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Finding thermal equilibrium of hot iron placed in water

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data

    You cool a 100g slug of red-hot iron (temperature 745 degrees C) by dropping it into an insulated cup of negligible mass containing 85g of water at 20 degrees C. Assuming no heat exchange with the surroundings, what is the final temperature of the water?

    2. Relevant equations

    Q=mcΔT
    Q=mL
    specific heat of water c=4190 J/kgK
    specific heat of iron c=470 J/kgK
    Heat of vaporization of water L=2256 x 10^3 J/kg


    3. The attempt at a solution

    I've looked at several other similar threads on the forums, but I'm not really looking for an answer, just some clarification of some points that I haven't quite grasped yet.

    My initial attempt at a solution entailed Q(water) + Q(heat) = 0, but this method gave me an answer of 104 degrees C, which implies some of the water changes phase, so I know I need to include Q=ml in my equation to account for that, but I don't really see how to do that.

    Some direction would be greatly appreciated. Thank you!
     
  2. jcsd
  3. Jan 13, 2012 #2
    There is something not right with this question or else it is a bit of a trick.
    I found that the heat energy lost by the iron cooling from 745C to 100C is greater than the heat energy needed to warm the water from 20C to 100C.
    To me it seems that the water boils and some water is lost as steam.
     
  4. Jan 13, 2012 #3
    You're right, the solution states the same thing. Specifically, it solves for Q(iron) and for the difference in temperature, it uses 745-100. Same thing for Q(water): 100-20. I understand that 100 degrees C is the boiling temperature of water, but I guess I'm just not understanding why we assume the iron cools from 745 to 100, and that the water heats to 100...

    ...hope that makes sense?
     
  5. Jan 13, 2012 #4
    Water can't be hotter than 100C under normal conditions.
    A sensible variation on this question would be to ask you the mass of water that is lost due to the boiling(evaporation)
     
  6. Jan 13, 2012 #5
    Ok, I understand that, but why is the difference in temperature of the iron 645 degrees C?

    Is no calculation really necessary to determine the final temperature of the water is 100C simply because we know it can't exist as liquid water past 100C, and since the iron is significantly hotter it will bring the water to boiling?
     
  7. Jan 13, 2012 #6
    Yes....that is basically it. The iron starts at 745C, once it is placed in water and the water gets to 100 C that is the end of the temperature change. The extra heat (I got about 2000J) that needs to be removed from the iron is used to convert liquid water into steam
     
  8. Jan 13, 2012 #7
    Ok, got it. Excellent explanation. Thank you so much!
     
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