SWMBO bought the pictured carafe because it was locally made and rather pretty. To our surprise, it keeps water about ten degrees (estimate) cooler than ambient temperature.

I mean, you can fill it with water that is 20 degrees warmer than ambient, and by the next morning it will be 10 degrees cooler, just sitting on the nightstand.

A little bit of crude testing suggests a possible reason for this phenomenon. I put it in the kitchen sink and filled it absolutely to the top of the spout, then covered the spout. After several hours, the water level had dropped by about an inch.

My hypothesis is that the material the carafe is made from acts something like GoreTex, and allows water to pass through it cooling the contents at a rate of 540 calories for each gram of water that evaporates. The outside of the carafe feels dry with no hint of moisture, and is startlingly cool to the touch.

Does this seem reasonable?

tanstaafl.

We have an earthenware wine cooler that is supposed to use the same effect, no idea whether it actually works.

These winecoolers work pretty well, we have one of those as well. Usually you have to submerge them in cool water, then dry them off before use. I've found it does a wonderul job of keeping the wine cooled, even on hot summer days.

That said, I have no idea of the 'technology' behind Doug's carafe. It's probably something as he himself states, but then it puzzles me how it can cool below room temperature, without any extra energy added. Normally all items in a room heat up or cool down until they reach room temperature, but this thing doesn't. Strange. I must say I'm intrigued!

"Do Ceramic Wine Holders Actually Cool?"

We've got one of those wine coolers as well. It seems to me that it's well-suited for those evenings when you're out in the garden, relaxing with friends. We've never used it: (a) the weather in Britain doesn't work like that; and (b) our garden's small enough that we're no more than 70ft from the fridge; and (c) any chilled wine near me doesn't get a chance to warm up.

It's also the case, in the particular experiment of leaving it overnight, that the large thermal mass of the water will take longer than the small thermal mass of the air, to heat up again from the night's coldest point once the sun rises.

Peter

Ah... but there IS extra energy being used.

The water level decreases even when the top of the carafe is sealed. It would appear that the water is evaporating through the semi-porous terracotta walls of the carafe. Every gram of liquid water that evaporates absorbs 540 calories of heat energy when it turns into water vapor.

Where I am hazy is the source of the energy. One would think it would come from the air surrounding the carafe (outside) in which case the carafe should act like a miniature swamp-cooler type air conditioner and cool the room, not the contents of the carafe. Yet apparently the energy to evaporate the water from the carafe is coming from inside the carafe itself by reducing the temperature inside the carafe.

A secondary issue is how does the evaporative process manage to function through the shiny and [apparently] non-porous paint covering the outside of the carafe.

C'mon, Roger and Peter, you thermodynamicists, can you elaborate?

tanstaafl.

Latent heat of vaporization -- it's amazing! :-)

The energy source is the remaining water, which as a liquid in an environment where the air is less than saturated (100% humidity) has excess energy. Because the "partial pressure" of the water vapor surrounding the liquid water is less than the saturation pressure, the water will evaporate, cooling it. You might say that the water has an excess of energy compared to the equilibrium state, which would be a state where no further evaporation would occur because the ambient air was saturated. Meaning: in super humid environments or in the fog (such as England much of the time), this thing won't work at all.

It probably evaporates out the top.

Uhhhh... no. I blocked the top and the water level still dropped.

A bit more research shows that the water does indeed migrate through the porous terracotta clay of the carafe where it evaporates, cooling the outer surface of the pot which in turn allows heat to flow by conduction from the inside of the carafe.

tanstaafl.

Yeah, ceramic is porous of course, but I would have thought the glaze was not.

I have a water mark on a wooden floorboard that is testament to the fact that seemingly solid glaze can be porous...

Most ceramic stuff I've seen is not fully glazed on the bottoms.

We used to keep our butter in a terracotta butter dish; kept it cool enough to "keep" but not so cold it'd be hard to spread (the butter sat in a glass dish above the water reservoir)...

Ha! An epiphany... as TigerJimmy said (and failed to register with me) the water evaporates inside the carafe (that's the partial pressure bit) absorbing the heat of vaporization and cooling the interior, and the water vapor is what migrates through the terracotta.

It takes me a while, but sometimes I get there... eventually.

tanstaafl.

I don't think it matters where the evaporation takes place - whether inside at the water/air boundary, on the outside surface of the jug, or somewhere within the jug wall itself. Regardless, heat is removed from the water in the evaporation process. If it's within or on the outside surface of the jug, the evaporation will cool the jug, and heat from the water will move into the jug. If it's at the water surface with the inside air, the water would cool, and absorb some heat from the jug.

If it works with the jug completely full, I suspect it's water going through the walls of the jug, as the air space would be minimal and reach 100% saturation relatively quickly. (Again, thermo was a long time ago!)

-jk

Yeah, it doesn't matter where the evaporation occurs. If it evaporates off of the outer surface, having seeped through the ceramic, then it cools the ceramic and the liquid water left behind. If it evaporates from the surface inside, then it cools the water inside directly. When a water molecule changes state from liquid to gas, it removes heat from whatever is in contact with it in order to make the phase change. The whole system reaches a thermal equilibrium over time. Note that the temperature depression achieved is a function of the *rate* of evaporation. The evaporation process sucks up 2260 J/g of water evaporated. So multiplying this by the amount of water evaporated per unit time gives one energy/time units (J/sec, for example). Energy/time is called "power". So as the water evaporates, there is energy being transferred FROM the remaining liquid water TO the ambient air (which is where the water vapor goes after it evaporates). This produces cooling. However, the ambient air, being a higher temperature than the remaining water, is exerting a heating affect on the remaining water by convective heat transfer, and also by conduction from the countertop into the base of the jug. So, then, if the evaporative cooling power is greater than the convective/conductive heat transfer back into the jar, then cooling will happen. The convective and conductive heat transfer back into the water is a function of the temperature difference -- the higher the "delta T" the faster the heat transfer. So what will happen is that evaporation will cause net cooling power up to a certain delta T. If you can slow that heat transfer back into the water from the environment, you can increase this equilibrium delta T, which is why the jug is made from an insulating substance like clay. If the jug were made of aluminum which would readily conduct the ambient heat of the countertop back into the water very easily, no net heat transfer would be established.

From all this we learn: if you stick this jar in front of a fan, you should be able to achieve a greater delta-T. And also, if you put the base onto a piece of styrofoam, you can probably squeeze out even a little more delta-T (by reducing conductive heat transfer from the base). Finally, the rate of evaporation is a function of surface area, so the fuller this jar is, the more damp clay there will be, and the greater the cooling power. It would be fun to use a thermocouple and see what kind of delta-T you can get for different conditions (with/without insulating base, with/without forced airflow, full/half-full). Ambient temperature is also a factor, but I can't remember to what degree (linear or not).

"Partial pressure" is the way of talking about the fraction of water vapor in the air. If the air has been dried, then it has no water in it and we say the "partial pressure of water vapor is zero" (in other words, none of the absolute pressure of the mixed gas called "air" is being caused by the presence of water vapor). Now if you put that dry air next to some liquid water, some of that water is going to evaporate into the air until an energy equilibrium is reached. The humidity where this equilibrium occurs is dependent on both temperature and pressure, with warmer air capable of holding more water vapor.