One reading, two temperatures: past its neutral point this couple is useless, because its property is no longer monotonic. The second demand is thermal. The probe reaches the bath's temperature by exchanging heat with it, so its heat capacity must be small next to the bath's or the act of measuring moves the answer — a 2 g bead dropped into 5 g of liquid reports the mixture, not the liquid. Sensitivity, range, monotonicity and size are the checklist, and no single instrument passes it everywhere, which is why a laboratory keeps several.
Key equations
Read the units, assumptions and limits before substituting.
Two-fixed-point empirical scale
t = 100 °C × (X − X₀)/(X₁₀₀ − X₀)
Turns any property into a Celsius reading by decree; two thermometers built this way agree only at 0 and 100 °C.
X is any thermometric property (in m, Ω, Pa or V); X₀ and X₁₀₀ are its values at the ice and steam pointsIdeal-gas temperature at constant volume
T = 273.16 K × lim (p/pₜₚ) as pₜₚ → 0
The limit is taken by refilling with less gas; it makes every gas report the same T.
p and pₜₚ in Pa: the bulb's pressure in the bath and at the triple point of water, at one fixed volumePressure line of a dilute gas at fixed volume
p = p₀ (1 + t/273.15 °C)
Every filling gives a line through p = 0 at t = −273.15 °C: that shared intercept is absolute zero.
p₀ in Pa is the pressure at the ice point; t in °C; holds while the gas stays dilute and V is fixedCelsius to kelvin
T/K = t/°C + 273.15
Shifts the zero and keeps the size of the degree; 273.15 is the ice point, 273.16 the triple point.
T in K, t in °C; an interval is the same size on both: ΔT/K = Δt/°CCallendar–Van Dusen, platinum above 0 °C
R(t) = R₀ (1 + A t + B t²)
The small B is the curvature that any linear two-point platinum scale ignores.
R₀ in Ω at 0 °C; A = 3.9083 × 10⁻³ °C⁻¹, B = −5.775 × 10⁻⁷ °C⁻²; t in °CError of a linear scale on a quadratic property
δ(t) = B t (t − 100 °C)/(A + 100 °C × B)
Extreme at 50 °C, where a Pt100 read on a two-point linear scale is 0.37 °C high.
δ = tₗᵢₙ − t in °C, with A and B as above; zero at both fixed pointsInteractive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag sample A from 40 to 120 kPa and watch its line swing about −273.15 °C: the slope follows the filling, the intercept never moves. Then take the bath down to −250 °C and check that both pressures still share one ratio p/p₀, which is all the kelvin reading uses.
Live results
- pA AT t
- 114.6 kPa
- pB AT t
- 68.8 kPa
- RATIO p/p₀, EITHER SAMPLE
- 1.1464
- T = 273.15 K × p/p₀
- 313.15 K
Results follow this lesson’s model and assumptions. Displayed values are rounded.
Predict → change one variable → check
Plan a change before running it. Every other input stays at the saved baseline.
Saved baseline. Sample A pressure at 0 °C: 100 kPa; Sample B pressure at 0 °C: 60 kPa; Bath temperature t: 40 °C
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
Calibrating at two fixed points does not make a thermometer right everywhere between them. Two points fix a straight line in the property, and nothing obliges the property to be straight in temperature: a Pt100 read on a linear scale is 0.37 °C high at 50 °C, and mercury, alcohol and platinum thermometers built by the same recipe disagree with one another by tenths of a degree while each reads exactly 0 and 100 °C. Nor does the zeroth law rescue them — it guarantees that bodies in equilibrium share a temperature, not that different scales print the same number for it. Between the fixed points there is no right reading until a scale is chosen, and the dilute gas is chosen because it is the one property on which every substance agrees.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A mercury-in-glass thermometer's column stands 4.20 cm above the bulb at the ice point and 24.70 cm at the steam point. Using a linear two-point scale, find the Celsius temperature indicated by a column of 12.50 cm, convert it to kelvin, and say whether a platinum thermometer calibrated the same way must agree.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 4 steps revealed.
- Declare the scale: the thermometric property is column length L, with L₀ = 4.20 cm at 0 °C and L₁₀₀ = 24.70 cm at 100 °C, and the rule is t = 100 °C × (L − L₀)/(L₁₀₀ − L₀).
- Substitute: t = 100 °C × (12.50 − 4.20)/(24.70 − 4.20) = 100 °C × 8.30/20.50 = 40.49 °C. The centimetres cancel in the ratio, so the unit of L never mattered.
- Convert: T = 40.49 + 273.15 = 313.64 K. The column was read to about ±0.05 cm, which is ±0.2 °C on this scale, so quote 40.5 °C and 313.6 K.
- A platinum element calibrated by the same rule also reads exactly 0 and 100 °C, but between them it follows R(t), not L(t). The two agree only if resistance is linear in column length, which it is not; expect a difference of a few tenths of a degree.
