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Bomb Calorimetry

  1. Feb 29, 2008 #1
    Hey guys, I'm having a real tough time with this question:

    A 30.14-g stainless steel ball bearing at 117.82 degrees C is place in a constant pressure calorimeter containing 120.0 mL of water at 18.44 degrees C. If the specific heat of the ball bearing is 0.474 J/g x C, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity.

    qsteel +qwater = 0
    qsteel = -qwater

    qsteel = ms(deltaT)
    =(30.14 g)(0.474 J/g x C)(18.44 degrees C - 117.82 degrees C)
    =-1420 J

    Therefore qwater =1420 J
    1420 J = (120 g)(4.184 J/g x C)(Tfinal - 18.44 degree C)
    Tfinal = 1420/502 + 18.44

    However, the answer in the book is 21.19 degrees celsius. I don't think that I'm doing this one properly. Can anybody set me straight?
  2. jcsd
  3. Feb 29, 2008 #2
    I think that going off the assumption that [tex]\Delta T = T_{ball} - T_{water}[/tex] might be where the deviation is arising, remember, the ball is heating up a limited amount of water up until a point when [tex]T_{ball} = T_{water}[/tex], your case would be true if there was a massive amount of water relative to the ball, because the final temperature of the system would be pretty close to the waters original temperature anyway.

    I would go about finding the final temperature as:
    [tex] q_{steel} = mC(x-117.82)[/tex]
    [tex] q_{water} = mC(x-18.44)[/tex]

    Since the change in heat energy must be equal to zero as you stated you find:
    [tex]q_{steel}= -q_{water}[/tex]
    which is what you stated

    [tex]mC_{steel}(x-117.82) = -mC_{water}(x-18.44)[/tex]

    [tex]mC_{steel}x - mC_{steel}\times 117.82 = mC_{water}\times 18.44 - mC_{water}x[/tex]

    [tex]mC_{steel}x+mC_{water}x = mC_{water}\times 18.44 + mC_{steel}\times 117.82[/tex]

    [tex]x(mC_{steel}+mC_{water}) = mC_{water}\times 18.44 + mC_{steel}\times 117.82[/tex]

    [tex]x = \frac{mC_{water}\times 18.44 + mC_{steel}\times 117.82}{mC_{steel}+mC_{water} }[/tex]


    [tex] x = \frac{120\times 4.184\times 18.44 + 30.14\times 0.474\times 117.82}{30.14\times 0.474+120\times 4.184} = 21.1896^{\circ}C[/tex]
  4. Feb 29, 2008 #3
    Thanks again AbedeuS!!!
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