Adding salt to water makes it freeze at a lower temperature. This fact is being used in two different ways in the two scenarios you mention. Dissolving sodium chloride in water is slighly endothermic, but this effect is small and to the best of my knowledge isn't important in the drink cooling process.
Putting salt on the highway is quite straightforward: we don't want ice to form, so we put salt in the water to prevent that. This doesn't just change the amount of time it takes ice to form, it actually completely prevents ice from forming, unless the temperature gets so low that the water can freeze even with salt in it.
Cooling your drink is a bit more complicated, because in this case the rate at which things happen is important. You don't want your drink to be less than $0^\circ C$ because it would freeze; instead you want to cool it down to a few degrees Celsius nice and quickly.
The rate at which it cools depends on two things: the temperature of its surroundings (the colder the better) and the heat conductivity between it and them. You could try to cool it by putting it in a bowl of ice at $0^\circ C$, but the problem is that the ice is solid and will only touch the bottle at a few points. This results in a poor thermal conductivity, so the drink will only cool slowly.
To get around this, you could try mixing the ice with some water. Now the bottle is touching the liquid over a large surface area, and the liquid itself has a higher thermal conductivity than solid ice due to mixing, so heat will be transferred much more quickly. But the problem is that the water won't be at zero degrees any more, at least not at first (I'm assuming the water comes from a tap, so it's not chilled initially). You have to wait for quite a bit of the ice to melt before the water's temperature will drop. Also, once you put your warm drink into the water it will heat the water up as the drink cools down, so again you have to wait for the ice to melt in order for the water to cool again.
The solution to this is to make the ice melt faster. You can do this by adding salt. This lowers the freezing point, making the water less "happy" about being in the liquid state, so it melts more quickly. This means firstly that the tap water you've added to the ice will cool to close to $0^\circ C$ much more quickly, and secondly that once you've put your drink in there the water will stay cold as the ice continues to melt.
It's also possible that, with the salt added, the water can go to below $0^\circ C$, but this will only happen if the ice is quite a bit colder than $0^\circ C$. This could be the case, but my intuition is that the rate of cooling due to the ice melting faster is more important here than the final temperature. You could easily test this by putting a thermometer in the salty ice water and seeing if it goes much below freezing.
There's also the fact that dissolving the salt is endothermic, as you mentioned. To test whether this is important, you could try adding salt to some chilled water without any ice, and see if the temperature drops a lot. My feeling is that it will only drop by a tiny amount that will be hard to measure with a normal kitchen thermometer, but you can always try the experiment.
I will not answer your question directly; only give you some tools that should help you answer the question (in practice) yourself.
To focus the attention, find below a typical heating/cooling diagram for a frozen pure substance.
The vertical axis marked $T$ represents temperature (in degrees Celsius).
Three significant temperatures are indicated on the $T$ axis:
$T_c$: this is the temperature of the cold pack while in your freezer.
$T_f$: this is the melting (fusion) point of the material in the pack.
$T_H$: this is the temperature after the pack has been carrying out its function for some time. Often it will be room temperature.
The horizontal axis marked $H$ represents Enthalpy. Enthalpy is what is known in physics as Heat Energy. $H_1$, $H_2$ etc are various heat energy contents of the pack as it slowly heats up from $T_c$ to $T_H$.
You’re quite correct that the best pack (there are other, practical criteria, of course) would be the one that on heating would extract the highest amount of Enthalpy from its surroundings (typically a cooled picnic box or such like) to heat up from $T_c$ to $T_H$. So let’s analyse this.
The overall Enthalpy the pack has to absorb is simply $\Delta H = H_4 - H_1$. We can calculate this as follows.
The heating diagram is divided in three areas marked $I$, $II$ and $III$.
Area $I$: the frozen pack is heated from $T_c$ to $T_f$. The Enthalpy change for area $I$ is given by:
$\Delta H_I=mC_s(T_f – T_c)$ where $m$ is the mass (in $kg$) of the pack and $C_s$ is the specific heat capacity of the solid pack material (in $J kg^{-1} K^{-1}$).
