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Grade 11 Thermodynamics Question

  1. May 3, 2016 #1
    1. The problem statement, all variables and given/known data
    You add 50 g of steam at 150 degrees centigrade to 50 g of ice at negative five degrees centigrade. What is the final temperature?

    2. Relevant equations
    Q = mLfv
    Q = mcΔT

    3. The attempt at a solution
    Qg = Ql
    Q1 + Q2 = -Q3 - Q4
    mLf + mcΔT = -mLv + mcΔT [ΔT will be negative, making the symbol positive in front]
    (0.05)(3.4x105) + (0.05)(4200)(Tf+ 5) = (0.05)(2.26x106) + (0.05)(4200)(Tf - 150)
    210Tf + 18 050 = 210Tf -154 600

    And I think that here you can see the issue. I asked my instructor since I previously had ...= -mLf - mcΔT, but that gave -232 degrees centigrade.

    I'm hoping someone on here can help me. I always get stuck with these problems and the test is in two days, meaning I need to really understand these types of exercises or I'm screwed [these are ranked as the highest marked].

    Thanks for any help you can offer! :smile:
     
    Last edited: May 3, 2016
  2. jcsd
  3. May 3, 2016 #2

    Simon Bridge

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    You can try doing it by stages ... The energy from the steam is used to heat the ice.
    How much energy would be released lowering the steam temperature to 100deg?
    How much energy is needed to raise the ice temperature to 0deg?
    Is the first number bigger or smaller than the second one?

    You can also try sketching a graph of energy vs temperature ... you will have two lines, one for energy released by the steam and the other for energy needed by the water. The two lines will have sloping and horizontal sections ... and they should cross somewhere.
     
  4. May 3, 2016 #3
    I sketched a graph, which my instructor looked at. He said it's good and that I'm on the right track, but I don't know what it is I'm doing wrong. I tried fiddling with the positive and negative signs to see if I could at least get a reasonable answer with no luck. I just tried to do it in some stages and it didn't work as well. I have never been stuck on a problem so long before in my life.
     
  5. May 3, 2016 #4

    Simon Bridge

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    OK - show me the stages ... please answer the questions in post #2.
     
  6. May 3, 2016 #5
    I did each one individually, as my teacher taught me to do. I'm not sure if this is what you meant, but I'll show you anyway.

    Q1 = mLf = (0.05)(3.4x106) = 17 000
    Q2 = mcΔT = (0.05)(4200)(Tf+5) = 210Tf + 1050
    Q3 = -mLv = -(0.05)(2.26x106) = -113 000
    Q4 = mcΔT = (0.05)(4200)(Tf -160)

    Then I subbed them in. Is this what you meant? Do all the calculations separately then combine them? Or did I misinterpret?
     
  7. May 3, 2016 #6

    Simon Bridge

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    No, when I say please answer the questions in post #2 I mean answer the questions in post #2.
    None of those lines answer the questions in post #2?

    Lets do it one at a time:
    1. How much energy is released lowering the steam to 100deg?

    The idea here is to make sure you understand the process.

    Aside: check your values for specific and latent heats ... check units etc.
    i.e. I have L_f = 334 kJ/kg = 3.34x105 J/kg
    from http://www.engineeringtoolbox.com/water-thermal-properties-d_162.html
    Compare with yours.

    The numbers in post #5 seem to assume the initial temp of the steam is 160deg (that is not what post #1 says) and that the specific heats are the same across phase changes, and that all the steam and all the ice ends up as water.
     
    Last edited: May 3, 2016
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