The simple answer is probably not. That's because the sun involves a special type of fire that is able to "burn" water, and so it will just get hotter and six times brighter. Here's why.

Sun's overall density is 1.4 times that of water. So if you have a bucket of water equal to the volume of the Sun, that water will have 1/1.4 = 0.71 times the mass of the sun, or 0.71 solar masses. After you pour that water (which is basically hydrogen and oxygen) onto the sun, you end up with a 1.71 solar mass star with a composition of 48 percent hydrogen, 37 percent oxygen, and 14 percent helium (with 1 percent heavier elements).

Now, will such a star burn? Yes, but not with the type of proton-proton fusion the sun uses. A star 1.7 times the mass of the sun will heat up and burn almost entirely by the carbon-nitrogen-oxygen (CNO) fusion cycle. So with CNO fusion and that mass you end up with a type F0 star with about 1.3 times the radius and 6 times the luminosity of the present Sun, and a temperature somewhat hotter than the Sun (7200 K vs. the Sun's 5800 K). It will be bluish-white, with more UV. That, along with that 6 times heat input, will fry the Earth's biosphere and probably cause the oceans to boil.

Incidentally, these F0 stars get their oomph because the CNO fusion cycle makes more light than the proton-proton fusion. The luminosity of CNO fusion is related to the star's temperature (T) to the 17th power (T^{17}), rather than just T^{4} for proton-proton stars like the sun. So that little bit of extra temperature from the water's extra mass here pushes the sun almost entirely into CNO territory, translating into a lot of extra "star power."

Advertisement

Such a star will sit on the main sequence until it burns most of its hydrogen, then expand to become a red giant, and finally shrink to being a white dwarf, just like our sun will. I can see no way for it to become a supernova—all these stars put together this way with no white dwarf to start, must be 8 times the Sun's mass, or more. This is only 1.7 times.

### What Happens Immediately After You Pour the Water

I thought I'd go ahead and answer the natural next question of whether a "bucket" of water the size of the Sun would even "bank" the Sun a little, so its light output went down for a time. And the answer to that is also "no." The tremendous gravitational energy of water falling on the Sun would heat the water to at least the temperature of the Sun's core. By the time the water (oxygen and hydrogen plasma) reached the core of the Sun (about half an hour after pouring it in), the water itself would be plasma at a temperature of 33 million K. At least.

Advertisement

Here's the math, very much simplified, as we only want the answer to order of magnitude.

The escape energy per kilogram from the Sun's "surface" (photosphere) is given by the quantity GM/R, where G is Newton's constant (6.67e-11), M is the solar mass (2e30 kg), and R is the solar radius (7e8 m). This works out to about 2e11 J/kg, or 200 billion joules per kilo. Of course, going the other direction, this is also the energy per kilogram of water (think of a one-liter water bottle) it obtains if it fell from the outer solar system to the Sun's surface. The original question didn't specify if we pour our water from way out in space or at rest, from a point suspended above the Sun's surface. You lose 2e11 J/kg if you pour it from a standing start at the surface vs. letting it fall in from far outer space.

Advertisement

This quantity is also on the same order of magnitude as the gravitational potential energy for a kilogram falling into the Sun, starting at the surface. This is harder to calculate, since the Sun has a complicated density structure and I'm not going to calculate it exactly. However, to get an estimate, we can calculate what energy we'd get if we had an object free fall to the solar core to 20 percent of its radius, which holds 1/3 of the Sun's mass. That's still GM/R, but now the M is 1/3, and the R is 1/5, so you get 5/3 times the potential at the solar surface, or 3e11 J/kg.

We didn't count the energy from accelerating through the outer 2/3rds of the Sun's mass, so this is an underestimate. And as we add another 70 percent to the Sun's mass, the additional mass will increase the binding energy of the mass that goes later, so that's underestimating it, too. But the real answer is not likely to be an order of magnitude more than 3e11 J/kg, so let's see what this amount of energy does to a kilogram of water:

First of all, the water is turned to steam and then plasma. The ionization energy of water is almost entirely due to the oxygen, and the ionization energy of oxygen to bare nuclei and electrons is 197 kJ/mole. and since a mole is 16 grams, that's 1.2e7 J/kg oxygen. This is roughly the same as for water, which is mostly oxygen by weight, and where all the ionization energy is taken up by ionizing the oxygen (oxygen is fully ionized above about 10 million K). So we have more than 20,000 times the energy we need for ionization. Ionizing the water costs us 1/20,000th of our energy budget, and we can totally neglect it. The rest goes into heating the plasma that is the product.

Advertisement

The heat capacity of plasma (constant volume) is about 3/2 R per mole of particles. For water, we have 3 atoms ionized to 3 nuclei and 10 electrons, or 13 particles. There are 1000/18 = 55.6 moles of water in a kilogram. So a kilogram of water has 55.6 x 13 = 722 moles of particles. Its heat capacity as plasma is about 3/2 * R * 722, and since R is 8.3 J/K/mole, we get 9,000 J/kg/K. Of course the process is not constant volume, and we'll get some heat as the kilogram is squeezed by a factor of 150 from the density of liquid water to the solar core, but let's ignore that for now and note that it makes us underestimate the temperature a little.

What's our final temperature? We take our energy budget of 3e11 J/kg (from gravity) and divide by our heat capacity of 9,000 J/kg/K and we get
**33.3 million K**. That's more than enough to completely ionize oxygen, and hotter than the core is now (**15.7 million K**). So even with all the approximations in favor of a lower temperature, we get a higher temperature than the sun, and there's no way this water can do anything but heat up, from the very beginning as we allow it to fall in from the solar "surface."

Advertisement

How long does this take, incidentally? Since it takes an object about 2.8 hours to orbit the sun at the surface, we can take a very rough guess that the time to fall through near vacuum into the sun core, is on the order of the orbit time divided by the square root of 32— about
**half an hour.** In as quick as half an hour, the falling gas/steam/oxygen plasma from the water reaches regions dense enough to slow and heat it.

Or course half an hour is for falling in a vacuum. The real time will be delayed due to terminal velocity reached from friction of falling gas hitting gas already there. In fact this friction is what generates the high heat and temperature from the gravitational potential, as with meteors and reentering spacecraft. So the actual time is very hard to pin down. We might guess hours or days. And the heat generated might take much longer— as it takes hundreds of thousands of years for heat to diffuse from the center of our Sun already. These long heat diffusion times are all that save the Earth from being fried immediately.

Advertisement

*If we pour water on the sun with a bucket as big as the sun, will the sun be extinguished?** originally appeared on Quora. You can follow Quora on Twitter, Facebook, and Google+.*

Advertisement

*This answer has been edited for grammar and clarity.*