Wort Boils Above 212 °F (100 °C) (I: Theory)


Wort being boiled in a modified (legally obtained) Sanke keg.

We all know that water boils at 212 °F (100 °C). Chemists would clarify that this refers to pure water at standard pressure (100 kPa). Many brewers assume, given that wort is mostly water, that it also boils at 212 °F (100 °C). This isn’t the case, however. Wort boils above 212 °F (100 °C) — the exact temperature depends on the gravity of the wort.

This article has quite a bit of chemistry in it. Don’t feel like you need to follow every little bit. I’ll explain the take home message in the concluding paragraphs.

Boiling Point Elevation

If you dissolve any non-volatile solute into any solvent, the boiling point of the solution is increased. (A solute is a material dissolved into a liquid. A solvent is a liquid capable of having the solute dissolve into it. The solution is the mixture of the solute in the solvent. In wort, malt sugars, proteins, minerals from the brewing liquor and other compounds are solutes and water is the solvent.) This phenomena, called boiling point elevation, works with any solute and any solvent. For example, let’s say you had a flask of boiling mercury, and you added table salt (NaCl) to it. The melting point of salt (801 °C) is much greater than the boiling point of mercury (357 °C), so salt wouldn’t be volatile in boiling mercury. Therefore adding salt to the boiling mercury would raise the boiling temperature of the solution (mercury plus salt) above the boiling temperature of the pure solvent (just mercury).

One way to understand why this works is to consider the vapor pressure of the solution. When a pure solvent is boiling, the vapor pressure of the liquid matches the atmospheric pressure pressing down on it. Some molecules on the surface of the solvent gain enough energy to escape the liquid and enter the gas above it. When a (non-volatile) solute is added to the solvent, it lowers the vapor pressure above of the solution. It does this simply by taking up space on the surface of the solution. Some percentage of the molecules with enough energy to escape the liquid will collide with molecules of solute. lose energy in the collision, and not enter the gas. Thus, more energy needs to be added to the solution to increase its vapor pressure and the boiling point of the solution is raised.

Interestingly, adding solute to a solvent also lowers its freezing point. This phenomena is called freezing point depression. So, wort not only boils at a temperature higher than water, but freezes at a temperature lower than water.


How Much is the Boiling Point Increased?

Here is a simple equation to describe boiling point elevation.





∆T = change in temperature (°C)

i = van’t Hoff factor (unitless)

Kb = ebullioscopic constant (°C kg mol-1)

m = molality of solute (mol kg-1)


The “i” in the equation, called the van’t Hoff factor, is the average number of ions into which the solute breaks while in solution. For sugars, it is 1, because sugars don’t ionize. For salts with two ions that completely dissociate, it is 2. (For sodium chloride, which almost entirely dissociates into Na+ and Cl ions, it is 1.9.) For salts with three ions that completely dissociate, it is 3. (For calcium chloride, which mostly dissociates into one Ca2+ ion and two Cl ions, it is 2.3.)

Kb is the ebullioscopic factor of the solvent. “Ebullioscopic” is just a big word that essentially means boiling. For example, an ebullioscope is an instrument that measures the boiling point of a solution. (The first ebullioscopes were invented to measure alcohol content.) For water, the ebullioscopic factor is 0.512 °C kg mol-1 [or 0.512 (°C kg)/mol]. I’ll stop saying “ebullioscope” now.

The “m” in the equation is the molality of the solute — the number of moles (a unit of mass) of solute per kilogram of solvent. [Molality is similar to molarity, which is moles of solute per volume of the solution (L). For small amounts of solute, the values for molarity and molality are very close.]

Maltose — the most abundant solute in wort — has a molecular weight of 342.30 g/mol, or, equivalently, 0.3423 kg/mol. So to calculate its molality in wort, you would first take the weight of the maltose (kg) and divided it by its atomic weight to yield the number of moles. Then, you would divide that by the weight of the liquid (in kg) it was dissolved in. [Recall that, at standard temperature and pressure (STP), a liter of water weighs a kilogram.]



So let’s say we had 3.00 kg (6.6 lbs.) of pure maltose, and we dissolved it in 19.0 L (which weighs 19.0 kg) of pure water. We’d end up with a solution a little greater than 19 L (5.0 gallons), with a specific gravity around 1.060. What temperature would this wort-like solution boil at?


