
Optional extension · Interactive lesson
Temperature and thermal equilibrium
Temperature is the one number that decides whether energy will flow between two bodies in contact. This lesson builds it from the ground up: the zeroth law that makes a thermometer trustworthy, the calibrated property that turns a reading into a number, and the gas extrapolation that fixes where zero actually sits.
UPII-E2.01 · Lesson 1 of 10 in this extension. Position is not a completion score.Optional extension. Use this unnumbered support sequence where your university’s syllabus or prerequisite needs require it. It is not an additional numbered unit. Optional work is not a required completion gate.
Build the model
Equilibrium first, then the number that labels it.
The key idea
Temperature is the single number shared by every system in thermal equilibrium with a given body, so that two systems at the same temperature exchange no net energy when they are placed in thermal contact.
Example. A thermometer reading 37.67 °C in a bath is strictly reporting its own temperature; the zeroth law is what licenses the further claim that the bath is at 37.67 °C too, which is 310.82 K.
The complete physical model and assumptions
Put two bodies in thermal contact and their measurable properties drift until they stop; that final state is thermal equilibrium. The zeroth law is the experimental fact that makes it useful: if A and B are each in equilibrium with a third body C, then A and B are in equilibrium with each other. Equilibrium is therefore transitive, it sorts every system into classes, and a single number can label each class — that number is temperature, and C is a thermometer. To read it you pick a thermometric property that varies smoothly and monotonically with hotness (a column length, a resistance, a gas pressure, a thermocouple emf), declare the scale linear in that property, and pin it with two reproducible fixed points. But the scale you then get belongs to the substance: two thermometers forced to agree at 0 °C and 100 °C disagree by tenths of a degree in between. The escape is the dilute gas, whose constant-volume pressure extrapolates to zero at the same −273.15 °C whatever the gas, defining an absolute scale on which ratios finally mean something. That scale is the kelvin, and its zero is a limit approached by extrapolation, never a temperature reached.
1Thermal contact, and what stops changing
Two systems are in thermal contact when energy can pass between them without work being done — through a conducting (diathermal) wall rather than an insulating (adiabatic) one. Press a warm copper block against a cool one and their measurable properties change: lengths, resistances, pressures. After long enough they stop. That final, unchanging macroscopic state is thermal equilibrium, and it is a state, not a process. Nothing has stopped at the molecular level; molecules still collide across the boundary and energy still crosses it in both directions. What has vanished is the net flow. The whole point of temperature is that a single number, assigned to each body separately, decides in advance whether that net flow will be zero, and if it is not, which way it goes.
2The zeroth law is what licenses a thermometer
Suppose block A is in equilibrium with thermometer C, and block B is separately in equilibrium with that same C. Nothing so far logically forces A and B to be in equilibrium with each other — and yet every time it is tested, they are. That experimental fact is the zeroth law of thermodynamics, and it does real work. It says thermal equilibrium is transitive, so it sorts all systems into disjoint classes, each class holding everything that would sit unchanged beside everything else in it. One label per class is therefore enough, and that label is temperature. Without transitivity a thermometer would be useless: you could only ever learn about the object it was touching, never compare two objects by comparing two readings. Recognised last but numbered zeroth, the law is logically prior to the first and second laws, which is exactly how it got its name.
3A scale is a property plus two fixed points
To attach numbers you need a thermometric property X: something measurable that changes smoothly and monotonically with hotness. A mercury column's length, a platinum wire's resistance, a gas pressure at fixed volume, a thermocouple emf and the spectral radiance of a hot surface all qualify. Monotonic matters — a property that rose and then fell would hand one reading two temperatures. Now declare the scale linear in X and pin it with two reproducible fixed points, the ice point at 0 °C and the steam point at 100 °C at 1 atm, so that t = 100 °C × (X − X₀)/(X₁₀₀ − X₀). A mercury column standing 24.0 mm at the ice point and 212.0 mm at the steam point reads 118.5 mm in a bath, giving t = 100 °C × (118.5 − 24.0)/(212.0 − 24.0) = 50.27 °C. Notice what just happened: the linearity was chosen, not discovered.
4Two good thermometers still disagree in between
Because linearity was a choice, the scale belongs to the substance. Calibrate a platinum resistance thermometer and a mercury thermometer at the same two fixed points and they agree exactly there, by construction, and disagree everywhere in between. Platinum's resistance really follows R = R₀(1 + At + Bt²) with A = 3.9083 × 10⁻³ °C⁻¹ and B = −5.775 × 10⁻⁷ °C⁻², so a scale that pretends R is linear reads 50.37 °C when the true temperature is 50.00 °C. The discrepancy is only 0.37 °C, but it is systematic, it vanishes at both fixed points, and it is largest halfway between them. Adding a third fixed point patches this but does not cure it. What is needed is a thermometric property whose linearity is not a convention about one particular material.
