Water mixture at different temperatures Let's say that we mix homegeneously and instantly cold water at $t^\circ $ C and hot water at $T^\circ$ C (like in a water tap) in ratio $p:1$. My question is the following: What is the instant temperature of this mixture? Is there a law in that sense? 
I am a mathematician, not a physicist;all I know about that is Newton's cooling law. However I can't see how to apply it here.
 A: The solution to your problem can be found by applying the law of conservation of mass and (thermal) energy.
Assuming the two water volumes at different temperatures mix instantaneously and there is no heat loss to the surroundings, applying conservation of energy yields:
$$m_1 c_{p,1} T_1 + m_2 c_{p,2} T_2 = m_f c_{p,f} T_f$$
where $m$ is the mass of the volume of water, $c_p$ is the specific heat capacity and $T$ the temperature. The indices must be understood as $f$ is the final mixed state and $1,2$ are the respective initial states of the water volumes.
Applying conservation of mass yields:
$$m_1 + m_2 = m_f$$
Substituting in the energy equation and rearranging for $T_f$ yields:
$$T_f = \frac{m_1 c_{p,1} T_1 + m_2 c_{p,2} T_2}{(m_1 + m_2)c_{p,f}}$$
Now generally, the specific heat capacities are temperature dependent but over the range of temperatures that generally come out of the tap we can safely assume it a constant which simplifies our equation to:
$$T_f = \frac{m_1 T_1 + m_2 T_2}{m_1 + m_2}$$
For your case we have $m_1=p$, $m_2=1$, $m_f=p+1$, $T_1=t$ and $T_2=T$ so:
$$T_f = \frac{p t + T}{p + 1}=t+\frac{1}{p+1}(T-t)$$
which is exactly the equation given by Chester in the other answer.
As a validation case, if $p=1$, i.e. a 1:1 ratio, then:
$$T_f = \frac{t+T}{2}$$
and the final temperature is the average of the temperature of the two volumes of water as expected.
The nice thing about approaching it this way is you can now modify the equation to account for any number of bodies at different temperatures.
A: If you mix cold water at temperature $T_c$ and hot water at a temperature $T_h$ in a ratio $x:1$ then the mixture is going to end up at a temperature:
$$ T_{mix} = T_c + \frac{1}{x+1}(T_h - T_c) $$
If we graphed the temperature as a function of time we'd get something like:

I've represented the part of the graph where the liquids are mixing by a cloud because during this time the mixture won't be homogeneous. There will be regions of different temperatures that depend on the exact mixing conditions. Under these circumstances the inhomogeneous mixture doesn't have a single temperature so there is no single figure we can plot on our graph.
Now what you are suggesting is that we make the mixing instantaneous i.e. reduce the mixing time to zero, and in that case our graph looks like:

But now we have a discontinuity in the temperature at the moment of mixing, and that means at the moment of mixing the temperature is not defined. So the answer to your question is that the temperature is not defined at the moment of mixing.
Newton's law of cooling describes how the temperature of an undisturbed system changes with time. It isn't applicable in this context.
A: The above answers are correct at normal pressure, where all is linear. At high pressure and temperature the heat capacity of water strongly depends on temperature and simple averaging no longer works.

A: If the mixture is homogeneous and instantaneous, the temperature is simply determined by the ratio p. The law is the law of mixtures. 
Newton's law of cooling is not applicable even if the instantaneous homogeneity is not assumed. That assumes heat loss form a body to the surrounding. 
