I have been curious about this question for a while. If you want to warm up a large amount of water, is it feasible to do this by adding a substance that has an exothermic reaction with the water? What is the relevant chemistry one has to know to work out how much the water would be heated up by?
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2$\begingroup$ This is an interesting question. What do you want to do with the hot water? If you want it to be potable (fit for drinking), that would limit some answers. $\endgroup$– Ben NorrisCommented Jun 6, 2013 at 15:38
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3$\begingroup$ Well... Swim in it :) $\endgroup$– marshallCommented Jun 6, 2013 at 17:20
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$\begingroup$ Theodore Gray, in his book ("Theo Gray's Mad Science", Black Dog & Leventhal Publishers, Inc, NY, ©2009, pp. 234-237) uses 500 pounds of CaO to heat 700 gallons of water in a hot tub, predicting that the temperature would rise from 60 to 100 °F. He cheated a bit: the CaO was incrementally added to water in an inner tub, heating the water (already hot at the start because the outside temperature was 10 °F) in the surrounding tub, where he and his 2 kids were. Details in the reference plus two nice photographs. $\endgroup$– Ed VCommented Mar 12, 2020 at 2:03
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3 Answers
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From the comments:
What do you want to do with the hot water? Swim in it. I was thinking of thousands of liters
That's an interesting idea, but unfortunately, I don't think adding chemicals to a pool in order to heat it is a good idea (especially yellowish chemicals). The water temperature would drop in a few hours, tops, and you'd have to be constantly adding more chemicals to it. At some point, you'd have a pool of chemicals instead of a pool of water. Not to mention that the chemicals would probably end up being more expensive than the electricity or the gas bill to heat up the pool to the same temperature the conventional way.
A nice way of heating up pools is using solar water heating systems, and coupling them with the conventional heating systems, if necessary.
So... Is it feasible? I'd say it isn't. But since you said this was more of a thought experiment... The only scenario I can imagine where the best solution to heat up a pool is to add chemicals to it is: You have a small pool, very little time to heat it up, you want to heat it up to sauna-like temperatures, and although you'll only use it for a couple of hours you are willing to spend a lot of money to do so. In this case, I'd add some water to the pool, and then add the following reagents to it, making the following reaction:
$$\ce{NaOH + HCl -> NaCl + H2O} \\ \Delta H = -56\,200\ \mathrm{\frac{J}{mol}}$$
You can do the math, but I assure you it heats up pretty quickly. We used to neutralize concentrated acid in ~10 L plastic buckets, and if we "accidentally" neutralized too much at a time the bucket would start to heat up to a point you couldn't even touch it.
NaCl is kitchen salt, so no issues with toxicity if you do the math correctly. Salt water also makes you float better, which I think is a nice quality in a pool. There would be salt on the bottom, but I think that's nice too, makes it feel like the sea.
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You’re looking for materials that either react with water with a large reaction enthalpy $\Delta H_\mathrm{r}$, or have a large enthalpy of solution $\Delta H_\mathrm{sol}$ (which really amounts to the same thing but with a slightly different scope).
The problem with this as a general method, though, is that water has a really high heat capacity: take a look at this table of heat capacities, which gives it as approximately 4.2 joules per gram per kelvin.
A common method of producing self-heating coffee cans is to combine anhydrous calcium chloride with water in a closed vessel surrounding the can. According to a vendor of calcium chloride, 100 % anhydrous $\ce{CaCl2}$ has a solution enthalpy of 737 joules per gram when dissolved in water.
Assuming an approximate density of water of 1 gram per millilitre, that means to heat a litre of water by 10 K, you’d need about:
$$ m_{\ce{CaCl2}} = \frac{m_\text{water} c_p \Delta T}{\Delta H_\text{sol}} = \frac{1000\ \mathrm{g} \times 4.2\ \mathrm{J\ g^{-1} K^{-1}} \times 10\ \mathrm{K}}{737\ \mathrm{J\ g^{-1}}} = 57\ \mathrm{g}$$
So, that’s feasible, but you may have a different idea of a large amount of water. And, let’s not forget, you're still adding stuff to the water, so it may not be suitable for what you wanted it for anymore.
Other notable reactions that can heat water include reacting it with lithium metal (~220 J/g), or dissolving sulfuric acid (~90 J/g). You’re looking for materials that don’t hold together particularly strongly (e.g. if they’re a crystal, they have a relatively weak lattice), because you have to break those bonds to dissolve/react it, and also bind to water very strongly, because that’s what gives you the exothermicity. Because of that, ionic materials and materials that form at least doubly charged ions when reacting with water are likely to be better candidates.
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$\begingroup$ Thank you! I wonder if 737 is as high as you can get. You are right that I was thinking of thousands of liters although I only need to heat it by a few degrees. In fact it's mostly a thought experiment but an interesting one I hope. $\endgroup$– marshallCommented Jun 6, 2013 at 6:50
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$\begingroup$ @marshall: Well, calcium chloride is available in large quantities -- while more rare than NaCl or MgCl, it is occasionally used for gritting roads in winter in some places, for example. Getting it anhydrous in large quantities will likely be trickier, though. $\endgroup$– AesinCommented Jun 6, 2013 at 9:28
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How about mixing calcium carbonate and sulfuric acid with the water? There will be a lot of heat released (and carbon dioxide, of course), and the resulting calcium sulfate will, for the most part, just precipitate out.
