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Thermodynamics (final temperature, ice mixed with steam)

  1. Feb 11, 2013 #1
    1. The problem statement, all variables and given/known data

    If 10-kg of ice at 0 degrees Celsius is added to 2-kg of steam at 100 degrees Celsius, the temperature of the resulting mixture is? Use Joules. ANSWER: 40


    2. Relevant equations

    Specific Heats:
    Ice = 2060
    Steam = 2020

    Latent Heats:
    Water melting = 3.33x10^5
    Water boiling = 2.26x10^6


    3. The attempt at a solution
    Q1=mL
    Q1=10(3.34x10^5)
    Q1=3.34x10^6

    Q2=(2)(2.26x10^6)
    Q2=4.52x10^6

    Q2-Q1=1.18x10^6

    Q=MCΔT
    1.18x10^6 = (2+10)(4180)(T-0)
    1.18x10^6 = 5.0x10^4T
    T=23.6

    Sadly 23.6 =/= 40
    I've been trying to do this problem for over an hour now using every method I remember, stumped.
     
  2. jcsd
  3. Feb 11, 2013 #2
    I'm confused by the question. Steam is a suspension of water droplets in air, basically a cloud. Presumably you mean water vapour not steam?

    In this system there will a phase change from ice to water that costs an amount of latent heat. There will also be a phase change from vapour to liquid, which releases an amount of latent heat.

    Perhaps it is easier to think about in too stages. 1) The vapour releases its latent heat and becomes water. This heat is sufficient to provide the latent heat required to melt the ice to water at 0C, and then heat up the mass of water that was ice. So finally you're left with two blobs of water, 2kg at 100C and 10kg at some temperature above 0C.

    The second stage (that you missed out): the temperature of the two water masses equilibrate.
     
  4. Feb 11, 2013 #3

    collinsmark

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    Don't forget your units.
    This is correct so far, :approve: except for the lack of units.
    Your approach *would* be correct, *if* the entire 12 kg of [liquid] water started from 0 deg C, and increased its temperature from there. But that's not the case for this problem.

    Here, there is only 10 kg of [liquid] water, starting at 0o C, rising to temperature T, absorbing energy as it goes.

    There is also 2 kg of [liquid] water, starting at 100o C, falling to temperature T (making a temperature change of 100 - T), releasing energy as it goes.

    You'll need to re-set up your equation to account for both of these things. :wink:

    That's really a matter of semantics. But commonly, water in gaseous form is called steam. (In which case -- you are correct insofar that -- H2O is transparent when in truly gaseous form.)
     
    Last edited: Feb 11, 2013
  5. Feb 12, 2013 #4

    SteamKing

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    It would also be nice if some units were included in the calculations.
     
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