Technopedia

The technologist’s encyclopedia

Zeroth Law and Temperature

The zeroth law defines thermal equilibrium and the temperature variable. It is the logical prerequisite for every other thermodynamic law: without a well-defined temperature, the first and second laws have no state to describe. Use it when you need to justify that a thermometer reading is meaningful, or when you need to define what “same temperature” means operationally.

Authoritative sources

Thermal equilibrium

Definition. Two systems in thermal contact are in thermal equilibrium when no net heat flows between them. The relation “is in thermal equilibrium with” (∼\sim) is an equivalence relation: reflexive, symmetric, and transitive. Transitivity is the zeroth law proper.

Details

Step 1 — Operational definition and diathermal walls.

To test whether two systems are in thermal equilibrium, you place them in thermal contact: they are separated by a diathermal wall (a wall that permits energy transfer as heat but not matter). If, after a sufficiently long time, no macroscopic change occurs in either system — no net heat flow, no change in any measurable variable — the two systems are in thermal equilibrium.

The entire operational definition rests on the existence of diathermal walls. Without a wall that allows heat to flow freely, you cannot bring two systems into thermal contact, and the equivalence relation has no physical meaning. In practice, a thin metal wall or direct contact between solids approximates a diathermal wall well.

Step 2 — The equivalence relation.

The zeroth law states that the relation ∼\sim satisfies all three properties of an equivalence relation:

Because ∼\sim is an equivalence relation, the set of all systems partitions into equivalence classes. Each class can be labeled by a single real number — temperature. Without transitivity, no single labeling variable would exist; you would need a separate parameter for every pair of systems.

Step 3 — Visualizing the partition.

The following diagram shows five systems and the thermal-contact relations between them. Systems connected by a solid line are in thermal equilibrium. The dashed line between A and C is not a direct contact; it is the consequence of transitivity. A, B, and C all belong to the same equivalence class (same temperature), even though A and C are never placed in direct contact. D and E form a second, distinct equivalence class.

A B C D E Class 1: T₁ (A ∼ B ∼ C) Class 2: T₂ (D ∼ E) direct contact transitivity

Step 4 — Worked example.

Suppose you have three metal blocks: A at 400 K, B at 300 K, and C at 350 K. You place A in contact with B through a thin copper wall (a diathermal wall). Heat flows from A to B until both reach a common temperature. You then place B in contact with C. Heat flows from B to C until both reach a common temperature. By transitivity, A and C are in thermal equilibrium with each other, even though they were never placed in direct contact. All three blocks now share the same equivalence class and the same temperature.

Step 5 — Thermometers.

A thermometer is a system C whose one macroscopic variable (e.g., volume of mercury, resistance of a wire) depends only on the equivalence class it shares with the system being measured. The zeroth law guarantees that this reading is independent of the thermometer used: any two thermometers in the same equivalence class give the same reading (up to calibration).

Authoritative sources

Example problem

You have four objects: a cup of hot water (A), a metal spoon (B), a glass of cold water (C), and a room-temperature table (D). You place the spoon in the hot water and wait until equilibrium. You then remove the spoon, place it in the cold water, and wait until equilibrium. Is the hot water in thermal equilibrium with the cold water? Explain using the zeroth law.

Solution

No. After the first step, A ∼ B (hot water and spoon share a temperature). After the second step, B ∼ C (spoon and cold water share a temperature). But the spoon’s temperature changed between the two steps: it was at the hot-water temperature in step 1 and at the cold-water temperature in step 2. The transitivity of ∼\sim requires that B be in the same equilibrium state in both relations. Since B is in different states, transitivity does not apply, and A and C are not in thermal equilibrium.

Temperature as an intensive variable

Definition. Temperature TT is the intensive variable that equalizes when two systems are placed in thermal contact:

TA=TB⟺A and B in thermal equilibriumT_A = T_B \quad \Longleftrightarrow \quad \text{A and B in thermal equilibrium}
Details

Step 1 — Intensive vs. extensive.