∆T = 1*0.512 (°C kg mol-1)*[3.00 (kg)/0.342 (kg mol-1)]/19.0 kg =  0.236 °C


∆T = 1*0.512 (°C kg mol-1)*0.462 (mol kg-1) =  0.236 °C


So the temperature changed by +0.236 °C, giving us a boiling temperature of 100.24 °C (or about 212.5 °F). Not a huge change, but certainly within the precision of most thermometers to measure. Actual wort, of course, contains more than just maltose including some minerals with higher van’t Hoff factors. However, at a homebrew scale, these are added in gram quantities, so we would predict that their contribution to boiling point elevation would be much less than the sugars. In addition, atmospheric pressure also modifies the boiling point of a solution, as people who live at high elevations (and correspondingly lowered atmospheric pressures) can attest.

The idea of boiling point elevation is mainly interesting from a scientific standpoint. For brewers near sea level boiling average strength worts, changes in atmospheric pressure will likely cause larger deviations in boiling temperature than boiling point elevation will. These small changes in boiling temperature are unlikely to have any big effect on any of the chemical reactions occurring during the boil. (I suppose that the elevated temperature from very dense worts made from malt extract could increase the production of Maillard products.)

Soon I’ll present some data showing how closely this simple equation predicts reality. I’ll take the temperature of boiling water at different salt concentrations within the reasonable range of strike waters and repeat the experiment with wort at a couple different densities and added salt concentrations.


(Thanks to Adam G. Fisher and Jon Loeliger for comments on this manuscript.)


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  1. It’s amazing to visit this website and reading the views of all friends about this post, while I am
    also keen of getting know-how.

  2. What about the elevation. Here you are talking about the contents of the solution affecting this.
    Does elevation come into play. For example, if a brewer is at see level while the other is in the mile high city, do the actually boil at different temperatures?

    • Chris Colby says

      Yes, elevation plays a fairly large (and well-known) role. At higher elevations, because they have lower atmospheric pressures, water boils at lower temperatures.

  3. Just curious…. How long does it typically take to bring 3-4 gallons on wort to a boil? We experience very long times even at the highest temperature settings. Gas or electric, doesn’t seem to matter for us.

  4. This is a different calculation altogether. The time it takes to boil depends on the amount of heat you’re putting in the fluid. Heat in the British units is given in BTU, and heat flow in BTU/min (sorry for the metric folks, but in my mind it’s easier to work heat transfer problems with BTU than with Calories, Joules or Watts).

    So, let’s do some quick math:
    The formula to calculate the heat needed to elevate the temperature is:
    Q = mCpDt
    Q: Heat needed [BTU]
    m: Mass of fluid to be heated [lb]
    Cp: Heat capacity of the fluid being heated [BTU/lb˚F]
    Dt: Change in temperature [˚F]

    So, for your case, you have 4 gal of water, which equals to 33.4 pounds of water.
    The heat capacity of water is 1 in British units, but as this article says, when you add your materials this will change. You can measure it or calculate it, or just add a bit and be done with it.

    So, lets say the Cp of wort is then ~1.02

    Now your Dt. Let’s say you start at room temp of 70˚F and you want to bring to a boil at sea level, which by this article would be 212.5˚F. Your Dt is 212.5-70 = 142.5˚F

    Therefore, your heat is:
    Q = 33.4*1.02*(142.5) = 4854 BTU

    Now we can calculate the time it takes.

    Say you want to heat up this mass of wort in 1hr. Then you need 4,854 BTU/Hr to get it there.

    The average gas stove can put out 7,000 BTU in an average burner (https://www.thekitchn.com/whats-a-btu-and-how-many-shoul-112070)

    So dividing the heat needed by the BTU of your average stove burner (4,845 BTU/Hr / 7000 BTU), we get that your 4gal of wort should start boiling at around 42 min (0.69hr)

    How far is this from what you’re experiencing? If your stove is taking way longer, then it’s probably because your pot is losing lots of heat by radiating to the surrounding air. You could put a lid on the pot (and risk overspill), or insulate your pot… or keep your room really warm 😉

    Hope this helps!

  5. Chuck Dvorsky says

    “Boiling” isn’t the issue. The issue is the temperature needed to crack/break the proteins.

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