I'm trying to make this whole bunch of ice bottles to cool down my little aquarium. What I want in the ice bottles is that, fter freezing the whole bottle full water into ice, I need the ice to melt as slowly as possible. My question is, will the ice melt slower if the ice is made up of:

  1. Pure tap water
  2. Salt water (Salt mixing with water)

The reason I ask this because I was taught when I was young that sprinkling salt onto ice cubes will make it melt slower (it lowers the freezing point too.)

But I'm not sure if it is the same thing because this time the salt is mixed with the water.

Anybody could help me?

  • $\begingroup$ So, would the salt significantly alter the enthalpy of fusion? I'm having a hard time seeing how it would, but am willing to be proven wrong. $\endgroup$
    – Jon Custer
    Commented Sep 8, 2015 at 15:52
  • $\begingroup$ Generally mixing impurities reduces both specific heat capacity as well as enthalpy of fusion, which are energy required to increase temperature and melt solid respectively. So if ice is of same temperature ( say -4 degree centigrade) pure water melts slowly. If pure water ice is at 0 degree and salt water at its respective freezing point again pure water melts slowly (higher enthalpy of fusion). $\endgroup$ Commented Sep 8, 2015 at 16:32
  • $\begingroup$ icebergs are mostly fresh water thoughtco.com/fresh-or-salt-water-icebergs-609402 $\endgroup$
    – anna v
    Commented Sep 7, 2019 at 10:33

5 Answers 5


No, salt water does not (in that situation) seem to melt slower. In more detail:

When you freeze the water, you make ice. When you freeze salt water, you make ice, and (depending on details a bit) either concentrated brine or ice and salt crystals. There is no (or very little) salt in the ice, it separates out into brine or salt.


Then, if you supply outside heat, the water will quickly warm to 0 C, and then slowly melt. The water plus salt will start to melt at a lower temperature, as you suggest. Since it starts to melt colder, heat will flow into it more quickly. There will be a little less heat adsorbed per gram of water: there will be both the heat of fusion of water less the heat of solution and heat of mixing of the water with the salt. This will also cause it to heat up more quickly, but "less effectively" - there will be a little less cooling of the rest of the world.

  • $\begingroup$ Wow, someone who actually knows what they’re talking about, unlike the other answers. $\endgroup$ Commented Dec 18, 2017 at 13:10

By shifting the melting point to a colder temperature, the surface of your salt water/ice will be below $0C$. Assuming the environment is warmer than that, the temperature difference will be greater with the salt water ice bottle than the pure water ice. This implies greater heat transfer and faster melting.

If you want it to melt more slowly, the simple answer would be to put a layer of insulation around the bottle. The rate of melting is related to the rate of heat transfer. This means your desire to melt as slowly as possible is equivalent to wanting it to cool the environment as slowly as possible. Insulation will accomplish this.

You can think of your bottles as having almost a set amount of cooling power, rather like the amount of energy in a battery. Adding salt doesn't increase the amount of cooling power. You can slow down how fast the battery is used, but that doesn't make it more useful. Here you can slow down how fast the ice melts, but that might not make it cool your aquarium more efficiently.


General principles

You should probably just run the experiment over the weekend, but here are the relevant bits of physics:

Typically, saltwater freezes at a lower temperature, so you will probably reach a point where your freezer no longer fully changes the state of the water to solid. You probably want the freezing to happen no matter what.

The reason that you probably want freezing no matter what is because the colder the bottle gets, the faster it will exchange heat with the aquarium. The heat flow will be proportional to the difference in temperatures. Ice has this really nice property that it forms and melts all at one temperature, so you do not have to increase the rate at which cold is pouring out of the system in order to store more coldness in the system. That's a big win.

Whether the water should be salted or not depends heavily on the temperature of your freezer, among other things. Water has about twice the specific heat of ice, so that suggests that you want as much water-stage as possible. So you have two competing things here: first off that the heat transfer is greater for a lower freezing temperature, pushing the freezing temperature up; second, that the water holds more heat than the ice will, pushing the freezing temperature down.

Calculating the maximum time.

The model that you want is something like: from temperature $T_0$ to $T_f$, we have ice; then from temperature $T_f$ to $0~^\circ\text{C}$ we have liquid, then we assume that the water and saltwater will heat up the same amount from then on. Your aquarium is at some temperature $T_1$ and so the heat exchange will go like $\frac{dq}{dt} = k (T_1 - T)$ for some unknown $k$, leading to temperature changes due to $q = c m T$ where $m$ is the mass and $c$ is the specific heat capacity per unit mass. Curves are then looking like $T_\tau(t) = T_1 - (T_1 - \tau) e^{-kt/(cm)}.$