A variable is extensive if it scales with the amount of material (volume, internal energy, entropy) and intensive if it does not (temperature, pressure). If you split a system in two, each half has the same temperature and pressure, but half the volume and internal energy. Temperature is the intensive variable that equalizes at thermal equilibrium.

Step 2 — Thermal contact and the equilibrium condition.

Two systems are in thermal contact when they can exchange energy as heat through a diathermal wall but not matter or volume. At equilibrium, their temperatures are equal. This is the operational content of the zeroth law: the condition TA=TBT_A = T_B is both necessary and sufficient for thermal equilibrium.

Step 3 — Temperature scales.

The zeroth law justifies the existence of a temperature scale but does not fix it. Any monotonic function of the equivalence-class label is a valid temperature scale. The Kelvin scale is fixed by the second law (Carnot efficiency) and, in the modern SI (since 2019), by the Boltzmann constant kB=1.380649×10−23k_B = 1.380649 \times 10^{-23} J/K.

Step 4 — Statistical definition (pointer).

At the microscopic level, temperature is defined by 1/T=(∂S/∂U)V,N1/T = (\partial S / \partial U)_{V,N}. This makes the operational (zeroth-law) definition and the fundamental (statistical) definition coincide. The full derivation is developed in Statistical Foundations of Thermodynamics.

Step 5 — Worked example.

Two identical copper blocks, each of mass 1 kg, are at 400 K and 300 K. The specific heat of copper is c=385c = 385 J/(kg·K). When placed in thermal contact (isolated from the surroundings), the final equilibrium temperature is:

Tf=mc T1+mc T22mc=T1+T22=400+3002=350 KT_f = \frac{m c \, T_1 + m c \, T_2}{2 m c} = \frac{T_1 + T_2}{2} = \frac{400 + 300}{2} = 350 \text{ K}

Both blocks end at 350 K regardless of their mass — this is the intensive character of temperature. If you doubled the mass of both blocks, the equilibrium temperature would still be 350 K.

Authoritative sources

Example problem

A 2 kg block of iron (c=450c = 450 J/(kg·K)) at 500 K is placed in thermal contact with a 1 kg block of aluminum (c=900c = 900 J/(kg·K)) at 300 K. The system is isolated. Find the equilibrium temperature.

Solution

Energy conservation (first law, no work): mFecFe(Tf−500)+mAlcAl(Tf−300)=0m_{\text{Fe}} c_{\text{Fe}} (T_f - 500) + m_{\text{Al}} c_{\text{Al}} (T_f - 300) = 0.

2×450 (Tf−500)+1×900 (Tf−300)=02 \times 450 \,(T_f - 500) + 1 \times 900\,(T_f - 300) = 0 900 Tf−450000+900 Tf−270000=0900\,T_f - 450000 + 900\,T_f - 270000 = 0 1800 Tf=720000⟹Tf=400 K1800\,T_f = 720000 \quad\Longrightarrow\quad T_f = 400 \text{ K}

The equilibrium temperature is 400 K. The iron and aluminum have equal heat capacities (mc=900m c = 900 J/K each), so the result is exactly the midpoint of the two initial temperatures weighted by heat capacity.

Empirical temperature

Definition. Before the second law, temperature is defined empirically: choose a reference substance and a measurable property XX, and set

Temp=T0 XX0T_{\text{emp}} = T_0 \, \frac{X}{X_0}

where T0T_0, X0X_0 are the values at a fixed reference point.

Details

Step 1 — The empirical construction.

Pick a reference substance (e.g., a gas at low pressure) and a measurable property XX (pressure, volume, resistance). Fix a reference point (e.g., the triple point of water at T0=273.16T_0 = 273.16 K). For any other state, measure XX and compute Temp=T0 X/X0T_{\text{emp}} = T_0 \, X / X_0. This gives a number that labels the equivalence class.

Step 2 — The ideal gas thermometer.

