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Equilbrium temp of ice and steam mixture

  1. Feb 12, 2014 #1
    1. The problem statement, all variables and given/known data
    You are asked to mix equal masses of of ice and steam at 0.00°C and 100.00°C respectively, What will the final temperature of the mixture be? The mixture is in a perfect insulator.

    ms=mass of steam
    mi=mass of ice
    LHV,s=Heat of vaporization of steam=2257J/g
    LHF,i=Heat of fusion of ice=334J/g
    Ti,s=100°C=373K
    Ti,i=0.00°C=273K
    cw=4.19J/(g*K)

    2. Relevant equations
    Q=mL
    Q=mcΔT


    3. The attempt at a solution

    I am fairly confident that my thought process is correct. I just keep making algebra mistakes(I assume) but I can't find what my error is. Please let me know where my process breaks down please.

    I did this in 3 parts

    Part 1)
    So first I consider the situation.
    Since the mixture is in a perfect insulator, that means no heat is lost to the surroundings.

    Q=0

    That means that the heat lost by the steam is gained by the ice. Q for the steam will be negative since it is losing heat.

    Q=0=Qgained+Qlost=Qgained+(-Qlost)

    Qlost=Qgained

    From here, the Qlost is given by the sum of energy used to turn steam into water, then to bring the steam-turned-water to thermal equilibrium

    Qlost=mscw(Ti,s-Teq)+msLHV

    and Qgained is given by the sum of energy used to melt the ice, then to bring the ice-turned-water to thermal equilibrium.

    Qgained=micw(Teq-Ti,i)+miLHF

    Again, since there is no heat lost to the surroundings, and the mixture is allowed to reach thermal equilibrium then...

    Qlost=Qgained

    Part 2)
    Now we plug in to be able to solve for Teq

    mscw(Ti,s-Teq)+msLHV=micw(Teq-Ti,i)+miLHF

    mscwTi,s-mscwTeq+msLHV=micwTeq-micwTi,i+miLHF

    Moving all terms containing Teq to one side

    mscwTi,s+micwTi,i+msLHV-miLHF=micwTeq+mscwTeq

    Factor out Teq

    mscwTi,s+micwTi,i+msLHV-miLHF=(micw+mscw)Teq

    and finally solving for Teq results in

    Teq=mscwTi,s+micwTi,i+msLHV-miLHF/(micw+mscw)

    We can do one more thing, since the masses are equal, we can factor out the mass.

    Teq=mcwTi,s+mcwTi,i+mLHV-mLHF/(mcw+mcw)

    Teq=m(cwTi,s+cwTi,i+LHV-LHF)/m(cw+cw)

    Teq=(cwTi,s+cwTi,i+LHV-LHF)/2cw


    Part 3)
    Plug in known values

    Teq=[(4.19J/(g*K)(373K)+(4.19J/(g*K))(273K)+(2257J/g)-(334J/g)]/[(2*4.19J/(g*K))]

    Teq=552K Obviously incorrect


    Please let me know where I am making my mistake. I seperated my work into 3 parts to make it easier to point out where i made an error. Thank you I have a test on Friday I NEEEEEED to understand how to do this properly. Thank you
     
  2. jcsd
  3. Feb 12, 2014 #2

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    First, there is no need to bring absolute temperature into this problem. A 1 degree centigrade change in temperature is equal to a 1 degree Kelvin change in temperature.

    Since you have equal masses of steam and ice, you know the equilibrium temperature of the mixture must lie somewhere between 0 C and 100 C.

    In problems of this type, it's sometimes easier to pretend you are doing bookkeeping with heat instead of money. Instead of writing out some long equation where it is easy to make a mistake, set up a table. You know how much heat steam @ 100 C contains per unit mass, and you know how much heat ice @ 0 C contains per unit mass. After accounting for the latent heats of fusion and evaporation, you should be able to write a simple equation to determine equilibrium temperature of the liquid formed after the phase change is complete.
     
  4. Feb 12, 2014 #3
    The only reason I did was so that whenever I multiplied any terms with the initial temperature of the ice it didn't zero out.

    I understand this, this would be easier. My goal is to be able to do this problem with all masses, temperatures, inside of containers of different material. I NEED to do this algebraically since our exam will consists of concept questions and solving for different terms where the values aren't given.
     
  5. Feb 12, 2014 #4
    Personally, I would prefer to work through the algebra rather than doing SteamKing's bookkeeping technique.

    You missed one major simplification that you could have done very early in your algebra. You noted that you can factor out an m from everything. However, you kept m around for a while. You can divide out m in your first algebra step and cut your pencil mileage in half.

