Above every liquid sits its own vapour, pressing down with a pressure that depends on one thing only: temperature. When that vapour pressure reaches the pressure of the surroundings, the liquid boils. So “water boils at 100 °C” is a statement about the atmosphere as much as about water. Drop the surroundings to 10 kPa and the same water boils in the forties — which is precisely how a vacuum deaerator strips oxygen out of feedwater without a burner, and how a condenser pulls the last work out of a turbine.

The handbook fit is Antoine: $\log_{10} P = A - \dfrac{B}{C + T}$. $P$ is the vapour pressure and $T$ the liquid temperature; $A$, $B$ and $C$ are three constants regressed for one substance over one temperature range. Here is the part that catches people: the constants carry their own units. The published set for water — $A = 8.07131$, $B = 1730.63$, $C = 233.426$ — is quoted for pressure in millimetres of mercury and temperature in degrees CELSIUS. Feed it kelvin, or read the answer as kilopascals, and the arithmetic is flawless and the answer is nonsense. One mmHg is 0.1333 kPa. Rearranged for temperature: $T = \dfrac{B}{A - \log_{10} P} - C$.

The physics-first alternative is Clausius–Clapeyron in its two-point form: $\ln\!\left(\dfrac{P_2}{P_1}\right) = -\dfrac{\Delta H_{vap}}{R}\left(\dfrac{1}{T_2} - \dfrac{1}{T_1}\right)$. Subscripts again: 1 is the point you know, 2 is the point you want, and either may be the hotter. $P_1$ and $P_2$ are vapour pressures in any consistent unit, because they enter as a ratio and their units cancel. $T_1$ and $T_2$ are absolute temperatures in kelvin, and they must be, because they enter as $1/T$ — an offset inside a reciprocal is a far worse crime than an offset anywhere else. $\Delta H_{vap}$ is the molar enthalpy of vaporisation in J/mol — 40 700 for water — and $R$ is the same 8.314 J/(mol·K) as ever.

Both relations are exponential in temperature, so estimates on them must be made by anchoring rather than by interpolating. Carry two numbers: water is one atmosphere at 100 °C and about half an atmosphere at 80 °C. Twenty degrees, half the pressure. That is the shape of the curve, and it will catch an inverted ratio before your calculator does.