5Dilute gas sets the absolute scale
The dilute gas supplies exactly that. Hold a fixed amount of gas at fixed volume and plot its pressure against Celsius temperature: the graph is a straight line. Take p₀ = 1.000 × 10⁵ Pa at the ice point, and the same sample reads 1.36610 × 10⁵ Pa at the steam point, a slope of 366.10 Pa per °C. Run the line backwards and it reaches p = 0 at t = −(1.000 × 10⁵ Pa)/(366.10 Pa °C⁻¹) = −273.15 °C. Repeat with helium, nitrogen or argon, at any starting pressure, and the intercept is unchanged: the lines fan out but share one zero. That common intercept is absolute zero, and putting the origin of the scale there gives T/K = t/°C + 273.15, on which p ∝ T outright. Absolute zero is where the extrapolation lands, not a state any gas survives to occupy — every real gas condenses long before it arrives.
6The reading is the thermometer's own temperature
A thermometer never reports the bath's temperature; it reports its own, and the zeroth law then transfers the number. Two practical consequences follow. First, the probe must be small in heat capacity compared with whatever it measures, or inserting it shifts the very temperature it was sent to find. Second, you must wait. Write the remaining gap as ΔT = Tprobe − Tbath, and let t stand here for elapsed time rather than a Celsius temperature: the gap closes exponentially, ΔT = ΔT₀ e(−t/τ), with a time constant τ set by the probe's heat capacity and by how well it couples to the bath. A probe with τ = 4.0 s starting 60 K below an 80.0 °C bath is still 60 × e(−2) = 8.1 K low after 8.0 s, and only after about 20 s — five time constants — has the gap shrunk to 60 × e(−5) = 0.40 K. A reading taken too early is an accurate measurement of the wrong thing.
Key equations
Read the units, assumptions and limits before substituting.
Linear two-point scale
t = 100 °C × (X − X₀) / (X₁₀₀ − X₀)
Turns one instrument reading into a number, and forces the scale to be exact at the two fixed points.
X is any thermometric property (Ω, mm, Pa, mV); X₀ at the ice point, X₁₀₀ at the steam point at 1 atmAbsolute temperature from Celsius
T / K = t / °C + 273.15
Convert before any ratio, product or proportionality — the Celsius zero is a convention, not an absence.
T is the absolute temperature in kelvin, t the Celsius temperature; the offset is exact by definitionTemperature differences
ΔT / K = Δt / °C
A 20.0 °C rise is a 20.0 K rise. Never add 273.15 to a difference.
The 273.15 offset cancels in a subtraction, so a kelvin and a Celsius degree are exactly the same sizeConstant-volume gas thermometer
T = 273.16 K × lim(p₃ → 0) (p / p₃)
The low-density limit deletes the gas itself, giving a scale that no substance can bias.
p is the gas pressure in Pa at T, p₃ its pressure at the water triple point, volume held fixedExtrapolation to absolute zero
p(t) = p₀ (1 + t / 273.15 °C), at fixed V
Locates the zero of the scale from measurements taken hundreds of degrees above it.
p₀ is the ice-point pressure in Pa; the straight line reaches p = 0 at t = −273.15 °CSI definition of the kelvin
kB = 1.380649 × 10⁻²³ J K⁻¹ exactly
Since 2019 the unit no longer depends on one particular sample of water.
Fixing the Boltzmann constant fixes the size of the kelvin; its units are J K⁻¹Interactive lab
Predict first. Explore the model. Compare the result.
Explore the model
Try this. Drag the ice-point pressure and watch the line pivot: every setting gives a different slope, and every one of them still crosses zero pressure at the same −273.15 °C. That shared crossing, not the slope, is what makes the kelvin scale independent of the gas.
Live results
- PRESSURE p
- 1.366 ×10⁵ Pa
- ABSOLUTE T
- 373.15 K
- p / p AT ICE POINT
- 1.366
- SLOPE dp/dT
- 366.1 Pa K⁻¹
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. Ice-point pressure p₀: 1.00 ×10⁵ Pa; Bath temperature t: 100 °C
This experiment is kept only while this page is open. It does not change saved course progress.
Catch the common trap
The trap is reading equal temperature as equal energy. A bathtub of water at 25 °C and a thimble of water at 25 °C are in thermal equilibrium: bring them into contact and no net energy flows, even though the bath holds thousands of times more internal energy. Temperature is intensive, indifferent to how much substance there is; internal energy is extensive and scales with the amount. The same slip runs in reverse when a student says the larger warm object must be at a higher temperature because it stores more energy. What decides the direction of the net flow is the temperature difference alone, never which body holds the larger store.
Practice & worked examples
3 authored examples. Make an attempt, then reveal the reasoning one step at a time.
