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Calorimetry with phase change

  1. Nov 1, 2014 #1
    1. The problem statement, all variables and given/known data
    An aluminum cup (of negligible mass) is filled with 0.5 kg of water with a temperature of 15◦C and 1.3 kg of ice (at -8 celsius) is added.

    A)What is the final phase or phases of the mixture?
    B)What is the final temperature of the mixture?
    C)How much heat is required to raise the mixture to 15◦C?

    2. Relevant equations
    Q = mcdeltaT
    Q = mL

    3. The attempt at a solution
    I guess my main confusion would be how to predict the phase it would end up in, how to tell if you were wrong , and if you are wrong what to do next. This is my solution for part A and B but I think my result is wrong.

    I assumed the ice melts to water and is brought to a final temp

    [itex] heat gained by ice = heat lost by water [/itex]

    [itex] Q_{ice to 0 C} + Q_{ice to water} + Q_{water to T_f} = Q_{water to T_f} [/itex]

    [itex] m_i c_i(0-(-8)) + m_i L_F + m_i c_w(T_f - 0) = m_w c_w(15-T_f) [/itex]

    [itex] m_i[c_i(8) + L_F + c_w(T_f)] = m_w c_w (15 - T_f) [/itex]

    mass of ice mi = 1.3kg, spec heat of ice ci = 2100 J/kgC, mass water mw = .5 kg, cw = 4186 J/kgC, heat of fusion of water = 33300 J/kg

    [itex] 1.3[2100(8) + 33300 + 4186T_f] = (.5)(4186)(15-T_f) [/itex]

    [itex] 1.3(50100 + 4186T_f) = 2093(15 - T_f) [/itex]

    [itex] 65130 + 5442T_f = 31395 - 2093T_f [/itex]

    [itex] 7535T_f = -33735 [/itex]

    [itex] T_f = -4.48 C [/itex]

    but if this was the case the ice would not have even started melting yet, it would just have lowering in temperature a few degrees. also I think it means all the water would have frozen? Im not sure how to fix this result
     
  2. jcsd
  3. Nov 1, 2014 #2

    haruspex

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    I'm not aware of any simple way to predict the final combination. You just have to make a guess and see if it works out.
    Right.
    You mean, raised a few degrees.
    Maybe not all.
    So what results, as combinations of states, are possible answers? Pick one and do the calculation.
     
  4. Nov 1, 2014 #3
    I think im doing something wrong. first i tried a calculation where the ice goes to 0 C but doesn't fully melt. the set up was like this
    mi ci (0 - (-8)) + miLf = mw cw (15 - Tf)
    but i ended up with an even larger negative number than before

    then i tried that it doesnt even get to 0 C
    mici(Tf + 8) = mwcw(15-Tf)
    but i ended up with positive 2 degrees.

    are those the right set ups for each situation?
     
  5. Nov 1, 2014 #4

    haruspex

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    Not quite sure what you are doing there. If all of the ice goes to 0C but not all melts you should have two different ice masses. And the final temp of the water must also be 0C.
    There is perhaps a slightly better way. First, suppose it all ends up as ice at 0C. Calculate the heat lost by the water and the heat gained by the ice, and see how much heat you have left over. If it's negative, chill the ice to match. If it's positive, melt ice to match. Only if the remaining mass of ice ends up negative do you have to try another scenario.
     
  6. Nov 1, 2014 #5
    so you mean try it assuming that the water freezes and all of it is ice at 0 C?
    mici(0-(-8) = mwcw(15 - 0) + mwLf
    1.3(2100)(8) = .5(4186)(15) + .5(33300)
    21840 = 48045
    so the water woud lose 26205 J more joules of heat than the ice gains
    so i have to melt more ice? am i understanding right?
     
  7. Nov 1, 2014 #6

    haruspex

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    Yes.
     
  8. Nov 1, 2014 #7
    so to find out how much ice to melt
    26205 = mi (33300)
    mi = .787 kg
    .787 kg of ice needs to be melted? im confused. what now? do i conclude that the final phase is a mix of ice and water and the final temp is 0?
     
  9. Nov 2, 2014 #8

    haruspex

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    Yes.
     
  10. Nov 2, 2014 #9
    thank you so much! i think i get what you are saying now. if the mass came out negative i would have to try a different scenario. as long as it was a positive mass it is in the middle of melting so the temp has to be 0 and the phase is a mixture of liquid and solid. Thanks so much my teacher never showed us a good method for solving these problems when phase changes are involved but now i think i am starting to get it.
     
  11. Nov 2, 2014 #10
    for part C the set up would be
    Q = heat for ice to get to 0 C + heat for ice to become water + heat for water to get to 15 C
    Q = mici(0-(-8)) + miLf + (micw(15-0)
    correct?
     
  12. Nov 2, 2014 #11

    BvU

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    I thought the ice is already at 0o ?
     
  13. Nov 2, 2014 #12
    yeah i was confused about that. idk if they want how much it would take to raise the whole thing to 15 from the beginning or how much it would take to raise it to 15 from part B
     
  14. Nov 2, 2014 #13
    if they mean from part b would it be valid to take the amount of heat exchanged in b and substract it from the heat exchanged for what i set up for c?
     
  15. Nov 2, 2014 #14

    BvU

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    That makes no difference for the amount of heat to be added!

    And therefore this amount has nothing to do with the amount of heat exhanged in b) either.

    Your setup for c) in post #10 is perfectly applicable: it's as if you bring the ice to 15o in a separate cup and then put it together with the water that started and ends at that temperature: all heat needed is needed for the ice.
     
  16. Nov 2, 2014 #15
    thank you!
     
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