Area $II$: in this area the pack is melting (during melting $T$ stays constant at $T_f$). The Enthalpy change for area $II$ is given by:
$\Delta H_{II}=mL_f$, with $L_f$ the latent heat of fusion (in $J kg^{-1}$). For melting $L_f$ is a positive value.
Area $III$: the molten pack is heated from $T_f$ to $T_H$. The Enthalpy change for area $III$ is given by:
$\Delta H_{III}= mC_l(T_H – T_f)$ where $C_l$ is the specific heat capacity of the liquid pack material (in $J kg^{-1} K^{-1}$).
We can now add it all up:
$\Delta H =\Delta H_I + \Delta H_{II} + \Delta H_{III}$.
Or:
$\Delta H = m[C_s(T_f – T_c) + L_f + C_l(T_H – T_f)]$
We can now conclude a few things, in order to maximise $\Delta H$:
- High mass $m$ is desirable.
- Large difference $T_H-T_c$ is desirable.
- Both $C_s$ and $C_l$ should be as large as possible.
- Large value of $L_f$ also increases $\Delta H$.
This general treatment was for a pack filled with a single pure substance but can as an approximation also be used for mixtures of substances, by adjusting the masses, latent heats and specific heat capacities.
Best Answer
Absolutely; all passive materials can be compressed. The bulk modulus, a material property with units of pressure, couples the applied pressure to a relative reduction in volume. The bulk modulus for ice at 0°C is around 8 GPa, which means that about 8 MPa or 80 bar pressure is required for a -0.1% volumetric change.
Here, a phase diagram for water is useful. The discussion in Powell-Palm et al.'s "Freezing water at constant volume and under confinement" includes a pressure–volume phase diagram:
From this, we can predict the equilibrium response when heating or cooling water at constant volume (by moving vertically) or compressing or expanding water at constant temperature (by moving horizontally). We find that at constant volume (moving vertically downward from 0°C and 1 g/cc), over 200 MPa and 20°C undercooling is required* to get even a 50% slush of water and ice.
Let's zoom out a little. From Powell-Palm, "On a temperature-volume phase diagram for water and three-phase invariant reactions in pure substances," we find that 209.9 MPa is ultimately required* for complete solidification, into a two-phase region (at equilibrium) of ice-Ih (ordinary ice) and ice-III:
(Note that "0.00611 MPa" should read "0.000611 MPa"—the authors missed a zero.)
We can interpret this as the compact structure of ice-III providing a solution to the problem of ice-Ih being anomalously voluminous. We find from the temperature–pressure phase diagram of water that this ice-III nucleates (at equilibrium) upon cooling to 251 K, or -22°C:
With further cooling, the ice-I–ice-III mixture transforms* to ice-I–ice-II, then to ice-IX–ice-II, and then to ice-XI–ice-IX. (How can this be determined, since the volume–temperature chart doesn't include any of that information? It's from the horizontal line on the temperature–pressure chart and the knowledge that ice-I and ice-XI have specific volumes of >1 g/cc and that ice-II, ice-III, and ice-IX have specific volumes of <1 g/cc; thus, a higher-density and lower-density combination is required to maintain a constant 1 cc/g, and we can't move an iota above or below that two-phase line upon cooling at constant volume.)
Note that no power can be generated under the condition of constant volume, as no displacement occurs. And although there's no thermodynamics prohibition about allowing the system to expand and do useful work, you would have to heat it up again to liquefy it to repeat the process, and this would use up the energy you gained.
*Note that this answer always refers to equilibrium phase predictions. Sufficiently rapid cooling involves kinetic limitations that delay or essentially even preclude phase transitions. For example, liquid water can be cooled fast enough that crystals essentially never form even though the thermodynamic driving force is large. Here, the solid water is said to be in a glassy or amorphous state.
(See also the fun rotatable 3D phase diagram of water here.)