EasyExample 1
A copper–constantan thermocouple is assumed to give an emf linear in temperature. It reads 0.000 mV with its measuring junction in melting ice and 4.277 mV with the junction in steam at 1 atm. Placed in a water bath it reads 1.611 mV. Find the bath temperature on this thermocouple's own scale, in °C and in K.
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.
- Identify the thermometric property: here X is the emf E, with E₀ = 0.000 mV at the ice point and E₁₀₀ = 4.277 mV at the steam point.
- Apply the two-point linear scale: t = 100 °C × (E − E₀)/(E₁₀₀ − E₀) = 100 °C × (1.611 − 0.000)/(4.277 − 0.000).
- The millivolts cancel, leaving a pure number: 1.611/4.277 = 0.37667, so t = 37.67 °C.
- Convert with the exact offset: T = 37.67 + 273.15 = 310.82 K.
- Check the sense: the reading sits 37.7% of the way from ice to steam, and 37.67 °C sits 37.7% of the way from 0 °C to 100 °C, as a linear scale demands.
Answer. t = 37.67 °C, which is T = 310.82 K
MediumExample 2
A constant-volume gas thermometer holds a fixed sample of helium. Its pressure is 9.600 × 10⁴ Pa with the bulb in melting ice, and 1.585 × 10⁵ Pa with the bulb in a furnace port. Treating the helium as ideal, find the furnace temperature in K and in °C, and predict the pressure the same bulb would show at 500.0 K.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- At fixed volume and fixed amount of ideal gas, p ∝ T with T absolute, so the quotient p/T is the same for both readings.
- The ice point is T₀ = 273.15 K, so T = T₀ × p/p₀ = 273.15 K × (1.585 × 10⁵ Pa)/(9.600 × 10⁴ Pa).
- The pressure ratio is p/p₀ = 1.65104, giving T = 273.15 K × 1.65104 = 450.98 K, which is 451.0 K to four figures.
- Subtract the offset from the unrounded value: t = 450.98 − 273.15 = 177.83 °C, or 177.8 °C to four figures.
- Run the proportionality forwards for 500.0 K: p = 9.600 × 10⁴ Pa × 500.0/273.15 = 1.757 × 10⁵ Pa.
- Check with the slope of the line through the origin: p₀/T₀ = (9.600 × 10⁴ Pa)/(273.15 K) = 351.5 Pa K⁻¹, and 351.5 × 451.0 = 1.585 × 10⁵ Pa, recovering the furnace reading.
Answer. T = 451.0 K, t = 177.8 °C; at 500.0 K the same bulb would read 1.757 × 10⁵ Pa.
HardExample 3
A Pt100 sensor has R₀ = 100.00 Ω at 0 °C and truly obeys R = R₀(1 + At + Bt²), with A = 3.9083 × 10⁻³ °C⁻¹ and B = −5.775 × 10⁻⁷ °C⁻², where t is the ideal-gas Celsius temperature. A datalogger ignores B and assumes its scale is linear in R between the ice and steam points. Find what the logger reports when the true temperature is 50.00 °C, and show where its error is worst.
Sketch your answer or reasoning, then compare it with the authored solution. Your writing is not automatically graded or saved.
0 of 6 steps revealed.
- True resistances: R(50.00 °C) = 100.00 Ω × (1 + 0.195415 − 0.00144375) = 119.397 Ω, and R(100.00 °C) = 100.00 Ω × (1 + 0.39083 − 0.005775) = 138.5055 Ω.
- The logger's scale: tₗᵢₙ = 100 °C × (R − R₀)/(R₁₀₀ − R₀) = 100 °C × (119.397 − 100.00)/(138.5055 − 100.00) = 100 °C × 19.397/38.5055.
- That quotient is 0.50375, so the logger reports tₗᵢₙ = 50.37 °C — high by 0.37 °C, even though the two scales are exact at 0 °C and at 100 °C.
- For the shape of the error, substitute R into the logger's formula. Writing K = 100A + 10⁴B = 0.385055, everything collapses to tₗᵢₙ − t = (100B/K)(t² − 100t), a parabola with roots at t = 0 and t = 100 °C.
- Its vertex sits at t = 50.00 °C, so the worst error is the one already found, 0.375 °C. At 25.00 °C and at 75.00 °C the deviation is only 0.281 °C.
- The moral: agreeing at the fixed points guarantees nothing in between, which is why the absolute scale is defined by the dilute gas and not by any solid.
Answer. The logger reports 50.37 °C, an error of +0.37 °C. The deviation is (100B/K)(t² − 100t): zero at both fixed points and maximal at 50.00 °C.
Quick check
Select an answer and explain it before checking.
Review & continue
Temperature is the single number shared by every system in thermal equilibrium with a given body, so that two systems at the same temperature exchange no net energy when they are placed in thermal contact.
- 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 II completion tasks.