The Thermodynamic Temperature Scale

Imagine two heat reservoirs, one at the temperature of boiling water and one at melting ice. Any reversible engine running between them has a fixed efficiency, no matter what gas, metal, or fluid sits inside it. That startling fact, guaranteed by Carnot's theorem, means efficiency ratios can define temperature itself. This matters because the everyday temperature scales, Celsius or even the gas thermometer, secretly depend on the quirks of mercury, gas, or platinum wires. Physical chemistry needs a temperature that means the same thing everywhere in the universe, hot star or cold flask. The thermodynamic scale delivers exactly that, built from work and heat alone. This article builds the thermodynamic temperature scale from Carnot's theorem. You will see why efficiency ratios must be universal functions of temperature. Then you will see how Kelvin turned that function into a linear scale. Finally, you will see how this scale sets up the thermodynamic definition of entropy.

Key takeaways

Carnot's theorem: all reversible engines between two reservoirs share the same efficiency, independent of the working substance.
This universality lets efficiency define a temperature scale, not the other way around.
Kelvin's choice makes $q_h/q_c = T_h/T_c$ , a ratio of heats equal to a ratio of temperatures.
Fixing the triple point of water at exactly 273.16 K anchors the whole scale.
The thermodynamic scale coincides numerically with the ideal-gas scale, but rests on no gas law at all.
Key idea: Reversible heat ratios between reservoirs are universal, so they can be used to define an absolute, substance-independent temperature.

The core insight

Because every reversible engine operating between the same two reservoirs has identical efficiency (Carnot's theorem), the ratio of heat exchanged with the hot and cold reservoirs, $q_h/q_c$ , must be a universal function of the two reservoir temperatures alone. Choosing that function to be the simple ratio $T_h/T_c$ turns Carnot efficiency into the operational definition of an absolute thermodynamic temperature, with no reference to any particular substance's thermal expansion or pressure behavior.

The physical picture

A Carnot engine is an idealized reversible heat engine operating in a four-step cycle between two reservoirs: an isothermal expansion at the hot temperature, an adiabatic expansion, an isothermal compression at the cold temperature, and an adiabatic compression back to the start. Carnot's theorem, established from the impossibility of building a perpetual motion machine of the second kind, states that no engine operating between two given reservoirs can exceed the efficiency of a reversible engine operating between them. Furthermore, all reversible engines between the same two reservoirs have exactly the same efficiency, regardless of whether the working fluid is an ideal gas, steam, or some exotic substance. This is a profound restriction. The efficiency of a reversible engine,

\varepsilon = 1 - \frac{q_c}{q_h}

cannot depend on the properties of the working substance. It can only depend on the two empirical temperatures of the reservoirs, $\theta_h$ and $\theta_c$ , measured on whatever thermometer you like (mercury, gas, resistance). That means $q_c/q_h$ must equal some universal function $f(\theta_h, \theta_c)$ common to all reversible engines. Kelvin recognized that this function has a special multiplicative structure: chaining two Carnot engines in series (one running between $\theta_h$ and an intermediate $\theta_m$ , another between $\theta_m$ and $\theta_c$ ) shows that $f(\theta_h,\theta_c) = f(\theta_h,\theta_m) f(\theta_m,\theta_c)$ . A function with this property must factor as a ratio,

f(\theta_h,\theta_c) = \frac{T(\theta_c)}{T(\theta_h)}

for some function $T(\theta)$ of the empirical temperature alone. Kelvin then simply defined $T(\theta)$ to be the thermodynamic temperature. The choice is a convention, but a spectacularly useful one: it makes the thermodynamic temperature track the ideal-gas temperature exactly, while remaining meaningful even where no gas thermometer could survive (inside a star, or near absolute zero where gases liquefy).

Figure

Carnot cycle between two reservoirs, defining T from heat ratios

The second useful picture is the scale itself: how the ratio definition, plus one fixed reference point, pins down every value of $T$ uniquely.

Figure

Anchoring the absolute scale at the triple point of water

The core relations

\varepsilon_{\text{rev}} = 1 - \frac{q_c}{q_h}

where: $\varepsilon_{\text{rev}}$ is the efficiency of any reversible engine, $q_h$ and $q_c$ are heats exchanged with hot and cold reservoirs (magnitudes).

\frac{q_c}{q_h} = \frac{T_c}{T_h}

where: $T_h, T_c$ are thermodynamic (absolute) temperatures of the two reservoirs, defined by this ratio holding for every reversible engine.

\varepsilon_{\text{rev}} = 1 - \frac{T_c}{T_h}

where: this expresses Carnot efficiency purely in terms of the absolute temperatures, with no reference to any working substance.

T_{\text{triple}} \equiv 273.16\ \text{K (exact, by definition prior to 2019 redefinition; now realized via } k_B\text{)}

where: the kelvin is now fixed by setting the Boltzmann constant $k_B = 1.380649\times10^{-23}\ \text{J K}^{-1}$ exactly, which reproduces 273.16 K for the triple point of water to experimental precision.

