Temperature Scales and Conversions

Temperature tells us how hot or cold something is. But the number we get depends entirely on which scale we choose. Celsius, Kelvin, and Fahrenheit are three different rulers laid against the same physical property. Each has its own zero point and its own step size. Converting between them is a routine but essential skill. It is closely related to the reasoning used in Dimensional Analysis and Unit Conversion. This article lays out the three scales. It explains why Kelvin is the only one tied directly to a physical minimum. It also gives the exact conversion formulas along with fully worked, checkable examples. Temperature conversions involve an offset as well as a scaling factor. Because of this, they behave differently from the simple multiplicative conversions used for length or mass. This is a distinction worth understanding carefully rather than memorizing blindly.

Key takeaways

Celsius (°C), Kelvin (K), and Fahrenheit (°F) are three temperature scales with different zero points and different degree sizes.
Kelvin is the SI base unit for thermodynamic temperature and starts at absolute zero, the coldest physically possible temperature.
A Celsius degree and a Kelvin equal the same size of temperature step; only the zero point shifts, so $T(\text{K}) = T(^\circ\text{C}) + 273.15$ .
Fahrenheit degrees are smaller than Celsius degrees, so converting between °C and °F needs both a scaling factor (9/5 or 5/9) and an offset (32).
Temperature conversions are affine (linear-plus-offset), unlike ordinary unit conversions, which are purely multiplicative.
Key idea: every temperature scale is a straight-line function of every other one, so converting is just solving $y = mx + b$ for the scale you want.

The core insight

Temperature scales differ in where they put zero and how big they make one degree. Kelvin anchors both to physics (absolute zero, and a degree size matching Celsius); Fahrenheit anchors both to historical convention. Every conversion is a straight line: multiply by a slope, then shift by an intercept.

The physical picture

Temperature is a measure of the average kinetic energy of the particles in a substance — this connects the temperature scales to the atomic-scale reasoning developed in SI Units and the Language of Measurement. But the number we assign to a given physical state depends on an arbitrary choice of reference points. Historically, different scientists picked different fixed points — the freezing and boiling points of water, or the coldest temperature a particular salt-ice mixture could reach — and divided the interval between them into a chosen number of degrees. Three scales dominate scientific and everyday use today:

Celsius (°C): defined so that water freezes at 0 °C and boils at 100 °C (at 1 atm pressure). This gives 100 equal degree-steps between the two reference points.
Fahrenheit (°F): defined historically so that water freezes at 32 °F and boils at 212 °F, an interval of 180 degree-steps. Fahrenheit degrees are therefore smaller than Celsius degrees.
Kelvin (K): the SI base unit of thermodynamic temperature. It uses the same degree size as Celsius, but sets its zero at absolute zero — the theoretical point at which particle motion contributes the minimum possible thermal energy, roughly $-273.15\ ^\circ\text{C}$ .

Because Kelvin and Celsius share a degree size, converting between them is a pure shift — no rescaling needed. Fahrenheit, having both a different zero and a different degree size, requires both a shift and a rescale.

Figure

Three scales lined up at water's freezing and boiling points

The core relations

T(\text{K}) = T(^\circ\text{C}) + 273.15

where: $T(\text{K})$ is the Kelvin temperature, $T(^\circ\text{C})$ is the Celsius temperature, and $273.15$ is the exact offset between the two zero points (no rescaling — one kelvin equals one Celsius degree in size).

T(^\circ\text{F}) = \frac{9}{5}\,T(^\circ\text{C}) + 32

where: $T(^\circ\text{F})$ is the Fahrenheit temperature, $\frac{9}{5}$ converts the size of a Celsius degree into Fahrenheit degrees (180 Fahrenheit degrees span the same interval as 100 Celsius degrees), and $32$ shifts the zero point to match water's freezing point in Fahrenheit.

T(^\circ\text{C}) = \frac{5}{9}\left[T(^\circ\text{F}) - 32\right]

where: this is simply the previous equation solved for Celsius — subtract the offset first, then undo the scaling.

