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Conservation of heat on a closed system

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data
    Hey folks, anyone who can help me with this one get a massive shoutout. Here goes: 75 kg of water at 45 degrees celsius is "mixed" with 70 kg of Aluminum at 2600 Degrees Celsius. What is the final state of both system and how much, If any, of liquid water will be left

    Assume the following conditions.
    Specific heat capacity :
    Water--4.18X10^3
    Aluminium-- 9.2X10^2

    Latent Heat of fusion
    Aluminium-- 6.6X10^5

    Latent Heat of Vaporization
    Water--2.3X10^6

    Boiling points
    Water-100

    Melting points
    Alumium--2519

    2. Relevant equations
    Q=mc(delta temperture)
    L_fusion or vaporavation=mL_f or mL_v

    I will represent the following as these variables
    Ti= temperture intial
    F=Final temperature of system
    w=Water
    a=alumium
    3. The attempt at a solution
    m_w*c_w(100-Ti_w)+m_w*Lv_w+m_w*c_w(F-100)+m_a*c_a(2519-Ti_a)+m_a*Lf_a+m_a*c_a(F-2519)=0

    Factoring out the F out of the two terms and isolating for it I get:

    F= [-m_w*c_w(55)-m_w*Lv_w+m_w*c_w(100)+m_a*c_a(81)-m_a*Lf_a+m_a*c_a(2519)]/(m_w*c_w+m_a*c_a)
    This results in a negative number. Any Ideas where I went wrong?
     
  2. jcsd
  3. Apr 15, 2015 #2

    SteamKing

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    We don't encourage "unit-free" calculations here at PF.

    You should show ALL units for your data and in your calculations.

    It avoids a lot of confusion and unnecessary back and forth between the OP and anyone wishing to help.

    You should also post your actual calculation results so it makes it easier to determine where your analysis went wrong.
     
  4. Apr 15, 2015 #3

    haruspex

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    With such questions you need to recognise that there are several possible final states: at one extreme, all the water vaporises and none of the aluminum soldifies; at the other, all the aluminium sets and no water vaporises. If you construct an equation according to the wrong final state you will get silly answers. (I haven't checked your arithmetic, so there are other possible explanations.)
    One approach is to go for a middling possibility, and if the answer is physically impossible it will show you which way to step. Another is to proceed in stages, first letting the aluminium cool and the water warm until one of them reaches its phase transition, etc.
     
  5. Apr 15, 2015 #4
    I see what you are saying, using the step by step approach, and getting each of the substances to their state changes and working from there, accordingly, I got something interesting happening. It so happened that when the water was in the Latent Vaporization state, The aluminium "ran out" of heat. I calculated this occurring at 87.07% of the way through the water vaporizing stage. However, I feel like the extreme difference in temperatures makes this scenario unlikely and almost improbable. With that being said, My calculation tell me that when The 70 kg of Aluminuim cools completely to 0 Celsius, 65.303kg of the water is in the vapour state (at 100*C) and the remaining 9.697kg is in liquid state (at 100* C). Does this seem probable or even realistic?
     
  6. Apr 15, 2015 #5
    Sorry about the confusion, SteamKing, Ill be sure to include units next time. However, all units above are in standard SI units.
     
  7. Apr 15, 2015 #6

    SteamKing

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    No, this is not possible. You are not going to have a bunch of icy cold aluminum surrounded by steam. The water started off at 45 °C, so it is impossible that the aluminum will wind up colder than this.

    You should work these types of problems knowing that heat flows from a hotter substance to a colder substance and stops completely when the temperatures of the substances are equalized.

    Since multiple phase changes are involved (and which occur at constant temperature), what haruspex recommended is apt: calculate the amount of heat transferred from the aluminum to the water at each step of the process, paying attention to the phases of the water and the Al.
     
  8. Apr 15, 2015 #7
    Ah, Yes, I see... I forgot to adjust for the initial temperature of the water. Would it then be more sensible to say that 70kg of 45*C Aluminium is surrounded by 10.957 kg of liquid water at 100*C and the remaining 64.043kg being water vapour at 100*C-- (this is after the adjustment for the 45 degrees) or would i have to go on use conservation of heat to find an equilibrium point between the 3 substances?
     
  9. Apr 15, 2015 #8

    SteamKing

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    There are only two substances: water and aluminum. The problem you have is that the water may exist in two different phases simultaneously.

    And, no, you are not going to have aluminum at 45 °C while the water is hotter. This is the same absurdity you had in the previous post.

    The heat transfer stops when all of the aluminum and all of the water, regardless of phase, are at the same temperature.
     
  10. Apr 15, 2015 #9

    haruspex

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    To elaborate a little on SteamKing's advice, there are three ranges of final temperature to be considered.
     
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