# Is it possible to heat up water using chemicals? (Meaning no outside source of heat like electricity, or gas) [closed]

I am asking in terms of heating something small like a 10oz cup of water with it also being safe to consume. It would have only need to heat the water for around 1-2 hour(s).

• Not sure what you mean exactly. There are heat packs involved to heating meals for military personnel (“MREs”) and chemical heat warmers for use in boots, etc. But adding a chemical substance to water, in order to heat the water, and then having the hot water be potable, is another matter. I did not downvote, by the way. – Ed V Dec 24 '20 at 18:07
• It is very easy to heat up water with chemicals. It is impossible to do so and still have drinkable water. – matt_black Dec 24 '20 at 21:04
• @matt_black I assume by "drinkable" you meant "safe to consume", since you were responding to the OP, who asked about the latter. If so, it's not impossible. Please see my answer below. – theorist Dec 24 '20 at 21:36
• Does this answer your question? Can you heat water with additives? – Mithoron Dec 24 '20 at 21:54
• Heat by how much? – Zhe Dec 25 '20 at 2:31

Technically, you could heat water by adding chemicals to it, and have something safe to consume. But there's two issues:

(1) In so doing, you wouldn't have pure water any more—you'd have water plus those chemicals.

(2) It's wildly impractical.

But here's how you'd do it:

(1) Add pure concentrated HCl to water. This is exothermic (i.e., thermal energy is released).

(2) Add just enough pure concentrated NaOH to neutralize the HCl. This is also exothermic. The result would be water with dissolved NaCl (table salt), which is safe to drink.

But the amount of HCl and NaOH you'd need to add to obtain significant heating might make the water too salty.

Also, once it's heated, just 10 oz. of water would not stay hot for 1-2 hours without an insulated container.

Here we calculate the amounts and concentrations that would be needed.

$$\Delta \text{H}_{solvation} \text{, HCl} \approx -57 \text{ kJ/mol}$$ $$\Delta \text{H}_{neutralization} \text{, HCl and NaOH} \approx -75 \text{ kJ/mol}$$

Thus for every mole of HCl that we add, and then neutralize with NaOH, about 132 kJ of thermal energy is released.

10 oz. water = 296 mL ~ 296 g

Now suppose we want to heat it from 21 C to 71 C (70 F to 160 F). The thermal energy needed is:

296 g x 4.184 J/(g C) x 50 C = 62 kJ

Thus we need .5 moles of HCl and .5 moles NaOH. To avoid the need for special handling, let's use lower concentrations of HCl and NaOH—say 6 M for each (6 M HCl is commonly seen in freshman chemistry labs). The quantity we would need of each would be:

(.5 moles/(6 moles/L))(1000 mL/L) = 83 mL, for a combined total of 166 mL of NaOH and HCl.

Since our goal is 296 mL (= 10 oz.) of hot water, we would add 83 mL of 6 M HCl and 83 mL of 6 M NaOH (both also assumed to be at 21 C) to 296 mL – 166 mL = 130 mL of water.

[Here, for the purposes of this rough calculation, I've assumed the volumetric heat capacity of salt water is about the same as that of pure water, i.e., ~4.2 J/(mL C).]

This would give a salt concentration of .5 moles/0.296 L = 1.7 M. That's about 3 x the 0.6 M salt concentration of seawater, and about 2/3 the 2.6 M salt concentration of soy sauce. So unless you plan to use it as a condiment, you would probably find it unpalatable.

• Hm, safety of drinking NaCl solution is conditional. If you are under conditions to control your daily sodium intake, for medical or dietary reasons.... – Poutnik Dec 24 '20 at 22:06
• This particular problem is amenable to a “back of the envelope” thermo calculation, especially for a theorist! :-) So, how much concentrated HCl and solid NaOH would have to be added? If the final NaCl concentration rivals seawater, then I would say it is closer to survivable than potable. Feel free to make any reasonable simplifying assumptions and Happy Holidays! :-) – Ed V Dec 24 '20 at 22:25
• How about using the "chemicals" dihydrogen and dioxygen, and letting them react? – Karsten Theis Dec 24 '20 at 22:42
• @EdV Yes, I was thinking I should do that very calculation and add it to my answer. Hopefully I'll have time later this weekend. – theorist Dec 24 '20 at 22:44
• Great! Then as a loophole cheat, reverse osmosis to get rid of the salt! Just needs high pressure for that and maybe get that as hydrostatic pressure from a largish dam. – Ed V Dec 24 '20 at 22:49