Is there an easy way to get water at roughly 70°C in our kitchen? Some green tea requires to pour water at 70°C. I have no thermal sensor or kettle with adjustable temperature with me. 
Do you know a way to get water at roughly 70°C like “boil water and wait for x minutes” or “mix x part of boiling water with 1-x part of fresh water” ?
 A: "Wait for $x$ minutes" is difficult because the cooling rate will depend on a lot of non-universal details (room temperature, the surface of the water, the material the pan is made of and its thickness).
However, there are two temperatures which are easy to attain in your kitchen. $100\,^\circ\mathrm C$ (don't need to explain this) and $0\,^\circ\mathrm C$ - for this, you can let ice cubes sit on the counter until they melt. Stir/mix the water/ice mixture and as long as they are some ice cubes left, you're guaranteed that it is at $0\,^\circ\mathrm C$.
Now the heat capacity of water stays almost constant if you go from $0\,^\circ\mathrm C$ to $100\,^\circ\mathrm C$, see for example this table - fifth column:
http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html 
Informally speaking, this means that you need the same amount of energy to get a kilogram of water from $2\,^\circ\mathrm C$ to $3\,^\circ\mathrm C$ as you need to get it from $92\,^\circ\mathrm C$ to $93\,^\circ\mathrm C$.
Because of this property, you can mix $30\,\%$ of the $0\,^\circ\mathrm C$-water and $70\,\%$ of the $100\,^\circ\mathrm C$-water and you'll end up with a liquid that is at $70\,^\circ\mathrm C$, up to a small error.
For liquids that have a heat capacity that varies strongly with temperature, you can do the same thing but the calculation is more difficult.
A: Assumptions: 


*

*we have a volume of water $(v_1)$ at $100^{\circ} C$ (boiling water)

*we have a volume of water $(v_2)$ at $20^{\circ} C$ (tap water)

*we want a volume of water $(v_1+v_2)$ at $70^{\circ} C$

*the pot that will finally hold the $70^{\circ} C$ water will consume a negligible amount of energy from the water (generally a bad assumption. Might have to account for an addition $5^{\circ} C$ of temperature loss)


The energy $(Q)$ required to heat a body of water is given by:$$ Q = m \cdot c \cdot \Delta T$$
The mass of water is the same as volume. When mixed, volume $1$ of water will need to increase by $50^{\circ} C$ and volume $2$ of water will decrease by $30^{\circ} C$. Using our assumption of no energy loss we can setup the equation $$Q_1 = Q_2$$
$$v_1 \cdot c (\text{Water}) \cdot 50 = v_2 \cdot c (\text{Water}) \cdot 30$$
Our ratios of tap water to boiling water will be: $$\frac{v_1}{v_2} = \frac{3}{5}~.$$ e.g., $1~$ L of boiling water needs $600~$ml of tap water and will result in $1.6 ~$L of $70^{\circ} C$ water.
A: You can start with water at $0^{\circ} C$, as suggested in Vibert's answer, measure the height of water in the container, and start heating the water. The density of water at $0^{\circ} C$ is $1 g \cdot cm^{-3}$, at $65^{\circ} C$ it is $0.981 g \cdot cm^{-3}$, at $70^{\circ} C$ it is $0.978 g \cdot cm^{-3}$. So if at $0^{\circ} C$ the height of water in the container is 10 cm, then the difference in heights of water between $65^{\circ} C$ and $70^{\circ} C$ will be approximately $0.3 mm$ and can be detected using a ruler (or a beam compass). Thermal expansion of a solid container is typically much less than that of water. I would think that not much water will evaporate when the water is heated from $0^{\circ} C$ to $70^{\circ} C$, but maybe this issue should be studied in more detail.
One could say that what I suggest uses the principle of an ordinary thermometer, but I am not sure this is worse than the calorimetric principle.
