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A question about calorimetry

  1. Mar 26, 2005 #1

    chem_tr

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    I haven't found my textbooks on physical chemistry, so I want to ask the forum members.

    When we heat 4 liters of water from 10°C to 90°C, and add this to 16 liters of water at 10°C, what will the final temperature be?

    My brainstorming revealed the following, but I'm not sure if they are correct or not (my physical chemistry really sucks):

    I am almost sure it is wrong, so could anybody please show me the very logic of this type of calorimetric problem?

    Thank you.
     
    Last edited: Mar 26, 2005
  2. jcsd
  3. Mar 26, 2005 #2

    Bystander

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    You have a total of 20 liters, not 16. That help?
     
  4. Mar 26, 2005 #3

    GCT

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    [tex]mc(x-90)=m_2c(x-10)[/tex]

    -x represents the equilibrium temperature

    -heat lost by the 90 degree water is gained by 10 degree water

    I'm pretty sure this is correct
     
  5. Mar 26, 2005 #4
    Let me try:

    [tex](\mbox{heat lost})=-(\mbox{heat gained})[/tex]
    [tex](mc\Delta t)_{1}=-(mc\Delta t)_{2}[/tex]
    [tex](4\mbox{kg})(4.19\mbox{kJ/kg}\cdot^{\circ}\mbox{C})({t_{f}-90^{\circ}°\mbox{C})=-(16\mbox{kg})(4.19\mbox{kJ/(kg}\cdot^{\circ}\mbox{C})({t_{f}-10^{\circ}\mbox{C})[/tex]

    Solving for [itex]t_{f}[/itex] gives 26°C.
     
  6. Mar 27, 2005 #5

    Borek

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    I will try it slightly other way - let's assume 10 deg C is a 'basic' state. Your heating adds 4*1000*80 calories. Now you use that heat in the 20000 ml of water. So your delta T is 4*1000*80/20000 = 16 which gives final temperature of 26 and is exactly the same result Sirus posted.


    Chemical calculators for labs and education
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  7. Mar 27, 2005 #6

    chem_tr

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    Oookay, so I have made the error of not considering the total volume as Bystander suggested; my logic doesn't seem so wrong. Thank you all!
     
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