Is there a formula that shows how the density of water is affected by temperature?

For example, with the speed of sound it can be shown how temperature will affect the speed:

$$V = 331\ \mathrm{m/s} + \left(0.6\ \mathrm{\frac{m/s}{°C}}\right)T$$

Is there a similar formula that relates the density of a liquid to temperature? (in this case, water)


In general, we can derive the expression for the density of any substance as a function of temperature using the definition of thermal expansion.

$$\frac{\mathrm dV}{\mathrm dT}=\alpha V\tag1$$

By differentiating $\rho=m/V$, we can show that $\mathrm d\rho/\rho = -\mathrm dV/V$. So using this and equation $\text{(1)}$,

$$\frac{\mathrm d\rho}{\mathrm dT}=-\alpha V\tag2$$

Integrating this above expression and approximating the expression,

$$\rho = \rho^\circ\cdot\left(1-\alpha\Delta T\right)\tag3$$

where $\rho^\circ$ is the initial density at the initial temperature, $\alpha$ is the volumetric thermal expansion coefficient and $\Delta T$ is the temperature.
For water, $\alpha$ is typically $207\times10^{-6}\ \mathrm{K^{-1}}$ at $20\ \mathrm{^\circ C}$. Using this, we can observe how temperature affects the density of water.

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  • $\begingroup$ Good answer. It's worth mentioning that $\alpha$ is a function of temperature, which gives rise to nonlinear behavior. For example, liquid water (at atmospheric pressure) has a density maximum at 4 °C, not 0 °C. $\endgroup$ – Curt F. Feb 8 '15 at 14:08
  • $\begingroup$ Equation 2 is incorrect (V should be $\rho$). This leads to an exponential function as solution. Equation 3 is a linearized version for small $\Delta T$ $\endgroup$ – Buck Thorn Oct 26 '19 at 8:24

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