The core relations

Derivation AFrom Carnot's theorem to an absolute temperature scale

Goal. Show that Carnot's theorem forces $q_c/q_h$ to be a universal function of empirical temperatures alone, and that this function can be chosen to define an absolute, substance-independent temperature $T$ .

Step 1. State Carnot's theorem. All reversible engines operating between the same two reservoirs, at empirical temperatures $\theta_h$ and $\theta_c$ , have identical efficiency. If two reversible engines had different efficiencies, the more efficient one could drive the less efficient one in reverse, producing net work from a single reservoir with no other change. This would violate the Kelvin statement of the second law. Hence

\varepsilon_{\text{rev}}(\theta_h,\theta_c) = 1 - \frac{q_c}{q_h}

is the same function for every substance, so $q_c/q_h \equiv f(\theta_h,\theta_c)$ is universal.

Step 2. Chain two Carnot engines. Run one reversible engine between $\theta_h$ and an intermediate temperature $\theta_m$ , and a second between $\theta_m$ and $\theta_c$ . Heat $q_h$ enters the first engine, $q_m$ passes between the two engines, and $q_c$ leaves the second engine. By the definition of $f$ ,

\frac{q_m}{q_h} = f(\theta_h,\theta_m), \qquad \frac{q_c}{q_m} = f(\theta_m,\theta_c)

Multiplying these,

\frac{q_c}{q_h} = f(\theta_h,\theta_m)\,f(\theta_m,\theta_c) = f(\theta_h,\theta_c)

Step 3. Deduce the factorized form. The composition rule $f(\theta_h,\theta_c) = f(\theta_h,\theta_m) f(\theta_m,\theta_c)$ must hold for any choice of the intermediate $\theta_m$ . The only functions satisfying this for all $\theta_m$ have the separable form

f(\theta_h,\theta_c) = \frac{g(\theta_c)}{g(\theta_h)}

for some single-variable function $g$ . This follows because differentiating the composition rule with respect to $\theta_m$ and demanding consistency forces $f$ to depend on its arguments only through such a ratio.

Step 4. Define the thermodynamic temperature. Kelvin defined the thermodynamic temperature as $T(\theta) \equiv g(\theta)$ , choosing the multiplicative constant so that the scale matches the familiar gas-thermometer scale in size (100 kelvin between ice and steam points, historically). With this choice,

\frac{q_c}{q_h} = \frac{T_c}{T_h}

for every reversible engine, independent of substance.

Step 5. Fix the scale with one reference point. A ratio alone cannot fix absolute values, only ratios of temperatures. One fixed point is needed. By international agreement, the triple point of water was assigned exactly $273.16\ \text{K}$ (now realized through the fixed value of $k_B$ rather than a material property). Every other thermodynamic temperature follows from measuring $q_c/q_h$ of a reversible engine run between the unknown reservoir and the triple-point cell, then scaling by 273.16 K.

Boxed result.

\boxed{T_h/T_c = q_h/q_c \ \text{for any reversible engine, with the scale fixed by } T_{\text{triple}} = 273.16\ \text{K}}

This is a temperature scale built entirely from heat and work measurements, with no reference to the expansion of mercury, the pressure of a gas, or any other material property.

Worked examples

Example 1: Finding an unknown reservoir temperature from Carnot efficiency. A reversible engine operates between a hot reservoir and the triple point of water ( $T_c = 273.16\ \text{K}$ ) as the cold reservoir. In one cycle it absorbs $q_h = 850\ \text{J}$ from the hot reservoir and rejects $q_c = 610\ \text{J}$ to the triple-point reservoir. Using the defining relation,

\frac{T_h}{T_c} = \frac{q_h}{q_c} = \frac{850\ \text{J}}{610\ \text{J}} = 1.3934

T_h = 1.3934 \times 273.16\ \text{K} = 380.7\ \text{K}

Sanity check: this is above the boiling point of water (373.15 K), consistent with $q_h > q_c$ and a positive engine efficiency,

\varepsilon = 1 - \frac{610}{850} = 1 - 0.7176 = 0.282

that is, 28.2%, a physically reasonable value for a reversible engine working across roughly 100 K near room temperature. No gas law, no thermal expansion coefficient, and no thermometer material entered the calculation. Temperature emerged purely from a heat ratio.

Example 2: Comparing engines with different working substances to test universality. Two experimental reversible engines are built, one using helium gas and one using a two-phase steam/water system, both operated between a hot reservoir at unknown $T_h$ and a cold reservoir fixed at the ice point, taken as $T_c = 273.15\ \text{K}$ (0 °C, essentially coincident with the triple point to within 0.01 K, the small difference being a defined offset). Helium engine: absorbs $q_h = 1200\ \text{J}$ , rejects $q_c = 900\ \text{J}$ .

\frac{T_h}{T_c} = \frac{1200}{900} = 1.3333 \implies T_h = 1.3333\times273.15\ \text{K} = 364.2\ \text{K}

Steam engine: absorbs $q_h' = 500\ \text{J}$ , rejects $q_c' = 375\ \text{J}$ while operating between the same two reservoirs.

