This can be computed for small changes in the pressure by considering the partial derivative of the temperature w.r.t. pressure at constant entropy. If we suddenly raise the pressure a bit, then this is to a good approximation an isentropic process as no heat is exchanged and it is not a violent process causing large irreversible effects. So, we want to evaluate:
$$\left(\frac{\partial T}{\partial P}\right)_S \text{ (1)}$$
To express this in terms of the readily measurable quantities (such has heat capcity, thermal expansion coefficient, compressibility etc.) we need to get rid of the entropy in this expression. A direct application of a Maxwell relation will not work because then you end up getting the entropy inside the parial derivative. Instead, we consider dS expressed in terms of dP and dT:
$$dS = \left(\frac{\partial S}{\partial T}\right)_P dT + \left(\frac{\partial S}{\partial P}\right)_T dP$$
The derivative (1) is the ratio of dT and dP at constant S, so we can obtain this by putting dS = 0 and solving for that ratio. This yields:
$$\left(\frac{\partial T}{\partial P}\right)_S = -\frac{\left(\frac{\partial S}{\partial P }\right)_T}{\left(\frac{\partial S}{\partial T }\right)_P}$$
In the denominator we recognize the heat capacity at constant pressure, T dS at constant P is the heat supplied to the system at constant pressure, therefore
$$\left(\frac{\partial S}{\partial T }\right)_P = \frac{C_P}{T}$$
The numerator can be simplified by using an appropriate Maxwell relation. This works as follows, you start with the fundametal thermodynamic relation:
$$dE = T dS - P dV$$
We can read off from this that the temperature is the partial derivative of the internal energy w.r.t. S at constant volume. Also the pressure is minus the partial derivative of the internal energy w.r.t. volume at constant entropy. Now, the second derivative of the internal energy w.r.t. S and V can then be computed by differentiating the temperature w.r.t. the volume at constant S or by differentiating minus the pressure w.r.t. S at constant V. The symmetry of partial derivatives implies that the two quantities are equal, such an equality is called a Maxwell relation.
To rewrite the partial derivative in the numerator, we manipulate the fundamental thermodynamic identity using Leibnitz rule for the differential of a product until we get an expression involving S and dP:
$$\begin{split}
&dE = T dS - P dV = d(TS) - S dT - d(PV) + V dP\Longrightarrow\\
&dG = -S dT + V dP
\end{split}$$
where $G = E - T S + P V$ is the Gibbs energy. The Maxwell relation derived from G is then:
$$\left(\frac{\partial S}{\partial P}\right)_T=-\left(\frac{\partial V}{\partial T}\right)_P$$
The thermal expansion coefficient $\alpha$ is defined as:
$$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P $$
Therefore the result is:
$$\left(\frac{\partial T}{\partial P}\right)_S = \frac{T V\alpha}{C_P}$$
In terms of the specific heat capacity (per unit mass) $c_P$ this is:
$$\left(\frac{\partial T}{\partial P}\right)_S = \frac{T\alpha}{\rho c_P}$$
where $\rho$ is the density. For water at 20 C this is about $1.45\times 10^{-8}\frac{K}{\text{Pa}}$, so 1000 bar pressure will raise the temperature by about 1.45 C.
