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Calorimeter problem

  1. May 5, 2006 #1
    I don't even know where to start with this problem, can someone please help.

    A block of ice at 0 degress C is added to a 150g aluminum calorimeter cup which holds 210g of water at 12 degrees C. If all but 2.0g of ice melt, what was the original mass of the block of ice?
     
  2. jcsd
  3. May 5, 2006 #2

    berkeman

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    What is special about the ending condition of the mix of ice and water? What is the equation for the temperature of a mixture that is in equilibrium now, but originally was a mix of two different masses at different temperatures?
     
  4. May 5, 2006 #3
    Its special because the ice didn't completely melt.
    And the equation I think is m(i)c(i)dT = m(w)c(w)dT ?
     
  5. May 5, 2006 #4

    berkeman

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    And what can you say about icewater that is in thermal equilibrium? Then you got it!
     
  6. May 5, 2006 #5
    I'm not sure. That the ice and the water would be at the same temperature?
     
  7. May 5, 2006 #6

    berkeman

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    Good. Now just do the math for the thermal equation, and that should give you your answer.
     
  8. May 5, 2006 #7
    I'm not sure that I'm doing this right. If the temperature is the same then ou only need m(i)c(i) = m(w)c(w)
    mi(2100) = 0.21(4186)
    mi = 0.004 kg

    I think I've missed something?
     
  9. May 5, 2006 #8

    berkeman

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    The energy content in the beginning (before the ice and water are combined) is equal to the energy content in the equilibrium condition at the end, when the ice and liquid are at the same temperature. You know the end temperature (what is it?), the starting temperatures, and the starting mass of the water. Use all that information to figure out the starting mass of the ice.
     
  10. May 5, 2006 #9
    ok so if i use 2g in the equation I get a end temperature of 11.9 deg C.

    And therefore mi(2100)(11.9-0) = 0.21(4186)(12-11.9)
    Solving for mi = 3.5 g

    Does this look right?
     
  11. May 5, 2006 #10

    berkeman

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    Icewater in equilibrium is not at 11.9C. If you need to, you can do a quick experiment with some icewater and a thermometer. But you should be able to figure this one out without the experiment....
     
  12. May 5, 2006 #11
    I kind of figured it wasn't right but plugging in the numbers thats what I kept getting. I'm assuming that the cup comes into play here, which I haven't used because I'm wasn't sure how.

    I tried something else and think that I may have it
    using m(i)c(i)(T-0) = m(w)c(w)(12-T) + m(c)w(c)(12-T)

    where (c) is the values of the aluminum cup.

    I found a temperature of 2.3 degrees C.

    This looks better to me, am I on the right track.
     
    Last edited: May 5, 2006
  13. May 5, 2006 #12
    I checked my math and actually got 11.9 degrees not 2.3! I'm missing something but I don't know what!!
     
  14. May 5, 2006 #13

    berkeman

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    The first thing we have to get straight is the final temperature of the icewater. How can water and ice coexist like that in equilibrium? Why doesn't the ice keep melting?

    Also, do you have any special things that you need to do in your equations to account for the fact that the ice that did melt went through a phase change? It takes energy to melt the ice, even though the temperature is not changing during the phase transition. Does your textbook talk about that?
     
  15. May 5, 2006 #14
    I'm thinking that in order for the ice to not melt any further both the water and the ice must be a 0 degrees. So the final temperature is zero?
     
  16. May 5, 2006 #15

    Hootenanny

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    Yes, the final temperature is zero. I don't know whether berkeman mentioned it already (I haven't fully read the thread), but when calculating the chnage in energy, you will have to take into account the calorimeter as it will also be in thermal equilibrium with the water / ice mixture.

    ~H
     
  17. May 5, 2006 #16
    so to bring the water down to zero I got m(w)c(w)(12-0) =10.5 kJ

    And for the final 2g of ice to stay in that state it would need

    m(i)L(f) = 0.002kg(333kJ/kg) = 0.666Kj

    Now I'm not sure where to go with this?
     
  18. May 5, 2006 #17

    berkeman

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    Thanks Hoot. I spaced the aluminum cup part. JG, you also need to take into account the heat energy that goes into the melting of the ice. Look up "heat of fusion" in your textbook, and merge that information with the equation that you've been working with so far.
     
  19. May 5, 2006 #18
    In my text there is only one example that combines these two things. So based on that this is what I've come up with.

    m(i)L(i) + M(i)c(i)(0-0) = m(w)c(w)(12-0) + m(al)c(al)(12-0)
    m(i)333 + 0 = (0.21)(2100)(12) + (0.15)(900)(12)
    m(i) = 20.8 kg

    This looks like too high of a number : (
     
  20. May 8, 2006 #19
    Ok I'm still working on this question and can't seem to figure out anything that is reasonable, if anyone out there can help it would be soooo appreciated.
     
  21. May 8, 2006 #20

    Astronuc

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