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Quantity of Heat/Specific Heat Water+Iron

  1. Feb 12, 2008 #1
    [SOLVED] Quantity of Heat/Specific Heat Water+Iron

    1. The problem statement, all variables and given/known data
    Exact Problem:

    What will be the final temperature of 100g of 20 degrees C water when 100g of 40 degrees C iron nails are submerged in it. (The specific heat of iron is .12 cal/g C. Here you should equate the heat gained by the water to the heat lost by the nails.)


    2. Relevant equations

    Q = cm(change in)t

    Q = quantity of heat

    c = specific heat
    (water = 4.184J/g or 1 cal/g)
    (iron = .502J/g or .12 cal/g)

    m = mass
    100g + 100g = 200g total substance(water+iron)

    t = temperature
    (40(final)-20(initial))

    3. The attempt at a solution

    I feel I slightly understand what I am supposed to do but the part where it says I need to find the heat lost by iron and the heat gained by water im totally confused about. Heres my attempt: My assumption is that I need to do two different calculations and difference them out. Also if someone could point me in the right direction to be able to post real equations instead of what I have shown that would be so awesome.

    Iron First:
    Q = cm(Tf-Ti)

    Q = .502J*100g(40-20)

    Q = 1004J of Energy Lost

    Water Second:

    q = 4.184*100g(40-20)

    Q = 8368J of Energy Gained

    Answer:
    8368J - 1004J = 7364J of Energy or 1760.04cal/g

    I have tried numerous ways to find the temperature difference, lost and gained but I am clueless. Also I cant figure out where to transfer back into Celcius. By my calc's it would have gained like 1700 degrees haha. >.> I feel dumb... Anyways TIPS are always appreciated in advance. Constructive critism is appreciated as well, and well anything else that can make me a better poster is welcome too. ^^ thanks!
     
  2. jcsd
  3. Feb 12, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Your only problem appears to be that you have not read the problem correctly and don't seem to know what is being asked! Your start is correct but why are you using 40 degrees and 20 degrees as "initial" and "final" temperatures for both water and iron?

    You also say
    "Answer:
    8368J - 1004J = 7364J of Energy or 1760.04cal/g "
    which you should realize makes no sense. The question was "What will be the final temperature" and that's not even a temperature.

    20 degrees is the initial temperature for the water and 40 degrees is the initial temperature for the iron. The final temperature of both will be the same, of course, so just call that "T".

    Your equations should be
    Iron: Energy lost Q = .502J/g*100g(T- 40) and
    Water: Energy gained q = 4.184J/g*100g(T-20) and they must be equal. Solve for T.
     
  4. Feb 12, 2008 #3
    I see the fault, Thanks a bunch Ivy. You have saved my day and made me feel all the more dumb o_O. That was a joke btw.

    Edit: I came out with 32 degrees C as the final for both. Any chance I can get confirmation on this? Ill wait awhile and if nobody posts Ill mark as solved. Thanks again Ivy, your my hero!
     
    Last edited: Feb 12, 2008
  5. Mar 19, 2010 #4
    Re: [SOLVED] Quantity of Heat/Specific Heat Water+Iron

    Oh my GOD

    Temperature will be 28 C in equilibrium with water and nail.
    If you have eqations ready is simple.
    Put them L one = P one and find answer for T.
    Nails will lose from 40C - 12C= 28C
    Water will gain heat from 20C + 8C = 28C also
    very simple and time consuming but some people are fast so it is their own negative to be not able to stop and think for a while to grab the answare in the balls
     
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