If I take a 1 watt heating element, put it in a glass of water and I put them both inside a sealed imaginary chamber that does not conduct heat outside (again, imaginary).
Will the water eventually boil?
If not, I don't understand why, since electricity keeps flowing through the heating element, generating more and more energy in joules, and since temperature is just an increase in joules per kilogram, in an ideally sealed chamber, heat would accumulate and slowly raise the temperature.
In the scenario you describe, with no heat loss to the environment (perfect insulation), the temperature of the water will rise without limit so long as you keep adding energy to the system, which in the scenario is at a rate of 1 W. Given enough time (and assuming your container doesn't break down) this will exceed the temperature of the sun. At some point there will probably be one or more phase changes, and the point at which these occur will depend on the pressure. You don't indicate if this is kept at atmospheric pressure (isobaric) or if the pressure is allowed to rise and the boundary is fixed (isochoric). This is relevant as water only boils at 100°C at sea level atmospheric pressure.
The temperature of the element is irrelevant because if you keep pumping energy into it, it will also continue to rise as the temperature of the water rises. The element and water system will never be in thermodynamic equilibrium so long as you continue to pump in energy. The relative temperatures of the element and water at any given moment will depend on the thermal properties of both materials.
Yes, if no heat is lost then the water will ultimately reach the same temperature as the heating element — presumably more than $100\,^\circ\mathrm C$.
Water has a heat capacity of $4200 \,\mathrm{J/(kg\,^\circ C)}$, so a $250 \,\mathrm{cm^3}$ glass of water being heated by $1 \,\mathrm W$ (i.e., $1 \,\mathrm{J/s}$) will heat it at a rate of
$$\frac{1\,\mathrm{J/s}}{0.25\,\mathrm{kg} \cdot 4200\,\mathrm{ J/(kg\,^\circ C)}} \approx 1/1000\,{^\circ}\mathrm{C/s}$$
You specified a perfect insulator, but it's fun to see what happens when we do allow for cooling.
We can imagine a cube of water with sides of length $l$.
A perfect black body will radiate heat at a rate
$P = \sigma T^4 A$ .
Real materials aren't perfect black bodies, so we need to multiply our result by a fudge factor called emissivity, which is about 0.96 for water.
Set power $P$ to our 1W supply.
Area $A = 6l^2$ where $l$ is the side length of our cube.
Temperature $T \approx 373K$ for boiling water.
$l = \sqrt{\frac{1}{\sigma T^4*6*0.96}} \approx 12$cm
Ignoring all other heat transfer, our water cube will boil if smaller than roughly 12cm on each side.
If you put a 1W heating element in a glass of water, it would not boil, because the glass with water would be cooled by air and radiate heat out. Once the temperature stabilizes at a certain level, the heat input from the element would be equal to the heat loss to the environment. If you increase the power of the element, the temperature would increase accordingly. Finally, at some specific power level, the temperature would reach the boiling point. This definitely would not happen at 1W and not likely at 10W. The exact power level required for boiling depends on a number of factors and would probably be closer to 100W, give or take generously.
To boil a glass of water with a 1W element, you'd need to prevent the heat loss. The easiest way to do so would be to put the element inside a thermos. If the thermos is high quality (double mirror walls with a vacuum in between and an energy efficient lid), then you should be able to boil water with a 10W element and, if the thermos is very good and efficiently closed, then potentially even with a 1W element.
