# Can a 1 watt heating element boil water? [closed]

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.

• Just for reference : your imaginary chamber is adiabatic. – Eric Duminil Jun 12 '18 at 11:58
• At one point your heating element will get too hot and melt, breaking the circuit. It would happen after the water boils, though — the melting point of tungsten is very high compared to the boiling point of water. – Anurag B. Jun 12 '18 at 12:16
• Funny, kind-of-related: what-if.xkcd.com/35 – Daan van Hoek Jun 13 '18 at 7:04
• Or just consider shaking a perfectly insulated thermos flask half-filled with water. – Count Iblis Jun 13 '18 at 11:09
• Not clear. Why do you think it will not boil? What is your difficulty? – sammy gerbil Jun 13 '18 at 18:44

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.

• To add to this, and please correct me if I'm wrong, it will become increasingly "difficult" to continue achieving 1 watt power input as the system temperature rises. For example, using an electric-resistive heating element, the increased temperature will increase resistance and thus require more voltage for the same power. So the "pump" (generator) putting energy in will need to keep getting transformed up to a higher voltage, similar to a car shifting up gears to keep up with the increasing speed of the wheels. Just a practical note (like "the heating element will melt" lol) – jnez71 Jun 12 '18 at 15:18
• @jnez71: It's not hard to make an electric heating element work just as well at +1000C as at -40C; the only difficulty (and it's generally only a problem at more extreme temperatures than those) is ensuring that the device won't melt and won't suffer damage from differential expansion and contraction. This is one of the advantages of electric heating for industrial applications. – supercat Jun 12 '18 at 19:54
• @supercat That's fair. The effect I described does happen (and shows itself in different ways across all constant-power-delivery problems) but as far as electrical heating goes, the added thermal resistance does not really limit practicality over a huge range of temperatures. I calculated that a Tungsten filament's resistance would only increase by a factor of 16 when heated from 0C to its notoriously high melting point, which means that one's voltage supply would only have to step up a factor of 4 to keep power constant. Melting / thermal fracture is far more limiting. – jnez71 Jun 12 '18 at 20:32
• @jnez71: Yup. And if the voltage goes up by a factor of four, the current required to hold that voltage will also go down by a factor of four. – supercat Jun 12 '18 at 20:41
• @jnez71 Some materials (eg metals) will increase resistance with temperature. Others (like semiconductors) work exactly the other way around, that's why electronics suffer from thermal runaway because the process has positive feedback. It's called "negative temperature coefficient". – Agent_L Jun 13 '18 at 10:36

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.

• "Ignoring all other heat transfer". It's a bit too simplistic. Convection and conduction will play a big role too, probably larger than your radiation losses. – Eric Duminil Jun 12 '18 at 11:59
• @EricDuminil But it is a nice little mental exercise. So if you are in space waiting for your ride and want to boil a cup of tea while you are waiting with the 2.5 W of the USB 2.0 port in your space suit you know ... oh wait ;-). – Peter - Reinstate Monica Jun 12 '18 at 16:34
• @PeterA.Schneider In that case, just do nothing, the water will boil on its own if there is no conduction in that situation (and thus no air, so no pressure) – Ferrybig Jun 13 '18 at 8:27
• @Ferrybig That was my "oh, wait" ;-). I'm afraid the tea will not be good; you'll be left with steamed leaves at a few K, and need a little food air lock. – Peter - Reinstate Monica Jun 13 '18 at 8:56

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.

• Yes, but my question is about an ideal imaginary chamber, one that does not allow heat to "leak out" – physicsnewbie Jun 12 '18 at 3:57
• @physicsnewbie Sure, it is essentially the same as a "very good thermos". My answer simply also includes a more general and realistic situation for completeness. In an ideal chamber the temperature would steadily rise until boiling at the rate calculated in Martin's answer or in about 24 hours starting from the room temperature. However, unlike he stated, the temperature of the heating element is irrelevant, because it would increase over time as well. – safesphere Jun 12 '18 at 4:06
• If you use a candle with a specific temperature, then, once water reaches this temperature, the heat transfer would stop. This would be a result of the fact that the candle is outside of your sealed chamber. Therefore, the power transfer from the element (candle) to the chamber would stop (drop from 1W to zero). This would violate the conditions of your question, specifically, a constant 1W heat transfer. To satisfy this condition you could use a hotter candle, but not necessarily. You could just use 1-Ohm resistor powered by 1 Volt and its temperature would simply increase with that of water. – safesphere Jun 12 '18 at 5:57
• @physicsnewbie As for candles, physics.stackexchange.com/questions/272893/… In short, you need a lot more magic to do this with a candle than with an electric heating element. If you could supply oxygen to the candle at the same temperature as the rest of the system, it could reach higher temperatures - it's just impossible in reality, since the carbon dioxide carries away a lot of heat, while the fresh oxygen will be rather cold (unless you pre-heat it - which means a lot of the heat no longer comes frmo the candle). – Luaan Jun 12 '18 at 6:02
• If hypothetically the candle is inside, as you describe, it's temperature would be th sum of its ambient temperature (of water) and the temperature of burning. So it would be burning hotter and hotter over time. In other words, the heat energy of the fuel burning would increase the temperature of the candle from its current temperature. The key here is the energy released by burning while the resulting temperature is a consequence of this energy added and thus the burning temperature of the candle would not be constant, but increase over time. – safesphere Jun 12 '18 at 6:03