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Question about humidity in milk

  1. Nov 14, 2004 #1

    chem_tr

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    Hello everybody,

    A friend of mine asked me a question regarding one of his milk-processing techniques. He wants to learn if a mathematical formula or a logical approach can be devised.

    Okay, the question involves a 1 liter-volume-bowl, whose outer (bigger) radius is 160 mm. He didn't say the inner (lower) radius of the bowl, as I think he considered it not important. He places hot milk at 80°C inside the bowl, and after three hours, he sees that the temperature is lowered to 40°C. He is wondering the amount of water vapor (humidity) evaporated, but in a theoretical basis. I suggested him that he can try simple weighing measures at different intervals, but he insists on seeing a theory-based calculation scheme.

    Any suggestions?
     
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  3. Nov 14, 2004 #2

    Gokul43201

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    Isn't this a Raoult's (or Henry) Law problem. If you know the percentage water vapor in the surroundings (ie : the dew point of the air in the room, at that particular time), then at any temperature, you can find the equilibrium percentage of water in the solution...can you not?

    No, it does not seem like this is the best way to approach the problem. On the other hand, I think it's more likely that water will be added to the milk rather than removed from it. The vapor pressure will decrease with lowering the temperature. Why does he think there will be a loss of water ? And does the milk equilibrate (sit for a long time) at 80C before it is put in the vessel ?

    The biggest problem here is that the situation is hardly ideal, nor is it an equilibrium process. Anyways, I need to think about this some more.
     
  4. Nov 14, 2004 #3

    russ_watters

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    Heat dissipated through evaporation is quantity of water evaporated times heat of vaporization.
     
  5. Nov 15, 2004 #4
    It's a common homework problem in transport phenomena classes!...unfortunately, I've forgotten too many of the formulas from that class ;) The basic transport principle is: Q = K*(F2-F1)/R, where Q is a flow rate, F2-F1 is a driving force, and R is the resistance to the flow (note this applies to just about anything; mass diffusion, heat flow, even charge--where it's called "Ohm's Law"). K is a proportionality constant that's sometimes needed to harmonize the units. In the case cited, F2-F1 would be the difference in concentration between the atmosphere in the "boundary layer," and at some point beyond, and R would be due to collisions between the molecules of gas. By integrating the expression with respect to time, you get the amount of mass that was transferred in that time...actually, I think there's a much neater formula than the one I gave, but maybe this will help with the general concept.
     
  6. Nov 15, 2004 #5

    chem_tr

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    I think chemical engineers rather than chemists (like me) are more involved with these things. But physicochemical approaches can be devised like russ_waters' one. Please go on giving feedback, and he'll be grateful to you.
     
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