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Finding the final temperature when steam and ice are mixed

  1. Nov 21, 2012 #1
    1. The problem statement, all variables and given/known data
    Suppose that 20.0g of steam at 110°C is mixed with 25.0g of ice at -40°C What will the final temperature be?

    Ice - Specific heat capacity 2.10 J/g°C
    Water - Specific heat capacity 4.19 J/g°C
    Water - Latent heat of fusion 334 J/g
    Water - Latent heat of vaporization 2268 J/g
    Steam - Specific heat capacity 2.08 J/g°C


    2. Relevant equations
    Q=mcΔt
    Q=mH

    3. The attempt at a solution
    Heat loss (steam) = heat gain (ice)
    mcΔt+mHvap+mcΔt = mcΔt+mHfus+mcΔt
    (20.0g)(2.08J/g°C)(10°C)+(20.0g)(2268J/g)+(20.0g)(4.19J/g°C)Δt = (25.0g)(2.10J/g°C)(40°C)+(25.0g)(334J/g)+(25.0g)(4.19J/g°C)Δt
    416J+45360J+83.8J/°C(Δt)=2100J+8350J+104.75J/°C(Δt)
    45776J+83.8J/°C(Δt)=10450J+104.75J/°C(Δt)
    35326J+83.8J/°C(Δt)=104.75J/°C
    [(421.5513126)(1/°C)]Δt=1.25Δt
    But then I get lost and confused as to what I need to do to isolate the variable. Any help or solutions would be greatly appreciated. Thanks in advance
     
    Last edited: Nov 21, 2012
  2. jcsd
  3. Nov 22, 2012 #2

    ehild

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    Do not use the same notation for different things. What is Δt in case of the steam and what is it in case of ice?

    ehild
     
  4. Nov 22, 2012 #3
    Well the first time /\t appears on the steam side of the equation it is equal to 10. And the first time it appears on the ice side of the equation it is equal to 40. But the second time it appears on both sides is the variable I'm trying to solve for because it is the final temperature of the mixture.
     
  5. Nov 22, 2012 #4

    ehild

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    Δt is not the final temperature of the mixture.

    ehild
     
  6. Nov 22, 2012 #5
    Then how would I find the final temperature?
     
  7. Nov 22, 2012 #6

    ehild

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    We use the symbol Δ for difference, for change of some quantity. ΔT is change of temperature, ΔT=T(final)-T(initial). You added the heat released when the steam cools down to 100 °C to the heat released when it becomes water at the boiling point, that is at 100 °C. That water cools down to the final common temperature. So what is the ΔT you need to use for the steam-water?


    ehild
     
  8. Nov 22, 2012 #7
    I'm not sure I understand what you're saying. How are you supposed to come up with the the final temperature? That's the whole point of the question. The steam water will have the same temperature as the ice water but I don't know how to solve for that.
     
  9. Nov 22, 2012 #8

    haruspex

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    Create an unknown variable, T say, for the temperature you're trying to determine. Then you can replace each Δt with the appropriate difference between two temperatures. Some of those will involve T.
     
  10. Nov 22, 2012 #9
    So for the steam it would be (110-t) and the ice would be (t-40), correct? But the rest of the equation posted above would stay the same?
     
  11. Nov 22, 2012 #10

    haruspex

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    No, that's wrong in a couple of ways. I presume you meant (t-(-40)) = t+40, but that would be wrong too. The delta t's you are replacing in the equation by these terms are the changes in temperature of what, exactly?
    Also, your equation does assume that the final state is all water. When you have solved it, how will you know whether that assumption was correct?
     
  12. Nov 22, 2012 #11
    Yes that is what I meant, with the t-(-40). These changes would be the final temperature. We know that the steam drops 10 degrees then condenses but we don't know the final temperature which would be the 100-t and we know the ice gains 40 degrees and melts and the final temperature would be t-(-40). Well the final temperature has to be the same, so when the equation is solved we will know the state because if the temperature is 0<t<100 it's water, and if t<0 then it's ice and if the t>100 then it's steam. But for sure it can't be steam, because the ice would have to gain a substantial amount of energy to become steam. Since we assume no heat was gained or lost to the environment, the only energy available is in the ice or steam.
     
