# Heat loss due to Evaporation in a closed environment

1. Aug 6, 2010

### kinga

I am working on an incubator/shaker for laboratory use. I am trying to work out a temperature failure and repair it, but that is besides the point here. I was looking through the user manual to try to get some clues about the failure and I came across this:

"Depending on various conditions within the chamber, such as flask placement and size, the heat produced by growing organisms, heat losses due to liquid evaporation from flasks, etc., the display temperature may differ from temperatures within the flasks themselves. "

I was confused by the idea that there can be heat loss due to evaporation inside a closed environment where everything within this unit is set to the same temperature for hours at a time. The point of the incubation unit is to warm everything up to a set point and hold it there for a set period of time. If the water in the flasks and the air around it is at a uniform temperature (37c) it seems that the heat energy would not be lost due to evaporation but simply moved around the closed chamber. Am I missing something?

thanks
-Kinga

2. Aug 6, 2010

### Petr Mugver

I don't know much about incubation chambers, but probably the heat loss you described is the so called "latent heat": Some energy is needed during the phase transition from liquid to vapor. This energy is not lost, but it is given back when the vapor turns to liquid again. Did I answer the question?

3. Aug 6, 2010

### kinga

Thank you for your fast response.

That does make sense, but it seems that there would not be an appreciable amount of heat loss in the rest of the fluid when the water changes states. This unit measures the temperature with a +/- .5 degree error margin, so the small amount of heat that is drawn away during the phase transition would be extremely small, therefore negligible. It seems like the only loss would be in fluid volume within the flask because the vapor would condense on the inside surface of the incubator and not return to the flask (the flasks are open to the chamber air).

I guess my question is:

Why would the manufacturer add the heat loss due to evaporation as a factor, if the effect is so negligible as to not really mater in the overall chamber temperature?

4. Aug 6, 2010

### Petr Mugver

As I told you, my little knowledge allows me to make hypothesis, nothing more. I imagine that you set a certain temperature, and a system of lamps/resistances together with a thermometer sets up a feedback mechanism that maintains that temperature... You say that in equilibrium conditions the temperature must be uniform, but maybe in presence of heat sources/wells this equilibrium is constantly disturbed, so that a gradient of temperature is always present, even if there's the feedback system. And it may happen that temperature measured in the location the thermometer is is different from that under an egg, for example. Every system has a certain relaxation time to reach equilibrium, and if the characteristic time of disturbances isn't long enough respect to the relaxation time, transient phenomena will occur.

Maybe I said a lot of stupidities... in that case I'm sorry.:uhh:

5. Aug 6, 2010

### kinga

That does make sense. I think also that with a full incubator, with several liquid filled containers all fluctuating slightly in temperature as the heaters cycle, will collectively remove heat from the rest of the liquid constantly. The vapor leaves the flasks and condenses on the cooler walls of the windows, so heat is not returned to the flasks. An incubator must be calibrated by measuring water temp inside the unit instead of just air temp. It is a slow process because of how slowly water absorbs and releases heat in comparison to air. The amount of heat lost by the transition to vapor is small and most likely doesn’t really matter, but the author must have wanted to sound like he had all his bases covered. I am going to have to just settle for that.

Your responses were very interesting and well thought out. I appreciate your input. You did help to answer my question.

thank you