T(\text{K}) = \frac{5}{9}\left[T(^\circ\text{F}) - 32\right] + 273.15

where: this composes the Fahrenheit-to-Celsius conversion with the Celsius-to-Kelvin shift, letting you go straight from Fahrenheit to Kelvin.

Derivation ABuilding the Fahrenheit–Celsius formula from two fixed points

Goal: derive the linear relationship between °F and °C using only the two shared physical reference points — water's freezing and boiling temperatures — without assuming the formula in advance.

Step 1. Identify two points that both scales agree describe the same physical state. Water freezes at $0\,^\circ\text{C}$ and at $32\,^\circ\text{F}$ . Water boils at $100\,^\circ\text{C}$ and at $212\,^\circ\text{F}$ (both at 1 atm).

Step 2. Since both scales are linear (equal degree-steps throughout), the relationship must have the form

T(^\circ\text{F}) = m \cdot T(^\circ\text{C}) + b

where $m$ is the slope (degrees-F per degree-C) and $b$ is the Fahrenheit value when Celsius is zero.

Step 3. Use the freezing point to find $b$ directly. Substituting $T(^\circ\text{C}) = 0$ and $T(^\circ\text{F}) = 32$ gives

32 = m \cdot 0 + b \quad\Rightarrow\quad b = 32.

Step 4. Use the boiling point to find $m$ . Substituting $T(^\circ\text{C}) = 100$ and $T(^\circ\text{F}) = 212$ gives

212 = m \cdot 100 + 32.

Step 5. Solve for $m$ :

m = \frac{212 - 32}{100} = \frac{180}{100} = \frac{9}{5}.

Step 6. Assemble the full formula:

T(^\circ\text{F}) = \frac{9}{5}\,T(^\circ\text{C}) + 32.

Step 7. Solve this for $T(^\circ\text{C})$ by isolating it algebraically: subtract 32 from both sides, then multiply by the reciprocal slope $5/9$ ,

T(^\circ\text{C}) = \frac{5}{9}\left[T(^\circ\text{F}) - 32\right].

This is exactly the inverse function of the line found in Step 6, confirming the two formulas are self-consistent.

Worked examples

Example 1 — A cold day in Fahrenheit, converted to Celsius and Kelvin. Suppose a weather report gives the outdoor temperature as $23\,^\circ\text{F}$ . Find the temperature in Celsius and Kelvin. Convert to Celsius using the inverse formula:

T(^\circ\text{C}) = \frac{5}{9}\left[23 - 32\right] = \frac{5}{9}(-9) = -5.0\,^\circ\text{C}.

Now shift to Kelvin:

T(\text{K}) = -5.0 + 273.15 = 268.15\ \text{K}.

Check by going directly from Fahrenheit to Kelvin using the combined formula:

T(\text{K}) = \frac{5}{9}(23 - 32) + 273.15 = -5.0 + 273.15 = 268.15\ \text{K}.

Both routes agree, as they must, since the combined formula is just the two individual steps chained together.

Example 2 — A laboratory heating experiment reported in Celsius, converted to Fahrenheit. A student heats a beaker of solution from $18.4\,^\circ\text{C}$ to $76.9\,^\circ\text{C}$ and wants both temperatures reported in Fahrenheit for a lab partner using U.S. units. Starting temperature:

T(^\circ\text{F}) = \frac{9}{5}(18.4) + 32 = 33.12 + 32 = 65.1\,^\circ\text{F}.

Final temperature:

T(^\circ\text{F}) = \frac{9}{5}(76.9) + 32 = 138.42 + 32 = 170.4\,^\circ\text{F}.

The temperature change in Fahrenheit degrees is

170.4 - 65.1 = 105.3\,^\circ\text{F},

while the change in Celsius degrees was $76.9 - 18.4 = 58.5\,^\circ\text{C}$ . Checking the ratio,

\frac{105.3}{58.5} = 1.80 = \frac{9}{5},

confirms the scaling factor correctly relates changes in temperature, even though the offset of 32 cancels out when only a difference is needed.

