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How much water boils off when a hot horseshoe is dropped in

  1. Dec 2, 2016 #1
    1. The problem statement, all variables and given/known data
    A blacksmith drops a 0.7 kg horseshoe of iron at a temperature of 1200 ̊C into a bucket of
    0.5 L
    of water at an initial temperature of 30 ̊C. How much of the water boils off? Assume the bucket
    absorbs none of the heat.

    2. Relevant equations
    I believe Q = mc delta(T) is what I'm aiming to use here, but I'm not entirely sure how

    3. The attempt at a solution
    I'm a bit lost, I know that the two objects (shoe and water) will ultimately reach the same final temperature, and I can write Q = mc delta(T) to reflect that, but I'm unsure how it corresponds to the water that boils off, and how I might find that value? I'm guessing it'd be the volume of steam which will share the same final T as the water and shoe? Not sure how to find it though.

    Thank You
     
  2. jcsd
  3. Dec 2, 2016 #2

    haruspex

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    No, the steam will come of at 100C and take no further part.
    Think of it in stages, and assume that at any given moment all the water is at the same temperature.
    What must happen before any water boils off?
     
  4. Dec 2, 2016 #3
    The water needs to rise to 100C first, right? I'm just generally confused how to work the equation up to that point, and then what I need to do to find how much water boils off.
     
  5. Dec 2, 2016 #4

    haruspex

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    Yes.
    What heat will the water have absorbed? What temperature drop occurs in the horseshoe?
     
  6. Dec 2, 2016 #5
    Do they both end up at 100C? So the horseshoe would drop 1100C? But then for the heat absorbed by the water, im not sure what it equals. Is it related to the horseshoe's Q = mc delta(T)? As in, I solve that with delta(T) = 1100C and thats how much heat the water absorbs?
     
  7. Dec 2, 2016 #6

    haruspex

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    Eventually, perhaps, but at this stage we are just considering the process up to the point where all the water is at 100C. The horseshoe may still be hotter.
    Answer my first question: how much heat has the water taken up at this point?
     
  8. Dec 2, 2016 #7
    Would that be = mc(70C)?
     
  9. Dec 2, 2016 #8

    haruspex

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    Maybe - you have not defined m and c.
     
  10. Dec 2, 2016 #9
    c is the specific heat of water and m is the mass, which, using density, would be .5kg
     
  11. Dec 2, 2016 #10

    haruspex

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    OK, so how much heat has gone from the horseshoe?
     
  12. Dec 2, 2016 #11
    Would its temperature change also equal 70C?
     
  13. Dec 2, 2016 #12

    haruspex

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    I didn't ask that.
    You calculated that the water absorbed mc (70C) of heat. So how much did the horseshoe lose in that time?
     
  14. Dec 2, 2016 #13
    What I meant by that was whether the horseshoe lost mc (70C) or if thats the wrong way to think about it
     
  15. Dec 2, 2016 #14

    haruspex

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    You need to base your thinking on sound principles, not wild guesses.
    It is a very simple question: if the water gained a quantity Q of heat from the horseshoe, what quantity of heat did the horseshoe lose to the water?
     
  16. Dec 2, 2016 #15
    I'd have to say I don't know
     
  17. Dec 2, 2016 #16

    haruspex

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    Heat is energy. Energy is conserved. If the total stays the same, but the water gained Q, how much did the horseshoe lose?
     
  18. Dec 2, 2016 #17
    Oh, then the horseshoe must have lost Q
     
  19. Dec 2, 2016 #18

    haruspex

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    Right. So what is the temperature of the horseshoe at this point?
     
  20. Dec 2, 2016 #19
    So, setting the two Q's equal, I found delta(T) of the shoe to be = -(mc 70C)/(MC)
    m - mass of the water
    c - specific heat of water
    M - mass of the shoe
    C - specific heat of iron
     
  21. Dec 2, 2016 #20

    haruspex

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    Ok. Time to turn that into a numerical temperature.
     
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