\frac{T_h}{T_c} = \frac{500}{375} = 1.3333 \implies T_h = 1.3333\times273.15\ \text{K} = 364.2\ \text{K}

Both substances, though physically utterly different (a monatomic ideal gas versus a condensing liquid-vapor mixture), give the identical ratio $q_h/q_c = 1.3333$ and hence the identical thermodynamic temperature, $T_h = 364.2\ \text{K}$ (about 91.1 °C). This numerical agreement is not a coincidence: it is the entire content of Carnot's theorem, now demonstrated with concrete heats rather than left as an abstract efficiency statement. If a third engine, using some novel refrigerant, gave a different ratio at the same two reservoirs, that would immediately signal an irreversibility (friction, unrestrained expansion, or a temperature mismatch at heat transfer) rather than a genuine substance dependence, since the theorem forbids the latter.

Limiting cases & common traps

Situation	What happens	Common error
$T_c \to 0$	Efficiency $\varepsilon \to 1$ ; unreachable in finite steps (Third Law)	Thinking a real engine can approach $\varepsilon = 1$ practically
$T_h = T_c$	$\varepsilon = 0$ , no work extractable, consistent with Kelvin statement	Forgetting that equal reservoirs forbid any engine, reversible or not
Irreversible engine	$q_c/q_h > T_c/T_h$ , so measured efficiency is below Carnot value	Using the ratio formula for a real, irreversible engine
Choosing the reference point	Any single fixed point suffices to set the scale's size	Believing the scale needs the freezing and boiling points of water, as in old Celsius definitions

The chemist behind it

William Thomson, later Lord Kelvin, recognized in 1848 that Carnot's theorem implied something no thermometer could offer: a temperature scale defined purely by work and heat, immune to the peculiarities of any real substance. Building directly on Sadi Carnot's and Rudolf Clausius's analyses of reversible cycles, Kelvin proposed that the ratio of heats exchanged in a reversible cycle be used to define temperature itself. This move, from measuring a property (like mercury's expansion) to defining a scale from a theorem, is one of the conceptual turning points of thermodynamics. It also directly paved the way for Clausius's introduction of entropy a few years later, since the same reversible heat ratios that define $T$ also define $q_{\text{rev}}/T$ as a state function.

Origins · 1 minWilliam Thomson (Lord Kelvin) and how this was discovered

William Thomson (Lord Kelvin)

1824–1907 · Defined the absolute temperature scale from Carnot efficiency, the direct ancestor of the modern kelvin

Ask the tutor: How does the relation $q_{\text{rev}}/T$ being cycle-invariant lead Clausius to define entropy as a state function?

Is the thermodynamic temperature the same number as the Celsius or ideal-gas temperature?

Yes, numerically. The kelvin was deliberately sized so 0 K to 273.16 K spans the same range as the ideal-gas scale gives between absolute zero and the triple point of water. So $T/\text{K} = t/{}^{\circ}\text{C} + 273.15$ holds exactly. The conceptual difference matters more than the numeric one: the thermodynamic scale is defined from Carnot's theorem and needs no particular substance, while the Celsius scale historically depended on the properties of mercury or a specific gas.

Why can't we just use the freezing and boiling points of water to build an absolute scale, as Celsius did?

That approach fixes two points and calls the interval 100 units, but it says nothing about why zero should be absolute zero, nor does it connect the scale to any physical law. The thermodynamic scale instead derives its entire structure from Carnot's theorem: the ratio $q_c/q_h$ automatically equals $T_c/T_h$ for any reversible engine. Only one fixed point, not two, is needed to set the overall size of the kelvin, because the ratio relation already fixes the zero at the point where $q_c \to 0$ .

Does the 2019 SI redefinition change any of this?

No, it refines how the scale is realized experimentally, not its physical meaning. Before 2019, the kelvin was defined by fixing the triple point of water at exactly 273.16 K. Since 2019, the kelvin is defined by fixing the Boltzmann constant $k_B$ at exactly $1.380649\times10^{-23}\ \text{J K}^{-1}$ . Both definitions reproduce the same Carnot-based scale; the triple point of water is now measured (very precisely) rather than defined, but its value remains 273.16 K within experimental uncertainty.

Can any real engine ever reach the Carnot efficiency exactly?

No. Real engines have friction, unrestrained expansions, finite-rate heat transfer across temperature differences, and other irreversibilities, all of which make $q_c/q_h$ exceed $T_c/T_h$ , lowering the actual efficiency below the reversible limit. The Carnot efficiency is a strict upper bound, approached only in the idealized limit of infinitely slow, frictionless operation. It remains an extremely useful benchmark for real power plants and refrigerators.

How does this connect to entropy, covered next?

Since $q_c/q_h = T_c/T_h$ for a reversible cycle, rearranging gives $q_h/T_h = q_c/T_c$ , meaning $q_{\text{rev}}/T$ is the same magnitude entering and leaving. Clausius recognized this as evidence for a new state function, entropy, defined by $dS = dq_{\text{rev}}/T$ . The thermodynamic temperature built here is exactly the $T$ that appears in that definition, so this scale is the essential prerequisite for entropy.

The Thermodynamic Temperature Scale

The physical picture

The core relations

The core relations

Worked examples

Limiting cases & common traps

The chemist behind it

Before this

Where this leads