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Specific Heat Capacity of Aluminium

  1. Apr 14, 2007 #1
    Hi all,

    I am currently in the process of writing up my AS coursework investigation of finding the specific heat capacity of Aluminium, and am a bit stuck at explaining some of my results.

    My results for the SHC of Aluminium are lower than the published values.

    My value: 700.93 JKgC
    Published: 918 JKgC

    I expected my result to be higher due to heat loss to surroundings, but why has my calculated value come out lower?

    Here are the calculations just to make sure i've done everything right:

    Current = 1.64
    Voltage = 5.71
    Time = 900
    Initial Temp = 21
    Final Temp = 33

    Energy supplied to block = power x time
    = V x I x T
    = 1.64 x 5.71 x 900
    = 8427.96 J

    Energy Received by block = mass x specific heat capacity x change in temp.

    specific heat capacity = (V x I x T) / (M x deltaT) JKgC
    = 8427.96 / (1.002 x 12)

    Specific heat capacity = 700.93 JKgC

    So why is my result lower than published 'true' value? Please if you can provide a detailed explanation so i can get my head around this as i really did expect a higher value than the 'true' one.

    Thanks for any help!
     
  2. jcsd
  3. Apr 14, 2007 #2
    Did you monitor current and voltage the entire time? Or are these initial values?
     
  4. Apr 14, 2007 #3
    Thanks for the reply!

    They are the average values for the voltage & current for the entire investigation.
     
    Last edited: Apr 14, 2007
  5. Apr 14, 2007 #4
    ok, probably only a couple of percent anyway. Do you know the type of aluminum, was this 100 percent pure?
     
  6. Apr 14, 2007 #5
    I don't, sorry. It was just a big 1kg cylindrical block. I'm guessing however that it wasn't 100% pure
     
  7. Apr 14, 2007 #6
    So should the value have been higher than the published one? In other words does heat loss make my calculated value higher or lower?

    its just something really confusing me at the moment, the reason to why my value is too low, or why it may be too high in other circumstances...
     
  8. Apr 14, 2007 #7
    well if your eqn is right, which it appears the larger the delta T, the smaller the heat capacity. with no loss, delta T would be maximal.
     
  9. Apr 14, 2007 #8
    I'm really confused :(, i seem to be getting more confused by the minute trying to get my head round this. . .

    What possible reasons are there behind my calculated result being lower than the 'true' value? Would you expect, if carried out correctly, the result to be a higher value than the 'true' one?

    Thanks for bearing with me!
     
  10. Apr 14, 2007 #9
    lets just make sure we are clear: delta t=applied heat/(mass*SHC)

    thus shc=k*applied heat/delta t
    the delta t at best can be that with no loss. That would lead to minimal number for SHC.

    So as you inferred, the error is in the other direction, so maybe there was error in applied heat calculation, which is why I asked, as with increasing t, resistivity becomes higher, so depending on setup, constant current v constant voltage, you might have delivered less heat than assumed. Other issue is whether SHC varies significantly with type of Aluminum? The value you have quoted might me for 99.99% crystalline aluminum and your sample varies from this considerably. I do not know. Doubtful SHC would vary that much, tho elecrical conductivity might. We need help of metallurgist, maybe?
     
    Last edited: Apr 14, 2007
  11. Apr 14, 2007 #10
    i noted that the initial current was 1.93A and the initial voltage was 5.71. The current dropped considerably (dont know why?) so i took this as an anomalous result... Don't know if this has much effect. . .
     
  12. Apr 14, 2007 #11
    I must have done something wrong....No further explanation surely....Can i just confirm that the value i should obtain in my practical should be higher than the published book values due to heat loss? and what other reasons?

    If so, i shall redo my investigation. . .
     
  13. Apr 14, 2007 #12
    well if the current dropped considerably, rephrase joule heating as I^2*R, my guess is that you have much less actual heat applied than we figured, which would explain result.

    edit: PS: have a little more time, If it were me, I'd redo the lab as it only takes 15 minutes of heating--the current should not have dropped by more than a few percent unless your power supply not very good. If you want to first consider what effect the heating of the block would do to resistance (and hence to current if you were applying fixed voltage), this can be looked up.

    First case V=constant. R is fx of T such that at Tfinal=Rf=R+deltaR
    then I droops be a factor V/(delta R).
    Second case I=constant in such case you would have not seen the big drop you mentioned. In one case the heat delivered would be more than expected if observing whatever is assumed to be constant, and the other less.
     
    Last edited: Apr 15, 2007
  14. Apr 19, 2007 #13
    Hi again,

    I repeated the experiment and obtained a value of 1045 JKGC. This is now higher than the published value (900).

    Now i need to think of reasons to which the value i obtained is higher than the published value. I know of heat loss, but how is the heat lost?

    Also, are there any other reasons other than heat loss of why my results obtained are higher?

    Thanks for your help!
     
  15. Apr 19, 2007 #14
    hey thats awesome, and a very reasonable result. I think 15 percent seems like a reasonable amt of heat loss and see no need to invoke other mechanisms. I assume it was just sitting out in the open with no efforts at insulation etc. If you wanted to really go extra mile, I don't think it would be too hard to set up eqns to estimate heat loss of rectangular solid with given current flux, or could do something simpler.
     
  16. Apr 19, 2007 #15
    Yes, i didn't insulate the block. I would like to go the extra mile, could you help me with deriving an equation to estimate heat loss? What can i do that is simpler also?

    Good to hear the result sounds reasonable also!
     
  17. Apr 19, 2007 #16
    oh and by the way, the block is cylindrical
     
  18. Apr 19, 2007 #17
    for simpler, maybe just trying something like newtons cooling. The other one can probably be looked up or better yet, maybe an applet where you can plug in some numbers. The other involves setting up set of pde's which is a bit of a chore--try googling heat transfer and see what you come up with. I'm off to play golf! Its like 75 here in Denver today:cool:
     
  19. Apr 19, 2007 #18
    lucky! Although the weather is on the up here in England!

    Thanks for all your help. Unfortunately i can't carry out any more experiments and have to write up the investigation now - the thing i struggle most with is explaining things, which is why i need help with explaining why there is heat lost etc.

    Thanks again

    Joe
     
  20. Apr 19, 2007 #19
    Well its a shame I didn't think of this before you went to the lab, had you measured a couple of temps and times as it cooled back to room temp would have made this a snap. Not sure why they didn't suggest it as part of the exercise.

    For heating anytime object T greater than surrounds t, you lose heat according to type of transfer. Conductive is what Newton' cooling is based on. In your study, it would grow with the difference between the rod and roomtemp or whatever it was supported by. If it was short and thick you can see where the effect might be smaller, verson long and skinny heat capacity is a function of volume while heat loss depends greatly on area. Hope this helps a little bit and good luck with your studies.
     
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