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Homework Help: Heat transfer by thermal radiation

  1. Aug 20, 2015 #1
    1. The problem statement, all variables and given/known data
    The ice is placed in water, we know the surface area A, emissivity of both ice and water, Stefan-Boltzmann constant and the temperature of both ice and water. What is the equation for heat transfer rate between water and ice?

    2. Relevant equations
    Heat transfer rate by radiation = [PLAIN]https://upload.wikimedia.org/math/9/d/4/9d43cb8bbcb702e9d5943de477f099e2.png*A*e*T^4 [Broken]

    3. The attempt at a solution
    I assume there are two options at least, one where we need to calculate radiation from both water and ice, and the difference will be output of whole system, and the othere one would be just water radiating the energy toward ice, so water directly surrounding ice would have the same surface area as ice.
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Aug 20, 2015 #2


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    Welcome to the forum.

    Ok, I didn't see a question in there. But I'm going to assume it is the following. Which of the two options should you use? Well, of course, ice has a temperature as well. So it will radiate also, though at a lower intensity. So you can't just ignore it. The ice will be at 0 degrees C, presumably. The water presumably at room temperature of 20 degrees C or so, presumably. So the difference of the fourth power of T_water and the fourth power of T_ice is what you need.

    As far as "directly surrounding" and such considerations, you probably have not done much on the transmission of thermal radiation through matter. But the thermal radiation you get from water at room-temperature-ish type temperatures will have only the very smallest ability to travel through matter. It will get absorbed very quickly indeed. If you think about how thermal radiation is emitted that will make sense. If the material can radiate a frequency, it is probably pretty good at absorbing it as well. So you probably can ignore anything except the surface in this calculation.

    Probably you then get to think about whether this radiation is important in the melting of ice. If you find the expected rate of energy transfer, you can estimate how much ice will melt per second. And then you can estimate how long it takes to melt an ice cube of some particular mass. Then you can estimate whether this is a reasonable comparison with how long it really takes to melt.
  4. Aug 20, 2015 #3
    Your last paragraph kinda describes what I am doing :D
    I checked conduction, convection is waay complicated for me, and radiation is left, so I am trying to get a good prediction equation. When you said to ignore everything except the surface area, did you mean all the constants? I don't see the point of that. Anyway, thank you very much for help.
  5. Aug 20, 2015 #4


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    Convection is way complicated for thermal hydraulics engineers with decades of experience. This is one reason, among many, why there are a lot of experiments done on complicated systems. If a system is at all complicated it is often cheaper and quicker to do an experiment and measure it than it is to try to calculate it accurately.

    When I said you should look at only the surface, I meant you should not try to worry about thermal radiation from inside the material. In some cases you might have to worry about radiation produced inside the material that makes its way out. But thermal radiation from water will get absorbed very quickly. In other words, the T^4 equation you have, which is based on the surface, is probably ok.

    In some situations the radiation produced is not thermalized, or can escape more easily. If it was such a case you might have to include radiation terms from the volume, not just the surface. If that were the case you would have to include the volume of the water the ice was floating in, not just the area.
  6. Aug 20, 2015 #5
    Thank you very much for your help, thanks to you I am able to continue my project. Interesting thing about convection you said, last time I checked there was a quite simple formula for it, only one constant was complicated as it was responsible for conveying the position of object in surrounding. But that constant being there means that it can be calculated for every single case, can't it?
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