The freezing point of pure water at standard pressure (1 atm) is 0°C (32°F or 273.15 K).
Explanation
Water freezes at 0°C under 1 atmosphere of pressure. That temperature can change if pressure or dissolved solutes change — common effects are freezing-point depression by solutes and slight melting-point change with pressure. The triple point of water (where solid, liquid, and vapor coexist) is at $0.01^\circ\text{C}$ and $611.657\text{ Pa}$.
Key equations
- Celsius ↔ Fahrenheit:
$T(^{\circ}\text{F})=\dfrac{9}{5}T(^{\circ}\text{C})+32$
- Celsius ↔ Kelvin:
$T(\text{K})=T(^{\circ}\text{C})+273.15$
- Freezing-point depression (colligative property) for a dilute solution:
$$\Delta T_f = i\,K_f\,m$$
where $i$ is the van ‘t Hoff factor (number of particles per formula unit), $K_f$ is the cryoscopic constant ($1.86^\circ\text{C}\cdot\text{kg}\cdot\text{mol}^{-1}$ for water), and $m$ is the molality (mol solute per kg solvent).
- Pressure dependence of melting/freezing (Clausius–Clapeyron form):
$$\frac{dP}{dT}=\frac{\Delta H_{\text{fus}}}{T\,\Delta V}$$
For ice/water this gives an approximate slope of about $-0.0074^\circ\text{C}$ per atm (melting point falls slightly with increasing pressure).
Examples (step-by-step)
1) Convert the freezing point to Fahrenheit and Kelvin.
- Celsius value: $0^\circ\text{C}$.
- Fahrenheit:
$$T(^{\circ}\text{F})=\frac{9}{5}\cdot 0 + 32 = 32^\circ\text{F}$$
- Kelvin:
$$T(\text{K})=0 + 273.15 = 273.15\text{ K}$$
2) Freezing-point depression for 1.00 molal NaCl (approximate; NaCl $\to$ Na$^+$ + Cl$^-$ so $i\approx 2$).
- Use $\Delta T_f = i K_f m$:
$$\Delta T_f = 2 \times 1.86^\circ\text{C}\cdot\text{kg}\cdot\text{mol}^{-1} \times 1.00\text{ m} = 3.72^\circ\text{C}$$
- New freezing point:
$$0^\circ\text{C} – 3.72^\circ\text{C} = -3.72^\circ\text{C}$$
(Real solutions deviate from ideality; $i$ and activity effects adjust the result.)
3) Pressure effect (qualitative):
- Increasing pressure by 100 atm would lower freezing point by roughly $100 \times 0.0074^\circ\text{C} \approx 0.74^\circ\text{C}$ (approximate).
Important notes
- Pure water can supercool below $0^\circ\text{C}$ and remain liquid if undisturbed or free of nucleation sites.
- Impurities (salts, alcohols, sugars) lower the freezing point; this is why salt melts ice on roads.
- The triple point at $0.01^\circ\text{C}$ is used to define the Kelvin temperature scale precisely.
If you want, I can: (a) compute freezing-point depression for a specific solution (type and concentration), (b) show a derivation of the Clausius–Clapeyron relation for solid–liquid equilibrium, or (c) give experimental examples of supercooling. Which would you like?