    If you do that, you should be able to find your algebra mistake. I found something, but I think you corrected it. WHen you solved for Teq, you divided by (micw+mscw), but you are missing parentheses around the numerator in the next step.
     
  6. Feb 12, 2014 #5

    SteamKing

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    Staff Emeritus
    Science Advisor
    Homework Helper

    I don't understand this statement. What terms 'zeroed' out?

    Still, if you are going algebra only, at least try for simple algebra. Writing one equation which covers half a page is hard to understand. At least find out how much heat exchange occurs to get the ice and steam both to liquid phase first. There's no requirement everything be done in one step.
     
  7. Feb 12, 2014 #6
    Yes I chose the long way to force myself to keep track of everything no matter how complicated but I re-did it. This time I cancelled out the masses from the beginning and I came up with the same equation for Teq.

    Teq=[cwTi,s+cwTi,i+LHV-LHF]/[2cw]

    Plugging the same values in gives me the same answer

    Teq=[(1562.87J/g)+(1143.87J/g)+(2260J/g)-(334J/g)]/[8.38J/(g*K)]
    Teq=(4632.74J/g)/8.38J/(g*K)
    Teq=552.83 K

    So I'll work out my algebra in detail here.....here goes

    Qlost=Qgained

    mLHV+mcw(Ti,s-Teq)=mLHF+mcw(Teq-Ti,i)

    Now from here I factor and cancel the masses

    m[LHV+cw(Ti,s-Teq)]=m[LHF+cw(Teq-Ti,i)]

    Then becomes

    LHV+cw(Ti,s-Teq)=LHF+cw(Teq-Ti,i)

    Distributing the temperature differences...

    LHV+cwTi,s-cwTeq=LHF+cwTeq-cwTi,i

    Moving all the terms containing the Teq to one side

    LHV+cwTi,s-LHF+cwTi,i=cwTeq+cwTeq

    Factoring out the Teq

    LHV+cwTi,s-LHF+cwTi,i=(cw+cw)Teq

    Now solving for Teq

    So I have

    Teq=[LHV+cwTi,s-LHF+cwTi,i]/(2cw)

    OR if it's clearer this way....

    In the numerator we have
    LHV+cwTi,s-LHF+cwTi,i

    Divided by (denominator)
    (2cw)

    Which is still the exact same thing. Did i make the same mistake?
     
  8. Feb 12, 2014 #7
    When I plug in values for the term cwTi,i, or anything multiplied by the initial temperature of the freshly melted ice, multiplying anything by 0.00°C will make that term 0. That term has been "zeroed out"
     
  9. Feb 12, 2014 #8
    You can see what's happening from your final equation. The heat of condensation of the steam is so high that the ice doesn't have enough oomph to even condense all the steam. Try the problem assuming that only a fraction of the steam condenses. Of course, the final temperature will have to be 100C. You will be solving for the fraction of the steam that condenses.

    Chet
     
  10. Feb 12, 2014 #9
    ah got it! you are correct. Bear with me as I am trying to do this without any notes or equations since I want to be able to do this properly during the exam.

    So let's say that the energy required to turn the steam into liquid is given by msLHV

    if I know that the melting of ice will contribute to the steam turning into water. So I could say that some fraction(α) of steam will have turned to water when the ice fully melts.

    αmsLHV=miLHF

    Knowing that the masses are the same will leave me with
    αLHV=LHF

    α=LHF/LHV

    That's a proportion. Is that right so far?

    So when the steam melts all the ice we have α% left.
    Next we have to calculate the rest of the steam turning into water.

    αLHV=micw(T2-Ti,i)

    The T2 is the temperature that the water is once the steam has fully converted to water.

    Lastly we calculate

    mscw(Ti,s-Teq)=micw(Teq-T2)

    mscwTi,s-mscwTeq=micwTeq-micwT2

    Solving for Teq

    mscwTi,s+micwT2=micwTeq+mscwTeq

    Teq(micw+mscw)=mscwTi,s+micwT2

    Finally

    Teq=(mscwTi,s+micwT2)/(micw+mscw)

    I haven't done the algebra by hand for it that's why I didn't cancel out the masses but was that the correct process?
     
  11. Feb 12, 2014 #10
    Unfortunately, no. If there is any steam left in the end, the final temperature is going to be 100C. You need to calculate the amount of heat it takes to melt all the ice and raise its temperature to 100 C. That will tell you how much steam can be condensed at 100 C. That will all become water at 100 C. The rest of the steam will remain steam at 100C.
     
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