Re: Elementary thermodynamics question
From: José Carlos Santos (jcsantos_at_fc.up.pt)
Date: 09/29/04
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Date: Wed, 29 Sep 2004 17:12:22 +0100
Paul Draper wrote:
>> > This is still not enough information. If the room is full of air, the
>> > heat transfer rate to an ice sphere is still actually quite low
>> > COMPARED TO the thermal conductivity of the ice, depending on the
>> > temperature of the air. Is the air warmer than the melting point of
>> > water?
>>
>> Yes, the air is warmer than the melting point of water.
>>
>> Best regards,
>>
>> Jose Carlos Santos
>
> OK, then you've got a complicated problem, much more than originally
> posed.
> 1. Heat will be transferred to the ice from the gas. You could use the
> specific heat of ice but this will only work if heat is being
> transferred VERY VERY slowly by lowering the pressure of the air or
> reducing the temperature difference to a very small number.
> 2. More realistically, heat will have to conduct through the ice from
> the outside to the inside, and so you'll have a temperature gradient
> across the ice, from surface to center.
> 3. However, as soon as the temperature of the surface of the ice gets
> to 0C, then the temperature at the surface will stop rising and some
> of the heat transfer to the ice will go to melting the ice at a rate
> calculable from the latent heat of fusion of water.
> 4. Meanwhile, however, the ice will still continue to approach thermal
> equilibrium, with the remaining heat transfer flowing inward to lower
> the temperature gradient.
> 5. Meanwhile, however, the ice that does change to water will drip off
> the sphere and fall to the bottom of the container, and now the mass
> of the ice will change, altering in real time your caculation in step
> (1). If you are operating in zero-gravity, at the most ideal, you have
> a sphere of ice inside a shell of water, and now you have to worry
> about heat conduction through two materials and a (different)
> temperature gradient across each one.
> 6. The ice that has melted continues to absorb heat from the
> surrounding gas, but with a different specific heat (that of water,
> not ice). Moreover, the temperature gradient in the water is
> complicated by the fact that it is a fluid and so convection is now a
> real contributor to heat flow within the water.
>
> Honestly, I don't know how to realistically calculate this without a
> stepping computer program that can simulate the processes
> incrementally and simultaneously.
>
> Now if you have a ball of ice at -15 C and a surrounding gas at -5 C,
> then I think my exponential solution earlier posted will approximate
> quite well.
OK. Thanks a lot.
Best regards,
Jose Carlos Santos
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