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Metal sheets of various mass left out in the Sun

  1. May 19, 2015 #1
    These are questions I've made up to try and wrap my head around the topic, so the theory within the questions themselves might be flawed.

    1. The problem statement, all variables and given/known data

    There are a number of metal sheets that are similar in every way except they each have slightly different masses. The temperature of the air in contact with the individual sheets will increase more with the heavier sheets, and less with the lighter sheets.

    a) What thermal properties describe this situation?

    b) If a small plastic square is placed in the middle of a sheet, how will this affect the average temperature of the air in contact with the sheet as a whole (the entire front surface including plastic)?

    2. Relevant equations


    3. The attempt at a solution
    a)
    I'm reasonably sure it has to do with the specific heat capacity of the metal. The sheets will absorb heat from the Sun via radiation. As the heat is absorbed the temperature of the metal will go up. The sheet will also be transferring heat to the air in contact with it, so the temperature of the air will go up. We can prove that the heavier sheets will heat up (and increase the temperature of) the air more from the specific heat capacity equation:

    Q = mcΔT

    So because the same amount of heat Q from the Sun is hitting each sheet, if the mass m increases then so will the temperature.

    b) For this part I'm a little uncertain. I think the thermal conductivity of the plastic will be the important bit here because it will be significantly lower than the metals'. This means that it is not able to effectively absorb the heat from the Sun - which means that it can't transfer the heat to the air in contact with it.

    Because the plastic is not contributing much to heating the air, we can say that it is reducing the amount of radiation hitting the surface of the metal sheet. Basically - the exposed surface area of the sheet has been reduced.

    This in turn means the sheet will be absorbing less heat, which means it won't be transferring as much heat to the air in contact with it, so the temperature of the air won't be as high as if there was no plastic square.


    Are those answers good logic? Especially for part b) is it the thermal conductivities that is key?
     
    Last edited: May 19, 2015
  2. jcsd
  3. May 19, 2015 #2
    Actually thinking about it a bit more, for a) if the sheets have the same dimensions therefore the same exposed surface area, would the amount of heat Q hitting the sheets be constant? Because the same amount of radiation will be hitting each one?

    Edit: I think I'm confusing myself. The amount of heat reaching the sheets will be the same, but the heavier sheets will be able to retain more heat due to their increased mass so Q wouldn't be constant in fact.
     
  4. May 19, 2015 #3

    CWatters

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    I'm not convinced that is true.

    That's not correct. When the sheets are first put out in the sun they will start warming up. If they are all the same except for their mass then that equation just proves the more massive sheet will take longer to heat up. Q is constant (same sun, same size sheet) so if M is larger delta T will be smaller. If the temperature gain is smaller for each unit of Q then it will take longer to heat up.

    As they warm up they will start to lose/emit more energy. It is possible/likely that they will all reach the same stable temperature after some time (if it doesn't get dark first).
     
    Last edited: May 19, 2015
  5. May 19, 2015 #4

    CWatters

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    Continued..

    If they do reach a stable temperature then at that temperature the heat gain from the sun equals the heat lost to the surroundings. So I would expect that air next to each sheet to be the same.

    If by the end of the day they haven't yet reached the same stable temperature then the more massive sheet will be colder than the lighter one. So the air next to it will be colder.
     
  6. May 19, 2015 #5
    So let's assume they have been left in the sun for a long time and are in equilibrium.

    Each sheet will be at the same temperature right? But the amount of heat hitting each sheet (therefore the amount of heat lost to the surroundings) is also the same. So the air on the surface of each sheet will be heated exactly the same meaning the mass has no impact?

    It may be the equation that is causing my confusion, because all the variables would be the same except for the mass.
     
  7. May 19, 2015 #6

    CWatters

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    Yes exactly.
     
  8. May 19, 2015 #7
    Ok great, thanks. That has cleared it up for me - but just to fully make sure I have another hypothetical for you if you don't mind:

    Now we have one thin sheet and one cube, again the same type of metal. The front surface areas of both are equal which means the same amount of radiation is hitting both (assume the sun is static so it is only ever the front face of the cube that the sun shines onto).

    Once they are both in equilibrium (assuming the sun shines on them for long enough) then like before they will be the same temperature and the air in contact with the front surfaces will be heated the same amount. Right?
     
  9. May 19, 2015 #8

    CWatters

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    Probably not...

    First, they may both reach an equilibrium (eg temperature is constant) but that temperature will probably be different for the sheet and the cube.

    That's because the heat loss is likely to be proportional to surface area. So the one with the larger surface area is likely to have a lower equilibrium temperature. If you are familiar with the inside of a computer...big heat sinks help the CPU to run cooler.
     
  10. May 19, 2015 #9
    Yes that makes sense, thanks again.

    You've been very helpful so I am going to push my luck and ask if you have any criticism of my logic for part b)?
     
  11. May 20, 2015 #10

    CWatters

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    If the plastic is small in area compared to the metal it won't have much effect if any significant effect on the air temperature. If it's larger...

    The specific heat capacity of plastic is about 2 to 4 times that of metal. So you might expect the plastic to take longer to heat up if they both absorbed the same amount of heat, however the emissivity (how easily a material radiates energy) depends on the material and the surface finish. Plastic has quite a high emissivity compared to polished metal.

    http://en.wikipedia.org/wiki/Emissivity



    So on balance I'd expect the plastic to take longer to heat up and to be in equilibrium at a lower temperature then the case with all metal.
     
  12. May 20, 2015 #11

    CWatters

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