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Determine the Specific Heat Capacity

  1. Dec 6, 2009 #1
    Hello, everyone.

    1. The problem statement, all variables and given/known data
    Now, I recently did an experiment in school to determine the Specific Heat Capacity of rubber using a calorimeter. We're using the mixing method to figure it out.
    The mass of the calorimeter, rubber and water was measured. And the S.H.C is already known for the water and calorimeter. The temperatures are as follows:

    Temp. of cold water = 26°C
    Temp. of hot object = 95°C
    Temp. of mixture = 32°C



    2. Relevant equations
    Mo co θo = Mc cc θc + Mw cw θw

    0 = object
    c = calorimeter
    w = water



    3. The attempt at a solution
    The problem I'm having is finding the temp. change. This is what I did.
    To find the temp. change of the rubber = 26 + 32 = 58. And from that 95 - 58 = 37.

    To find the temp. change of the calorimeter = 95 - 32 = 63.

    To find the temp. change of the water = 95 - 26.

    I'd like to know if this is accurate because, when I do the calculations I get a really weird figure for the S.H.C of the rubber.
     
  2. jcsd
  3. Dec 6, 2009 #2
    Re: Calorimetry

    I'm confused--what was hot before mixing and cold before mixing. I'm assuming the rubber and calorimeter were both hot and the water cold.
     
  4. Dec 6, 2009 #3
    Re: Calorimetry

    My apologies, for not stating it properly.

    The temp. of the mixture is the temp. of the heated rubber inside the calorimeter filled with water.

    The temp. of the cold water is the temp. of the water inside the calorimeter alone.

    The temp. of the hot object is the temp. of the heated rubber in a beaker of boiling water.
     
  5. Dec 6, 2009 #4
    Re: Calorimetry

    very well. So the temp difference between the rubber before and after is 95-32. The rise in temp of both the water and calorimiter by adding the rubber is 6 degrees.
    Try those and see what you get. The signs of the temp changes should both be positive BTW.
     
  6. Dec 6, 2009 #5
    Re: Calorimetry

    Okay, I understand. Reading through your post made me realize that what you said makes more sense than what I did. There is no need to use the 26°C for the calculation of the temp. change of the rubber, right? The 32°C already has the 26°C included.

    And, there's no need to use the 95°C for the calculation of the temp. change of the water and calorimeter because, the 32°C already has the 95°C included.

    Is the way I stated it correct? I just want to know if I understand the logic behind it so I won't have any problems in the future.
     
  7. Dec 6, 2009 #6
    Re: Calorimetry

    You're interested in the heat gain on the one hand, so thats final temp minus initial temp for the water/calorimeter combo and heat loss on the other hand, which is again the initial temp minus the final. Heat is neither created or destroyed if we are careful so the heat loss (temp times mass times capacity) of the rubber is equal to the heat gain of the H20/calorimeter. That help. What value did you get by the way>
     
  8. Dec 6, 2009 #7
    Re: Calorimetry

    Oh, okay. I see. Many thanks, for that explanation.
    I got 2162 J/kg K. The answer was rounded off. Was I supposed to convert the degrees Celsius to Kelvins before I made the calculations?
    Because that's what I did.
     
  9. Dec 6, 2009 #8
    Re: Calorimetry

    Degrees K is customary but the caclulation should be the same. The result falls squarely in the range cited for india rubber.
     
  10. Dec 6, 2009 #9
    Re: Calorimetry

    Is that right? Well, I have no idea what type of rubber it is. So, I guess I'll stick with that answer. On Monday, I'll see if it's correct or at least close to being correct from my lecturer.
    Many thanks, for your help.
     
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