  13. Nov 22, 2012 #12

    haruspex

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    No, the final temperature is t. 100-t is the drop in temperature of the water after it condenses. Note that this is the right quantity to multiply by the s.h. of the water, not (110-t) as you wrote before. And for the melted ice, what temperature difference should you use there?
     
  14. Nov 22, 2012 #13

    ehild

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    When the steam cools down to its boiling point Q1=c(steam)m(steam) (T(steam)-100) heat is released.
    During condensation, Q2=m(steam)L(heat of vaporization) is released.

    When the condensed steam cools down from 100 °C to the final temperature T, Q3=c(water)m(steam)(100-T) heat is released.

    q1=c(ice)m(ice)(0-(-40)) is the heat needed to warm up the ice to 0°C.
    q2=L(heat of fusion)m(ice) is needed to melt it and gain m(ice) water at 0°C.
    q3=c(water)m(ice)(T-0) heat is needed to warm up the ice-water to the common final temperature T.

    Q1+Q2+Q3=q1+q2+q3, if the final stage is water. But you can not be sure that all the ice is melt or all the steam condenses. So calculate the amounts of heat separately and see if the released heat of steam is is enough to warm up the ice, then melt it.

    Q1=416 J..............q1=2100 J
    Q2=45360 J..........q2=8350 J

    You see that Q1+Q2>q1+q2, so all the ice will be melt.

    It can happen that the released heat is enough to warm the ice-water to 100°0. How much is the heat q4 needed to warm up the ice-water to the boiling point?

    If the heat of the steam is enough to warm up the ice-water to 100°C then you have 20+25 g water at 100 °C and Q1+Q2-(q1+q2+q4) heat available to evaporate some water. Is it enough to evaporate all?


    ehild
     
  15. Nov 22, 2012 #14
    Ok so I tried this again
    Heat loss (steam) = Heat gain (ice)
    mcΔt+mH+mcΔt = mcΔt+mH+mcΔt
    (20g)(2.08J/g°C)(10°C)+(20g)(2268J/g)+(20g)(4.19J/g°C)(100-t) = (25g)(2.1T/g°C)(40°C)+(25g)(334J/g)+(25g)(4.19J/g°C)(t-0)
    416J+45360J+83.8J/°C(100-t) = 2100J+8350J+104.75J/°C(t-0)
    Combining like terms
    45776J+83.8J/°C(100-t) = 10450J + 104.75J/°C(t-0)
    35326J+83.8J/°C(100-t) = 104.75J/°C(t-0)
    35326J+8380J/°C-83.8J/°Cx = 104.75J/°Cx
    35326J+8380J/°C = 188.55J/°Cx
    187.356139(1/°C)+44.444444444444444444444444444444 = x
    231.8005834°C = x
    232°C = x
    But the final temperature can't be higher than the starting temperature of steam, because the steam is already the hottest substance.
     
  16. Nov 22, 2012 #15

    ehild

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    Yes, so the final state is not all water. Try the second part of my post. Is the heat of the steam enough to warm up all the ice to 100 °C?

    ehild
     
  17. Nov 23, 2012 #16
    Well this has already been answered by solving for the variable in the equation I posted before this post. That variable balances out the equation, because you can see that it was very unequal before. T is the final temperature which came out to 232 degrees. You will get the same answer trying to see if all the ice can become steam as you suggested. But if you think about this question logically, 20g of steam at 110 degrees isn't going to be able to change the temperature of a greater mass of ice by 140 degrees without condensing. Plus we know for a fact the the final temperature has to be between -40-110 degrees.
     
  18. Nov 23, 2012 #17

    ehild

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    Well, what is the final temperature?

    ehild
     
  19. Nov 23, 2012 #18
    Well I came up with 232 degrees which isn't the right answer
     
  20. Nov 23, 2012 #19

    ehild

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    You supposed that the final state is all water which results in an impossible final temperature. So the final state is not just water. I tried to explain it in post #13.
    The steam cools down to 100 °C and some condenses. The released heat is enough to warm up the ice to 0°C, melt it and warm up the ice-water from 0°C to 100 °C. How much heat is needed to that? How much steam need to condense to provide that heat?

    ehild
     
  21. Nov 24, 2012 #20
    Well according to the equation, the steam releases enough heat to melt the ice and raise the temperature of the mixture to 232 degrees. Something about the equation itself has to change, but I don't know what
     
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