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Homework Help: Specific Heat Capacity

  1. Jan 14, 2016 #1
    • Member advised to use the homework template for posts in the homework sections of PF.
    Equations: E=m*c* dTheta or T

    Experiment: I had two beakers of water with one thermometer in each beaker. One beaker had boiling water and the other beaker had room temperature water. I also had a 100g mass.

    I recorded T1 which was the intial temperature of the cold water.
    I then put the mass of 100g into the boiling hot water and waited 30 seconds. I recorded this temperature as Tm which I assumed was the temperature of the 100g mass.

    I then put the hot 100g mass into the cold water and recorded the maximum temperature rise and called this T2

    I then worked out that E=mass of water x 4200*T2-T1
    I then assumed that E of water = E of the mass
    So then worked out that E=0.1kg*C*Tm-T2 and found a value for C

    I then repeated this experiment for lower temperatures of Tm and therefore T2 and found a value for C... I found out that the lower temperatures of T got a lower specific heat capacity.. and the question is why?

    My take on it so far is that when I lift the 100g mass out of the boiling water and put it into the cold beaker it transfers heat energy to the surroundings and the higher the temperature of Tm the more heat energy it transfers. Is this correct?

  2. jcsd
  3. Jan 14, 2016 #2


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    Gold Member

    Specific heat capacity is actually temperature dependent, that's why in physics books you see a different value than chemistry books.in chemistry books they measure it at 25 C and in physics they average it from 0-100. Your explanation of heat energy being lost while moving through the air is reasonable. If you wanted you could calculate that energy loss.there are several factors you will need though, emmisivity of material, and I there are values for the convection loss and conduction loss but I forget what they're called.
  4. Jan 14, 2016 #3
    I doubt that the mass lost a significant amount of heat in moving from one beaker to the other. How long did it take?

    In the cold beaker, when you put the mass in and let the system equilibrate, you also changed the temperature of the beaker. Also, the table under the beaker? Did you take all that into account?
  5. Jan 15, 2016 #4
    Ok thank you.

    The transition between hot and cold beakers was only about 5-10 seconds. Aaah I didn't take into the account, yes maybe it lost some heat energy due to the transfer of energy to the table. But my teacher hinted it was something to due with the transition between taking the mass out of the hot beaker and placing it into the cold beaker.
  6. Jan 15, 2016 #5
    Also just a quick suggestion, do you think it could be due to the internal energy of the mass?
  7. Jan 15, 2016 #6
    With all due respect to your teacher, I doubt that. The beaker is going to have to heat up too, and that will be significant. And the table is a significant heat sink also. You should at least calculate the amount of heat that goes into the beaker. Maybe the heat loss to the table would be about the same as that. Put some sort of insulation under the beaker so it doesn't contact the table directly, or, better yet, have the beaker raised in the air by some sort of metal ring frame.

    If you want to test the heat-loss-in-the-transfer hypothesis, just do some additional tests where you do the transfer faster.
  8. Jan 15, 2016 #7
    The internal energy of the mass is the same as the heat capacity times the temperature, which is the thing you are measuring.
  9. Jan 15, 2016 #8
    Yeah this makes so much more sense, thank you so much chestermiller.
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