Limiting cases & common traps

Situation	What happens	Why
Converting a temperature difference (ΔT), not an absolute temperature	Use only the $9/5$ or $5/9$ factor — drop the $+32$ or $+273.15$	Offsets cancel when subtracting two temperatures on the same scale
$T(^\circ\text{C}) = 0$	$T(\text{K}) = 273.15$ , $T(^\circ\text{F}) = 32$	Celsius zero is water's freezing point, not absolute zero
$T(\text{K}) \to 0$	$T(^\circ\text{C}) \to -273.15$	Kelvin's zero is absolute zero; Celsius cannot go lower than $-273.15\,^\circ\text{C}$
$T(^\circ\text{C}) = T(^\circ\text{F})$	Both equal $-40$	The one point where the two linear scales cross: solving $\frac{9}{5}x + 32 = x$ gives $x = -40$

The chemist behind it

The modern temperature scale story is really three interwoven stories. Daniel Gabriel Fahrenheit, an instrument maker, built the first reliable mercury thermometer in the early 1700s and calibrated it using a brine mixture (his zero) and human body temperature (originally near 96 °F on his scale, later refined). Anders Celsius proposed a centigrade scale in 1742, originally with 0° as boiling and 100° as freezing — the scale we use today is actually the reversed version, adopted shortly after his death. William Thomson, who became Lord Kelvin, recognized in the 1840s that thermodynamic reasoning demanded an absolute zero, a temperature below which nothing could exist, and built a scale anchored there rather than to any particular substance's behavior. His scale is why the SI system chose Kelvin, not Celsius, as the fundamental unit of temperature.

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

William Thomson (Lord Kelvin)

1824–1907 · Established the absolute temperature scale from thermodynamic principles, showing that a lower limit to temperature exists independent of any specific substance, unlike scales tied to water's freezing or boiling points.

Ask the tutor: If a chemical reaction's rate doubles for every 10 °C rise in temperature, does it also double for every 10 K rise — and why does the answer depend on whether you're tracking an absolute temperature or a temperature change?

Why does Kelvin have no negative values, but Celsius and Fahrenheit do?

Kelvin's zero point is set at absolute zero, the theoretical lower limit of thermal energy in any physical system. Since nothing can be colder than absolute zero, there are no negative Kelvin temperatures in standard thermodynamics. Celsius and Fahrenheit instead place zero at arbitrary reference points (water's freezing point, and a historical brine mixture, respectively), so both scales freely extend below zero.

Why do the freezing and boiling points of water not have to be exactly 0 and 100 on the Kelvin scale?

Kelvin's zero is defined by absolute zero, not by any property of water. Water's freezing point happens to land at $273.15\ \text{K}$ and its boiling point at $373.15\ \text{K}$ under standard atmospheric pressure, purely because the Celsius scale (which does use water's freezing point as its reference) is shifted by exactly 273.15 to produce Kelvin.

Is it correct to write "273.15 K" as "273.15 °K" or "273.15 degrees Kelvin"?

No. By SI convention the kelvin is written without a degree symbol and without the word "degrees" — simply "K", as in "273.15 K". This reflects that the kelvin is treated as an absolute unit analogous to units like the meter or second, not a degree-based scale like Celsius or Fahrenheit.

Why is the conversion between Celsius and Fahrenheit not a simple multiplication, unlike converting between meters and centimeters?

Ordinary unit conversions (length, mass) share a common zero point — zero meters is the same physical state as zero centimeters — so multiplying by a conversion factor suffices, as discussed in Dimensional Analysis and Unit Conversion. Celsius and Fahrenheit have different zero points as well as different degree sizes, so the conversion needs both a multiplicative slope and an additive offset, making it an affine transformation rather than a pure scaling.

How would this look if a fourth, hypothetical linear scale were invented?

Any new linear scale would still be describable as $T(\text{new}) = m\cdot T(^\circ\text{C}) + b$ , with $m$ set by comparing degree sizes between two shared reference points and $b$ set by matching where the new scale reads zero. Every conversion in this article follows exactly this two-parameter pattern — the details differ, but the method (pick two reference points, solve for slope and intercept) is completely general.