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Mechanical equivalnet of heat

  1. Jun 13, 2006 #1
    Hello.
    I need to calc the mechanical equivalent of heat.
    I got a tube with some lead shot in it.
    Tube is 1 m long. Lead shot is 11.3 kg.
    It falls though 66 cm of the tube. (66 cm of the tube are empty).
    It was fliped 100 times
    temperature rose 10 degrees.
    SO we get:
    Q=11.3*130*10
    where 130 is the specific heat of lead.
    We got some number there.
    That is heat.
    Now we do:
    Joules of mechanical energy=11.3*0.66*9.8*100(number of times flopped)
    Now we get another number.

    The two number need to be close to each other right? Or am i missing something when i am relating mechanical energy with heat?
    the heat that was generated from Q=mcdeltat should equal the mgh#oftimesflipped, right?
    Sry for the dumb post, but i don;t have a textbook on hand and i haven;t done this in a long while...
    Thx
     
    Last edited: Jun 13, 2006
  2. jcsd
  3. Jun 14, 2006 #2
    the energy you have given there is potential energy of a falling object. the mechanical energy is the total energy which is KE+V. the loss of kinetic energy is usually converted to heat or sound, so the energy due to heat will be less than the potential energy.

    I think I am right in saying this
     
    Last edited: Jun 14, 2006
  4. Jun 14, 2006 #3
    y woudl it be less?
    Cuz the way heat is generated is when the lead falls the 66 cm shaft and hits the bottom, it generates heat. So the change from mgh to Ke to Heat energy shoudl all be equivalent. THere will be major loses due to sound and becuase the system is not adiabatic, but under ideal conditions should the q=mct and mgh*#offlips should be the same rite?
    Thx
     
  5. Jun 14, 2006 #4

    andrevdh

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    Eleven kilograms is quite a lot. Don't you mean 11 grams?
     
  6. Jun 14, 2006 #5
    The kinetic energy gained by the lead shot is going to be less than the potential energy because you lose some heat to sound.
     
  7. Jun 15, 2006 #6

    andrevdh

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    As you swing the lead from the bottom to the top you give it some kinetic energy. When you stop it at the top this energy needs to go somewhere (it does not simply vanish into thin air). You did work on the lead shot to give it this kinetic energy, it is not a result of a decrease in potential energy. Quantifying this change in kinetic energy will be challenging though.
     
  8. Jun 16, 2006 #7

    andrevdh

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    Doing experiments is quite a different process than doing calculations in Physics. When we do calculations we ignore details in order to get to the answer. In real life this is not possible. All effects are in operation and will influence the result. This makes it difficult to get to the bottom of things. This means that you are on your own when you do a experiment. Unless you can find someone who has done a similar experiment you have to draw your own conclusions. Having said that I confess that I have not done the experiment myself so these are suggestions as to what may be the cause of the discrepencancy in your results. It is up to you to investigte their validity or reject them on a theoretical basis.

    The first attachment shows some interaction forces arising on the lead shot as they hit the bottom of the tube. It is these interactions that cause the lead shot to heat up.

    The second attachment shows the same sort of interaction forces arising as the tube is turned sideways.

    In the third attachment one sees the same situation arising as the turning motion is stopped at the top.

    Could it be that an acceleration of the lead shot heats it up? If so it is experiences an acceleration all the way up due to a required centripetal force. Admittedly it is much smaller, but it acts over a longer time.

    One can envisage an experiment where this turning effect is eliminated with maybe a tube with a platform at the bottom and an attached plunger sticking out. The platform can then be raised with the lead on it and quickly lowered. The same interacting forces will arise when the lead shot is accelerated initially. This suggests that one should raise them very slowly in order to minimize the magnitude of this effect. So we conclude that it is not just the falling down that caused the lead shot to gain heat, but also the raising. Could this maybe explain the resulting temperature to be double what one would expect?
     
    Last edited: Nov 29, 2006
  9. Jun 19, 2006 #8

    andrevdh

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    I am not to happy to report the result of an investigation that I made regarding the increase in temperature of lead shot due to the conversion of mechanical energy. I mounted a bicycle wheel horizontally and checked it with a radially mounted spirit level. Next a bottle with 308 grams of lead shot was mounted to the rim at a distance of (35.0 +/- 1.0) cm from the axle and rotated from the one side to the other halfway along the perimeter 100 times. A temperature increase of only 0.7 degrees celcius was observed. My previous hypothesis that the rotation of the tube might be responsible for the additional temperature increase therefore seems to be wrong since the expected temperature increase due to the conversion of the potential energy should be
    [tex]\Delta T = \frac{9.79 \times 70}{128}\ =\ 5.4\ C^o[/tex]
    in this case.
     
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