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Lee's Disc experiment for Thermal Conductivities

  1. Sep 19, 2013 #1
    Here is the experiment in brief for those uninitiated :
    http://media.uws.ac.uk/~davison/labpage/leedisk/leedisk.html

    My question is, when the steady state has been obtained (heat flowing into Brass Disc = heat flowing out of it), why do we remove the upper Brass Base and then heat the disc? Why not with it? We need to find the value of rate of fall of temperature at the steady state temperature T1 ; if we remove the Brass Base, are we not changing our initial conditions, since now the glass will be in direct contact with air, an extremely poor conductor, while in the other case it would have been in contact with Brass.

    Is it because while obtaining the steady state, heat is flowing out of Base only in the downward direction, and we want the same to happen for the cooling curve? If so, why?

    Thanks!
     
  2. jcsd
  3. Sep 19, 2013 #2

    AlephZero

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    Without the lower brass disk, the temperature over the lower surface of the disk would not be uniform, because of the convective heat transfer to the air. The temperature is made uniform by the high thermal conductivity of the brass.

    Also, think about why the thermometers are "inside" the brass disks, and how you would measure the temperatures accurately without the disks. (Measuring the air temperature close to the bottom surface of the disk with a thermometer is no good).
     
  4. Sep 19, 2013 #3

    Philip Wood

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    What you're trying to find is the rate of heat loss from the sides and bottom of the lower brass plate (since this is equal to the rate of heat entry, through the insulating disc, from the heated top brass plate). This rate of loss of heat is found by heating the bottom plate and letting it cool without heat being lost or gained through the upper surface. Hence no top heated plate, but retain the insulating disc to inhibit heat loss from the top surface of the lower brass plate.
     
  5. Sep 19, 2013 #4
    I think my naming of the apparatus might have caused some confusion (I used the link). I am talking about removing the upper brass piece, not the lower one. Or maybe I don't follow what you are saying - can you dumb it down please?

    I understand that we need to find the rate of heat loss from the sides and bottom of the lower brass piece. But does adding an insulator on the top really stop heat loss from the top? Glass is a better conductor than air, so if heat can flow down into the air(a worse conductor than glass), it should be able to flow up into the glass as well, right?
     
  6. Sep 20, 2013 #5

    Philip Wood

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    Right. But leaving on the insulating disc will help. Without the disc, the top surface of the bottom plate will lose heat mainly by convection - far more important than conduction here, because air is a very poor conductor. Of course, with the disc in place, there will still be some convective loss, as the top of the disc will be warm, but being a poor conductor the disc will have a large temperature difference between its two flat faces, i.e. the top face will be cooler than the top face of the bottom plate, so the convective heat loss rate will be reduced. Having said that, I've always thought that the experiment would benefit by use (in the second half) of a better insulator than the insulator being investigated in the first half!
     
  7. Sep 20, 2013 #6
    Ah, I think I understand. Thanks Philip and Alpeh!
     
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