For an ideal gas at fixed volume, X=PX = P (pressure). The empirical temperature is proportional to pressure. In the limit of zero pressure, all gas thermometers (regardless of gas species) agree, defining the ideal-gas temperature scale. This is the most practical realization of the thermodynamic temperature.

Step 3 — Common empirical thermometers.

Three practical types are widely used:

  1. Constant-volume gas thermometer — measures pressure of a fixed amount of gas.
  2. Resistance (platinum) thermometer — measures electrical resistance of a platinum wire.
  3. Thermocouple — measures the Seebeck voltage between two dissimilar metals.

At finite pressure, current, or junction temperature, these three give slightly different readings for the same equilibrium state. They converge to a common value only in the ideal limit (zero pressure, zero current, ideal junctions), which is the thermodynamic temperature.

Step 4 — Convergence diagram.

The following diagram shows how the readings of three different empirical thermometers converge to the thermodynamic temperature as the non-ideal parameter (pressure, current, etc.) is driven to zero.

non-ideal parameter (P, I, …) → 0 reading T (thermodynamic) gas (P) resistance (I) thermocouple ideal limit

Step 5 — Worked example.

A constant-volume gas thermometer uses helium. At the triple point of water (T0=273.16T_0 = 273.16 K), the helium pressure is P0=100.0P_0 = 100.0 kPa. At an unknown temperature, the pressure reads P=120.0P = 120.0 kPa. The empirical temperature is:

Temp=273.16×120.0100.0=327.8 KT_{\text{emp}} = 273.16 \times \frac{120.0}{100.0} = 327.8 \text{ K}

In the zero-pressure limit this would equal the thermodynamic temperature. At finite pressure, a small correction (on the order of a few tenths of a kelvin) would be needed.

Step 6 — Limitations.

The empirical scale depends on the choice of reference substance. The second law (via Carnot) shows that there is a unique absolute scale — the thermodynamic temperature — that makes the efficiency of a Carnot engine depend only on the two reservoir temperatures. See Second Law and Entropy for the absolute temperature definition.

Authoritative sources

Example problem

A platinum resistance thermometer reads R=130.7R = 130.7 Ω at the triple point of water (T0=273.16T_0 = 273.16 K, R0=100.0R_0 = 100.0 Ω) and R=155.0R = 155.0 Ω at an unknown temperature. Using the linear approximation Temp=T0 R/R0T_{\text{emp}} = T_0 \, R / R_0, find the temperature.

Solution Temp=273.16×155.0100.0=423.4 KT_{\text{emp}} = 273.16 \times \frac{155.0}{100.0} = 423.4 \text{ K}

This is approximately 150.2 °C. In practice, the platinum resistance–temperature relation is not perfectly linear, and a higher-order polynomial (the Callendar–Van Dusen equation) is used for precision work. The linear approximation is adequate to within about 0.1 K over this range.

Quizz

Problem

A mercury thermometer reads 37.0 °C in a patient’s mouth. A resistance thermometer reads 310.15 K in the same mouth. A third system (a cup of coffee) is in thermal equilibrium with the patient. What can you conclude about the coffee’s temperature, and why is the answer independent of which thermometer you use?

Solution

The patient’s mouth is in thermal equilibrium with both thermometers, so all three (mercury thermometer, resistance thermometer, mouth) share the same equivalence class. The coffee is in thermal equilibrium with the patient, so by transitivity (the zeroth law) the coffee is in the same equivalence class as both thermometers.

Therefore the coffee’s temperature is 37.0 °C = 310.15 K. The answer is independent of which thermometer you use because the zeroth law guarantees that all thermometers in the same equivalence class read the same value (up to calibration). The mercury thermometer and the resistance thermometer agree here because 37.0 °C + 273.15 = 310.15 K; at other temperatures or with non-ideal instruments they might differ slightly, but the conclusion that the coffee is in the same equilibrium class does not depend on the instrument.