Question #95491

1 Answer
Mar 3, 2015

It depends on how much moisture is in the air; the relative humidity, so if the temperature of the pot is near the dew point, there will be no evaporative cooling in either pot.

Explanation:

Your question is a little confusing to someone in the western world. This Wikipedia article may help to explain some of what you are asking.

If I understand the principle of these types of pots, there are two layers. The outer pot is a porous earthenware container which allows some moisture from the inside to evaporate and keep the pot cool. Inside this is another pot which is glazed to seal in the water in the center. Between the two a layer of sand and water holds the pots apart. Evaporation at the surface of the outer pot keeps the inner pot cool.

One might expect a spherical pot to have the smallest cooling efficiency because the ratio of the volume of the pot to the surface area is the largest. A cylindrical pot would have a smaller volume in relation to its surface area. This would provide more cooling per liter of material inside.

How might we explain an observation that two different pots provide about the same amount of cooling? Hopefully this is getting close to answering the question you are asking.

Here are some guesses:
For the images of these pots that I am able to find, the shapes do not look very different. The long neck of the surahi does not change its shape very much. Most of the volume is still a sphere. The cooling effects on this part are possibly more important than any heat transfer through the surface of the long neck of the container.

Since the cooling depends on water in the outer pot, a pot which evaporates water more quickly may be able to cool faster. However, if the water evaporates too quickly, the surface of the outer pot will begin to dry out and the cooling effect will be lost. A pot which is known to cool water more quickly may, after some number of hours, actually be warmer than a pot which cools more slowly.

Both pots may be as cold as they are going to get. If the pots cool to the dew point, no further cooling will take place. The dew points is the temperature at which moisture in the air will begin to condense on cool surfaces. It depends on how much moisture is in the air; the relative humidity. If the temperature of the pot is near the dew point, there will be no evaporative cooling.

Other things to think about:
In the United States there has been a story going around for many years which claims that hot water freezes into ice more quickly in a freezer than cold water. Obviously, when the water is hot it looses heat more quickly. But that rate of cooling will slow as it reaches 0º C. Students often spend many hours speculating whether this statement is true, and, if it is true, how could it possibly happen that way?