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Specific Heat Capacity Question Mixture Q=MCdeltaT

  1. Nov 5, 2011 #1
    1. The problem statement, all variables and given/known data

    A mixture is made by adding 75g of an unknown liquid at a temperature of 25 degrees Celsius to 60g of water at a temperature of 90 degrees Celsius. The final temperature f the mixture is 65 degrees Celsius. Calculate the specific heat capacity of the liquid. What is the liquid? How are you sure that you are correct?

    2. Relevant equations
    Q=m*c*deltaT


    3. The attempt at a solution
    I don't even know where to start. You can't calculate Q for water because C is not 4.181 at 90 degrees C. You can't find the Q of the unknown liquid because we do not know it's c value...
     
  2. jcsd
  3. Nov 5, 2011 #2

    Ygggdrasil

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    You are correct that the heat capacity of a substance (C) varies with temperature. However C does not change very much with temperature, so unless we are doing very precise calculations, we usually just assume stays constant at all temperatures.
     
  4. Nov 5, 2011 #3
    Ok, so in that case/

    Is it correct to find Q for water with c=4.181, delta T = -25 degrees C

    set that equal to mcdeltaT for the unknown liquid, rearrange and solve for c of the unknown and then match it up with a table??
     
  5. Nov 5, 2011 #4

    Ygggdrasil

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    That sounds like a good plan.
     
  6. Nov 5, 2011 #5
    So c, specific heat capacity, changes only slightly between temperatures, so we can assume they are the same always? When does this not work?
     
  7. Nov 5, 2011 #6

    Ygggdrasil

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    Unless the material changes phase (i.e. liquid water turning into ice or vapor), it's a pretty good assumption that the heat capacity will not change significantly.
     
  8. Nov 6, 2011 #7

    Borek

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    Staff: Mentor

    See for example this table. Specific heat of liquid water between 0°C and 100°C doesn't vary by more than 1%.

    Apparently even without phase change differences can be substantial when temperature approaches critical point.
     
  9. Nov 6, 2011 #8
    Ok so
    Qwater=mcdeltaT
    =60g*4.181J/gdegree C*(65-90degree C)
    =-6271.5J

    Qunknown=6271.5J (flip signs since the energy is absorbed by the unknown liquid)
    c=-6271.5J/(75g)(65-25degree C)
    =2.0905J/gdegree C

    Now I don't know what it is, according to wikipedia tables for specific heat capacity it should be steam??? or is it ice? But the question says liquid...
     
    Last edited: Nov 6, 2011
  10. Nov 6, 2011 #9

    Borek

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    Staff: Mentor

    Do you have a table with specific heats in your textbook, notes, course materials? If so, look there.
     
  11. Nov 6, 2011 #10
    Ʃ
    I have, but it's pretty much split between ice or steam, which makes no sense since the temperature is too low to form steam and too high to form ice.
     
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