Answer. t = 40.5 °C and T = 313.6 K on the mercury scale; a platinum thermometer need not agree away from the fixed points.
MediumExample 2
A constant-volume nitrogen thermometer is used to find the normal boiling point of water. With three fillings the triple-point pressures are 100.000, 60.000 and 20.000 kPa and the corresponding steam-point pressures 136.675, 81.9858 and 27.3222 kPa. Extrapolate to zero filling, give the boiling point in kelvin and Celsius, and find the error of trusting the 100 kPa filling alone.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Form the ratio p/pₜₚ for each filling: 136.675/100.000 = 1.36675, 81.9858/60.000 = 1.36643, 27.3222/20.000 = 1.36611. The ratio falls as the filling falls, so the gas is not yet ideal at these densities.
- The correction to p ∝ T is proportional to density, so p/pₜₚ is linear in pₜₚ. Slope from the outer points: (1.36675 − 1.36611)/(100.000 kPa − 20.000 kPa) = 8.0 × 10⁻⁶ kPa⁻¹. The middle point, 1.36611 + 40.000 × 8.0 × 10⁻⁶ = 1.36643, sits on the line.
- Intercept at pₜₚ = 0: 1.36611 − 20.000 × 8.0 × 10⁻⁶ = 1.36595. This is the ideal-gas limit, the value every dilute gas would give.
- T = 273.16 K × 1.36595 = 373.123 K, so t = 373.123 − 273.15 = 99.97 °C: with 273.16 K fixed at the triple point, the boiling point of water is a measured quantity and is not 100 °C.
- A single reading at 100 kPa would have given 273.16 K × 1.36675 = 373.341 K, high by 0.22 K. Limiting-case check: as pₜₚ → 0 the line tends to its intercept, and a perfect gas would have shown no slope at all.
Answer. T = 373.12 K, t = 99.97 °C; the unextrapolated 100 kPa reading is 0.22 K too high.
HardExample 3
A Pt100 element obeys R = R₀(1 + At + Bt²) with R₀ = 100.00 Ω, A = 3.9083 × 10⁻³ °C⁻¹ and B = −5.775 × 10⁻⁷ °C⁻². A student calibrates it only at the ice and steam points and assumes a linear scale. Derive the error of that scale as a function of the true temperature, locate its extreme by differentiation, and state what the student reads in a bath whose true temperature is 50.00 °C.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 5 steps revealed.
- Fixed-point resistances: R₀ = 100.00 Ω and R₁₀₀ = 100.00 Ω × (1 + 0.39083 − 0.005775) = 138.506 Ω, so the linear rule is tₗᵢₙ = 100 °C × (R − R₀)/(R₁₀₀ − R₀).
- Substitute the true R(t): tₗᵢₙ = 100 °C × (At + Bt²)/(100 °C × (A + 100 °C × B)) = (At + Bt²)/(A + 100 °C × B). The error is δ = tₗᵢₙ − t = (Bt² − 100 °C × Bt)/(A + 100 °C × B) = B t (t − 100 °C)/(A + 100 °C × B).
- Check the limits: δ = 0 at t = 0 and at t = 100 °C, as a two-point calibration must give. Differentiate: dδ/dt = B (2t − 100 °C)/(A + 100 °C × B) = 0 at t = 50 °C, and with B < 0 the extreme is a maximum of the reading.
- Evaluate: A + 100 °C × B = 3.9083 × 10⁻³ − 5.775 × 10⁻⁵ = 3.8506 × 10⁻³ °C⁻¹, and δ(50 °C) = (−5.775 × 10⁻⁷)(50)(−50)/(3.8506 × 10⁻³) = 1.4438 × 10⁻³/(3.8506 × 10⁻³) = 0.375 °C.
- Directly: R(50 °C) = 100.00 Ω × (1 + 0.195415 − 0.00144375) = 119.397 Ω, and tₗᵢₙ = 100 °C × 19.397/38.506 = 50.37 °C, confirming the formula. The scale is right where it was forced to be and 0.37 °C high in the middle.
Answer. δ(t) = B t (t − 100 °C)/(A + 100 °C × B), extreme at 50 °C; the student reads 50.37 °C at a true 50.00 °C, an error of +0.37 °C.
Quick check
Select an answer and explain it before checking.
Review & continue
Temperature is the number shared by every body in mutual thermal equilibrium, made measurable by choosing a property that changes with hotness and fixing its scale at reproducible states; the kelvin takes that property to be the pressure of a dilute gas at constant volume.
- Explain the relationship without looking at the equation.
- Check a sign, a unit and a model assumption in your working.
- Return to the lab or worked example if a step is still unclear.
Optional supporting lessons
These chapter resources can help with earlier ideas; they are not additional required Physics III completion tasks.