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Thermodynamics: Steam is added to an ice cube

  1. Jun 17, 2017 #1
    1. The problem statement, all variables and given/known data

    Vapors of 100 C is added to an ice cube of 0 C.

    How much of the ice cube has melted and what's the final temperature if the masses of the steam and the ice cube are 10.0 g & 50.0 g respectively?

    2. Relevant equations

    Lw = 3.33 x 105 J/kg
    cw = 4186 J/kg*C

    Q = m*c*ΔΤ
    Q = L*Δm

    Qcold = -Qhot

    3. The attempt at a solution

    To be honest this part of the chapter has me confused a bit because I'm not sure what to do each time and what rules to follow. For example, I looked back to an exercise about vapors being added to water, and it had the vapors being liquified, then having their temp lowered, the water getting its temp raised, and then going with the Qcold = - Qhot equation to find the final temperature.

    Here though, I've tried all the combinations I can think of, and I still can't find the right answer.

    1) First, I tried with the vapors being liquified and having their temp lowered, the ice cube being liquified as well and having its temp raised, and then putting that in the Qc = -Qh. Wrong result.

    2) Then I tried with just the vapors getting liquified not the ice cube. Wrong again.

    3) Then I figured it happens instantly, and went straight for Qc = - Qh. Wrong one more time.

    Generally I don't understand how I can use my data to find both unknown factors. I don't know how to approach this, because I'm not sure if I should assume that the vapors and/or the cube start gettingliquified first and then do the "exchange" though Qc = -Qh, or if its happens instantly. Furthermore, I don't see how I can find two unknown variables from just one equation.

    Any help is appreciated!
     
  2. jcsd
  3. Jun 17, 2017 #2

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    Hi Darthkostis,

    Let's stick to (1) for now.
    The constants are incomplete.
    The given latent heat ##L_w## is for liquefying ice.
    We have a different latent heat for vapor condensation, which is ##L_{v\to l}=2257\text{ kJ/kg}##.
    Can we find the final temperature assuming that the final phase is liquid?
     
  4. Jun 17, 2017 #3
    Hello!

    Yeah, I noticed that after I saw your comment. It was all the way to the right of the book's "board" so I missed it.

    Well, my problem is that I don't know how to combine the various formulas/put the transformations in order. From what I gather, not all of the ice liquifies, so I have two unknown quantities, the amount of the ice cube that was liquified, and the final temperature. But at the same time, I only have one equation Qcold = - Qhot.
     
  5. Jun 17, 2017 #4

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    How much energy does it free up to condensate all of the vapor, and how much energy does it take to liquefy all of the ice?
     
  6. Jun 17, 2017 #5
    Alright, here goes:

    Qv-l = Lv*Δmv = Lv*(0 - mv)= - Lv*mv = ... = - 22,6 kJ

    Qi-l = Lf*Δmw = Lf*(mi - 0) = Lf*mi = ... = 16,65 kJ

    Note: mi = mass of the ice cube / mw = mass of the water / mv = mass of the vapors

    If I'm not mistaken, in the formula Q = L*Δm, Δm concerns the material of the highest phase, with the hierarchy being water > ice & vapor > water, right?
     
  7. Jun 17, 2017 #6

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    Right.
    So let's assume they end up at the same unknown temperature ##t## in the liquid phase.
    How much heat do we get from the condensated vapor going down to ##t##?
    And how much heat does it take from the liquefied ice going up to ##t##?
     
  8. Jun 17, 2017 #7
    Qiw = cw*mi*ΔT = (4186 J/kg*C)*50.0*10-3kg*(Tf - 0) = 209.3 JC-1*Tf

    Qvw = cw*mv*ΔT = (4186 J/kg*C)*10.0*10-3kg*(Tf - 100C) = 41,86 JC-1*(Tf - 100C)
     
    Last edited: Jun 17, 2017
  9. Jun 17, 2017 #8

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    Good!
    So what's the equation for the heats to be the same?
    Can we solve it for the final temperature?
     
  10. Jun 17, 2017 #9
    Well, the general formula is: Qcold = - Qhot where Hot: Qvl & Qvw / Cold: Qil & Qiw

    Since the vapors, and the liquid they turn to, get colder, and the ice cube, and the liquid it turns to, get hotter. So (I forgot to add the J at the last two Qs in my previous post):

    Qil & Qiw = - [ Qvl + Qvw ] <=> ... <=> Tf = 40.4 C (which is the book's answer)

    Question: Doesn't the question imply that not all of the ice cube melted? So how can I assume it in order to find the final temperature? That's where I got stuck.
     
  11. Jun 17, 2017 #10

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    There are 3 possible end states.
    1) All vapor condensated and all ice liquefied, and they meet at a temperature in between.
    2) Some ice remained, and we have a liquid at 0 degrees.
    3) Some vapor remained, and we have a liquid at 100 degrees.

    When calculating for (1) we found a solution didn't we?
    If not, we'd have tried (2) or (3) depending on the results for (1).
     
  12. Jun 17, 2017 #11
    So if I had, say, mi = 50 g & mv = 1.00 g, and obviously case (1) didn't come true, how would I tackle it? This time I couldn't assume that the whole ice cube gets liquified, so I'd have two unknown variables.
     
  13. Jun 17, 2017 #12

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    Then we'd have that all vapor condensated and went down in temperature to 0 degrees.
    And some unknown fraction ##\Delta m## of the ice liquefied.
    We'd solve for this ##\Delta m## instead of the final temperature.
     
  14. Jun 17, 2017 #13
    Ah, so Tf = 0 essentially. I put that in, change the numbers, use Qc = - Qh again and that's it, right?
     
  15. Jun 17, 2017 #14

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  16. Jun 17, 2017 #15
    Not to hijack the thread, but the next exercise is very similar. This time we have a 250 g ice cube at 0C and 600g water at 18C. I did the same thing as before (ice melts completely) and ended up with Tf = -10.[...] which is obviously wrong. So based on the logic from before, I should assume that Tf = 0C. Then I will go to the beginning, and instead of assuming that all of the ice melted, I'll put (mf - mi) with mf being the unknown quanity. Then put all that in Qc = - Qh, to find the final mass of the ice cube which remains, right?
     
  17. Jun 17, 2017 #16

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  18. Jun 17, 2017 #17
    Alright!

    Thanks a ton for the help, seriously! You were very patient and walked me through this in great detail!
     
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