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Specific Heat Capacity Answer Wrong, yet no error to be found

  1. Mar 18, 2007 #1
    1. The problem statement, all variables and given/known data

    My teacher instructed my class to conduct an experiment to determine the specific heat capacity of Brass, Copper, Aluminium, and Steel. The following setup was used:

    http://mypage.bluewin.ch/gutter/diagram1.JPG [Broken]

    The following method was used:
    1.) The electric heater (still off) and thermometer are placed into the metal and left for a while.
    2.) The start temperature is recorded, the stopwatch is started, and the power pack is turned on.
    3.) After aproximately a 10 degree C increase in temperature the timer and power pack are turned off.
    4.) As the temperature is still rising, sit and watch the temperature and record the highest temperature it reaches.
    6.) Repeat for rest of metals

    These were the results

    http://mypage.bluewin.ch/gutter/results1.JPG [Broken]

    2. Relevant equations

    I used the formula http://mypage.bluewin.ch/gutter/formula1.JPG [Broken] to plug in my results (calculating Q by multiplying the time in seconds by 50 (the power of the heater)) and came up with:

    748.9 J kg-1 C-1 for Brass
    831.2 J kg-1 C-1 for Copper
    1615.3 J kg-1 C-1 for Aluminium
    747.4 J kg-1 C-1 for Steel

    The acutal results were supposed to be (according to my teacher):

    370 J kg-1 C-1 for Brass
    385 J kg-1 C-1 for Copper
    913 J kg-1 C-1 for Aluminium
    420 J kg-1 C-1 for Steel

    As you can see, my results were about double what they were supposed to be

    3. The attempt at a solution

    I studied my results much with several people yet I could not find the correct error. My first reaction was to think that much heat had been given off, yet I had used an insulator and even without an insulator, this would not double your results! My second thought was that I had an error in my calculations (sorry I didnt post them, they were a bit long), yet redid the formula three times and everything worked out perfectly. My next suspicion was that my teacher had given me wrong results to confuse me yet a textbook soon disproved that theory. I finally looked at the experiment itself with my father to find an error, but all hope prevailed. I studied several other Methods posted on the internet, yet I noticed no large difference between mine and theirs (they were the same at the base).

    In conclusion, I belive the culprit was the heat being given off, even though metal was insulated and placed on a very inconductive surface.

    Any and all help would be of great appreciation
    (sorry about the lack of equations and such, I am only in high school so I have a limited knowledge of Math and Physics, although I try my best to learn more)
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Mar 18, 2007 #2


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    a quick question: you said that t he power was turned off when the temperature had increased by about 10 degrees. So does that mean that the power was not on for the entire time that you quote in your table? Wouldn't you have to use only the time during which the power was on to find the value of Q? Maybe I misunderstood.
  4. Mar 18, 2007 #3
    sorry my fault, I misworded the thing https://www.physicsforums.com/images/smilies/frown.gif [Broken].[/URL] What I ment was the power pack was left off so that the thermometer could adjust itself to the original temperature and then later turned on with the stopwatch, emmiting 50W until the temperature had been increased by about 10 degrees. It was then turned off, yet the thermometer was still observed until the experiment reched peak temperature, so as to let the rest of the heat in the heater into the metal.

    (I hope that clears it up a litte)

    Just a quick note: I forgot to mention that half the class also had the exact same problem.
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  5. Mar 18, 2007 #4


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    I am still not quite sure I understand (sorry). When you give 275.6 seconds for brass, is that the time the 50W was on? I amguessing that what you are saying is that this 275 sec is indeed the time during which the power was on, and after turning it off you waited until the temperature reached the max value, right? So the time it took to reach 39.4 degrees was actually larger than 275 seconds, right?

    Did half of the class get the (roughly) correct result? Was the power source AC or DC volts?
    Last edited by a moderator: May 2, 2017
  6. Mar 18, 2007 #5
    yup :) thats completely correct

    I made a mistake again (sorry), by typing half instead of whole. Everyone in the class was around +-100 what I got. The power was DC, 12V.

    (Oh, and thanks for your time and effort)
    Last edited: Mar 18, 2007
  7. Mar 18, 2007 #6


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    You are welcome. I am sorry, I can't see the problem :frown:. If the source had been AC I would have pointed out that the power must be an rms (root mean square ) value and that could possibly have introduced a factor of 1/2. Maybe you are right, that lots of heat was lost (was the top surface insulated? Did the air above it felt very warm?) Was it the first time the prof was doing this experiment?

    I hope someone else will be able to find the problem!
  8. Mar 18, 2007 #7
    Thanks anyway though. :)

    Further notes:

    The top surface was not insulated, yet it didn't feel warm. I doubt that it was the profs first time.
  9. Mar 18, 2007 #8


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    And you mentioned your problem to the prof? So I am assuming that the same setup had been used before and had given correct results? If that's the case, that would mean that there has to be a mistake in your procedure (as opposed to a problem with the setup itself). I am still bothered by the 50 W. Did the prof give you that figure or was it indicated on the power pack?
  10. Mar 18, 2007 #9
    the figure was written on the heater (something like 12V will cause the heater to emmit 50W). During class the prof just said that we would have something to go home and think about, he didnt want to tell us (although I doubt he knew). A mistake in the procedure would make the most sentence, yet I cant see one. I thought that it might have been that I would have had to read the temperature right when I turned off the power pack, not when it reached peak, yet when I took one of those results I used it with the formula, but it didnt work out. :(
  11. Mar 18, 2007 #10


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    No, it does make more sense to wait until the temperature is maximum.
    Sorry, I can't see :cry: . You are sure the power supply was DC? (it was a battery, it was not a transformer connected to an outlet?). The only thing I can think off that could introduce a factor of 2 is this AC vs DC aspect but other than that I have no clue (unless the mass was really 2 kg :wink: ).
    I hope someone else may come up with another idea.
  12. Mar 18, 2007 #11
    The power pack was connected to and outlet but I made sure that I connected the wires to the dc slot in the power pack. The mass could have been 2kg as the teacher told us the metals were made to be exactly 1kg yet noone checked. I doubt this though as the metals felt more around 1kg than 2kg. (Thanks for your help though)
  13. May 6, 2011 #12
    seems you ve done it without any calorimeter or any insulation.. heat loss through radiation and convection could add to erraneous situations..
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