crystalsphase-transitionphysical-chemistrythermodynamicswater
In my kitchen I have a vial with very salt water, and some food coloring. It's been left to slowly dry out for a year or so. Recently, salt (presumably) started growing up the walls of the vial.
Does anyone have an explanation for what is happening here?
(I'm assuming this is physics, because I don't think it's alive or due to chemical reactions.)
In real life situations like this there tend to be lots of variables, so it's dangerous to make predictions based on (over?) simplified physical models.
Having said this, the article makes a good argument. Newton's law of cooling tells us that the rate of heat loss per unit area from an object is roughly proportional to the temperature difference between the object and it's surroundings. For a large and a small pan, both containing boiling water and both in the same kitchen, the larger pan has the larger surface area and it will lose more heat per second than the small pan.
You ned to distinguish carefully between heat loss and temperature fall. The temperature of the big pan will fall more slowly than the temperature of the small pan because its surface area to volume ratio is smaller. However it will lose heat faster.
The point the article is making is that the burners on your stove supply heat at a constant rate that is independant of the pan size. Because the larger pan loses more heat per second the net heat flow from the burner is lower, and it will take longer to get it back to the boil.
Of course this assumes that the efficiency of heat transfer from the burners to the pan is independant of pan size, but then as I mentioned at the outset in real life there are lots of such variables.
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 $\Delta H$:
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
surface tension tends to draw the water up the sides of the jar. evaporation then removes the water, leaving the salt behind as a porous layer on the jar surface. the capillary lift of water in the salt network is greater than the capillary draw on plain glass so water with salt in it continues to get pulled upwards through the salt, only to be evaporated away, leaving behind more salt higher up the walls, and so on.