|MadSci Network: Chemistry|
These chemical recipes for making ice are fine for demonstration purposes with small quantities, but rather inefficient and expensive when scaled up to try to use them for any practical purposes. The Glauber's salt recipe that you have come across does specifically require that Glauber's salt be used. No other salt will do. The chemical process is Glauber's salt + sulfuric acid solution --> sodium bisulfate + water Na2SO4.10H2O (solid) + H2SO4 (aq) --> 2 NaHSO4 (solid) + 10 H2O (liq) This reaction will take in about 105 kJ of heat from the surrounding environment per mole of reactants. Scaling this up, we arrive at 330 kJ of heat per kg of Glauber's salt and 300 g of sulfuric acid. For the sulfuric acid to be truly "aqueous", it would need to be in at least 1 litre of water. Any less water, and heat would actually be given out when any residual sulfuric acid reacted with water released by the reaction. But of course too much water would lead to sodium bisulfate being formed in solution rather than as a solid. If you are operating in Puerto Rico, I would suppose that the water you can get hold of during a power failure would be at about 25 deg C, so the 1 liter (plus) of cooling solution would itself need to be chilled from that temperature to about -5 deg C in order to do any freezing. That would require at least 1 kg * 4.184 kJ/kg/deg C * 30 deg C = 125 kJ of heat to be removed, leaving up to 200 kJ maximum of potential heat removal to do your freezing. By the time we have allowed for heat losses in the apparatus, and the fact that it might be necessary to use slightly more water in proportion, a realistic figure might be 120 kJ. Another recipe that I am personally more familiar with, and that may well work better for your purposes is simply to take the coolest available water, and try to dissolve up some ammonium nitrate in it. Ammonium nitrate is a commonly available nitrogen fertilizer. It is very soluble. 1 kg of ammonium nitrate will dissolve in 1 litre of water. It will remove about 28 kJ of heat from the surroundings per mole of ammonium nitrate, which works out at 350 kJ per kg of ammonium nitrate -- very similar to the Glauber's salt system. Obviously you would need to get hold of your ammonium nitrate in a powdered form of some sort, rather than in a slow dissolving, slow release form. How much freezing will 120 kJ of heat removal do? If we take 1 gram of water at 25 deg C, we need to remove 4.184 * 25 = 104.6 joule to chill it to water at 0 deg C, and then 334 joule to freeze that water -- 449 joule per gram = 0.449 kJ per gram. 120 kJ is therefore enough heat removal to freeze 120/0.449 = 267 gram of water. 1 imperial gallon of water is about 4520 gram; 1 american gallon is therefore about 3620 gram. Puerto Rico sounds more American than imperial, so I'll guess that you mean the latter ;-) That would mean a scale-up by a factor of about 14: You need either 14 kg of Glauber's salt in 14-20 litres of previously made up dilute sulfuric acid (20 to 25% by volume of concentrated sulfuric in water, or about one part of water to 2 parts of battery acid. NB extreme care required -- a lot of heat is given off when sulfuric acid and water are mixed. Always add acid to water, slowly.) or 14 kg of ammonium nitrate fine crystals in 14 litres of water. A plastic jug is a bad way to go -- the plastic is an insulator, and the jug shape is bad for heat removal from the centre. You need at least one small dimension for rapid heat exchange. A number of smaller containers or a book-shaped thin rectangular container would be better. ------ The second part of your recipe about dry ice seems to me to be nonsense. Yes, it is possible to remove heat with a reaction that produces a gas, like the vinegar/carb soda reaction. But to make dry ice you need to achieve temperatures of -80 deg C, and there is no way you are going to do that in or near an aqueous environment (and especially not in Puerto Rico!;-)
Try the links in the MadSci Library for more information on Chemistry.