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I drink a lot of iced tea and live in a very humid part of the country. My drink accumulates so much water, I had to purchase coasters to collect all the water dripping from the sides.

Where does that water come from? I thought may the glass was porous and the water was leaking. More likely it is coming from the air... but is it that humid?

How much water am I collecting? With my pint full of ice cubes I seem to get around 2 ounces. Humidity here is about 75% and the temperature is 90F (feels like 98F).


It seems like my iced tea is behaving like a condenser. Is it possible to reduce humidity of my room (10 feet × 10 feet ) in a substantial way using ice and a glass?

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Removing humidity from air requires cooling it to the point of condensation, then removing the latent heat of evaporation so it actually condenses.

Ice, on the other hand, will remain at 0°C as you add heat, until it melts. So it's convenient to estimate the ability to remove moisture from the air in terms of the amount of ice it would take.

Latent heat of fusion of ice: 330 J/g (approx 80 calories)
Latent heat of evaporation of water: 2400 J/g (about 600 calories: approximate, function of T)

This means that you need to melt about 8 grams of ice to remove one gram of water from the air.

Let's estimate the water content of a room at 24°C next: the saturated vapor pressure is about 3000 Pa, meaning that 3% of the molecules in the air are water. From

$$PV=nRT$$

we can compute that $n = 1.2 mol/m^3$. Thus, at 80% relative humidity, there would be about 18 gram of water per cubic meter. If your room is a modest 20 $m^2$ and 2 m tall, dropping the humidity by 10% would require you to remove 4.8 mol of water, or roughly 100 gram. You would need about 1 kg of ice to drop the humidity by about 10%.

On the "feels like" scale, this would drop the apparent temperature of the room about 1 degree (because less humidity feels cooler to the body). Note that using the ice to remove moisture from the air means that the heat is not available to reduce the temperature. This leads to an interesting conundrum: if you have a really cold cooling element (like the ice) you will first remove water from the air, and have less impact on the temperature of the air itself; if you could somehow cool the air more slowly so no condensation takes place (using a "cold plate" at a temperature above the saturation temperature for the air), you could cool the air down. Most cooling systems will operate in the condensing regime - in fact, the "air conditioning" unit started life as a "moisture removing" unit in order to get more consistent printing quality (which depends on moisture in the paper: that's the industry where Carrier worked when he developed the first unit, and that is the problem he was trying to solve).

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At 75% relative humidity and 90 F, there is abound 26 grams of water per cubic meter of air. If you had a sealed room 10 ft x 10 ft x 10 ft, at 90 F, reducing the humidity from 75% to 25% would require removing 615 grams of water vapor from the air, or 327 grams at 70 F.

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  • $\begingroup$ How much ice would that take? Or how much work (in the thermodynamic sense) would that take? $\endgroup$ – john mangual Nov 1 '14 at 14:39
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According to Engineering Toolbox: the water vapor in air at 90°F and 70% RH is about 9.85 g per pound of dry air. The maximum amount of water vapor at 32°F is about 1.62 g per pound of dry air. So, depending on the air flow around your glass, it can condense a lot of water.

Is it possible to reduce humidity of my room (10 feet × 10 feet ) in a substantial way using ice and a glass?

There are commercial units called swamp coolers. Unfortunately, they only cool the air, they do not decrease humidity.

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