From $T_0$ to $T_f$ requires a time $t_0$ such that $T_{T_0}(t_0) = T_f;$ solving this yields: $$ t_0 = \frac{m ~ c_\text{ice}}{k} ~\log\left(\frac{T_1 - T_0} {T_1 - T_f}\right).$$ Then there is some heat of fusion $h_f$ with timescale $t_f$ given simply by $ m ~h_f = k (T_1 - T_f) t_f,$ so that:$$t_f = \frac{m~h_f}{k (T_1 - T_f)}.$$Finally, heating up to $0~^\circ C$ takes a time$$t_1 = \frac{m ~ c_\text{water}}{k} ~\log\left(\frac{T_1 - T_f} {T_1 - 0~^\circ C}\right).$$The total time is $t^* = t_0 + t_f + t_1$ which is maximized independently of $m/k$ by setting its derivative to zero,$$\frac{k}{m} \frac{dt^*}{dT_f} = \frac {c_\text{ice}} {T_1 - T_f} + \frac{h_f}{(T_1 - T_f)^2} - \frac {c_\text{water}} {T_1 - T_f} = 0$$This is solved by $T_f = T_1 - h_f / (c_\text{water} - c_\text{ice}).$

Now of course the addition of salt will change these parameters, but it's important to see the order of magnitude of what we're talking about here. This critical temperature $T^* = h_f / (c_\text{water} - c_\text{ice})$ is something like (300 kJ/kg) / (200 kJ (kg °C)) ~= 1.5 °C: it is tiny in comparison to $T_1$.

This means that the extra heat capacity gotten from shifting the temperature at which the water freezes is tiny in comparison to the higher heat transfer: you want to push this temperature as high as possible rather than as low as possible. So you want no salt.

What else can you do?

The absolute simplest way to make the ice last longer is to lower the parameter $k$ above. The easiest way to do this is to get some air between the ice and the water. Those foam pads that one wraps around beers might work best for this, but they might also pollute the water and/or be eaten slowly by fish. You might instead try a zip-lock bag or even a combination of the two. Of course, then you will need to weight the bag down so that it stays in the water.


To a simplified, less dichotomous version of the question, that is, "Which melts faster? Ice from mostly pure water or ice from a salt water solution?"

Would a layman-friendly answer not be:

"It depends a great deal on the temperature outside of the ice"? (This is the same as saying the two different solutions have different specific freezing/melting temperatures and therefore different rates of melting at different temperatures, only less technical and more applicable for the layman).

Examples of the scope and scale involved if needed:

If it is just barely over approximately 21 degrees fahrenheit, ice from relatively pure water does not melt as it has a melting point of 32 degrees, while ice from "ideal" (lowest possible salt water freezing temperature) salt water will start to melt. Therefore, under these circumstances, salt water melts faster than "pure" water. Between 21 and 32 degrees is an ideal time to apply salt to ice on roads and sidewalks. Applying it at lower temperatures will be a waste of salt unless the temperatures increase soon while the salt remains. Applying at higher temperatures than 32 could still help with specific patches of ice where temps locally reach 21 to 32 but could otherwise be fairly inefficient, used to melt what is already melting.

If it is well over 21 degrees, say room temperature, there may well be a difference, in either direction, in part dependant on the volume and shape of the ice, in part dependant on the specific solution being frozen and temperature respective solutions are frozen at, but it could be relatively insignificant as it is well over the freezing/melting points of both "pure" water and salt water. Therefore, here, they could melt seemingly at the same rate or either/or depending on other factors.

If it is under 21 degrees, both will stay frozen solid. Therefore, neither is melting, so in a sense they are both (not) melting at the same rate. (You can experience this in the winter when attempting to melt ice with salt but it is about 10 below...doesn't really work as expected given salt's reputation!).

If the salt solution is frozen at just under 21 degrees but the "pure" water is frozen at only just under 32 degrees, and both are out in, say, 40 degree temperatures at the same time, it may very well be that the salt water melts considerably slower than the "pure" water, but having more to do with their respective temperatures at beginning of melt (21 and 32 -- easy to gauge if you check the respective solutions in the freezer every 10 minutes waiting for them to become solid ice) than their melting/freezing rates assuming same temperature of frozen solution at beginning of melt (say 20 and 20 -- hard to gauge without precise thermometer and thermostat control, as "pure" water ice at 32 and "pure" water ice at 21 will both be frozen solid and thus mostly indistinguishable). Therefore, occassionally "pure" water melts faster than salt water.


Fast and slow are rate of temperature changes But most cooling situations we want the most heat (energy) removal values and that is in Joules or BTU for a given mass. That value is based on the heat of fusion for phase change from solid to liquid, plus the heat for heating the liquid above freezing, and heating the solid below freezing. For water, the phase change energy is about 75 times the specific heat of water. For simplification we can neglectg the small amount of energy involved in heating the liquid or the solid in the narrow range of typical cooling (refrigerating foods etc) and look at the energy in the phase change.
I believe that potable water will have a heat of fusion of 333 J/gm. So most of the cooling will occur at a steady temperature of 0C. I also believe that gel and other forms of package ice, have a freezing temperature of less than 0C....which means the temperature plateau will occur at a lower than 'water freezing'. Also the additives generally lower the heat of fusion, meaning that they are have less cooling capacity per unit (mass). Two issues here. A. From a purely theromdynamic cooling need, Ice is better than gels, for a given mass. It would be the cheaper form considering shipping weight. There may be chemical systems that provide more capacity. B. If you have a product with water, that should not be frozen, then do not use gels as the temperature where the phase change energy absorbtion will occur below 'water freezing' temperature, possibly causing the product with water to freeze or solidify. That is an issue in transporting medications such as Insulin with water content, that can be damaged by 'freezing"


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.