Law for tap water temperature I was wondering if anyone put together a law to describe the rising temperature of the water coming out of a tap.
The setup is fairly simple: there's a water tank at temperature T, a metal tube of length L connected to it and a tap at the end where temperature is measured. The water flows at P l/s.
Given that the metal tube is at room temperature initially, what law describes the temperature of the water at any instant? What is the limit temperature of the water?
Thanks.
 A: We can consider the following model: a tube of constant temperature $T_e$ of lenght L, radius $r$ where water is flowing uniformly at a speed $v$ (that you can obtain from your flow $P$).
A "slice" of water travels an interval $dx$ in a duration $dt = \frac{dx}{v}$.
The tube will contribute to the "heating" of the water by $\frac{dQ}{dt} = (T-T_e) k 2 \pi r dx$ where $k$ is the conductivity and where we use a very simple model (in particular for the radius, we do not distinguish external and internal radii).
During this interval the temperature $T(x)$ of the water will vary by $dT = -\frac{dQ}{c \rho dV}$ where $C$ is the heat capacity at constant pressure of water, and where $dV = 2 \pi r dx$.
Replacing we have $\frac{dT}{T-T_e}=-\frac{k}{\rho C v} dx$ whose solution, if the temperature in the tank (ie x = 0) is $T_t$ :
$T(x) = (T_t - T_e) e^{(-\alpha x)}+T_e$ where $\alpha = \frac{k}{\rho C v}$.
Depending on the lenght of the tube you have the temperature at the tap.
A: The answer is going to depend on the heat transfer coefficient between the tube and the surrounding room (unless you specify that the tube is held at constant temperature), the heat transfer coefficient between the tube and water, the outside & inside diameter of the tube, and the length of the tube. This is a moderately involved heat transfer problem, unless additional constraints are provided to simplify it.   
An excellent resource for understanding the mathematics behind this sort of heat transfer problem can be found over here. It's very similar to the solution posted above by @Cedric, but may be a little easier for some to follow.
