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Heat Transfer(ice in water)

  1. Mar 4, 2008 #1
    1. You have a tub of water that you are placing ice into. There is 300kg of water at 70 degrees celsius. You add 10kg of ice at -3 degrees C. What is the final temperature of the water? The latent heat of fusion for water is 33.5 x 10^4 j/kg

    I have worked out the best I can the problem as I understand it and I come up with an answer that does not quite make sense o_O

    Q= heat
    Qgained = Qlost;
    Qgained - Qlost = 0

    3. ice is gaining ...water is losing

    mc(changeintemp) + mLfusion - mc(ChangeinTemp) = 0

    mc(start - final) + mLfusion - mc(start - final)=0

    mc(-3 - final) + mLfusion - mc(70-final)=0

    ((10*4186(-3-final)) + 33.5x10^4) - (300*4186(70-final))=0

    (41,860(-3-final) + 33.5 *10^4) - (1,255,800(70-final))=0

    -125,580 - 41,860final +33.5*10^4 - (87,906,000 - 1,255,800final)=0

    -125,580 - 41,860final + 335,000 - 87,906,000 + 1,255,800final = 0

    -41,860final + 1,255,800final + 209,420 - 87,906,000= 0

    -41,860final + 1,255,800final -87,696,580 = 0

    -41,860final + 1,255,800final = 87,696,580

    final(-41,860 + 1,255,800) = 87,696,580

    final(1,213,940) = 87,696,580

    final = 87,696,580 / 1,213,940 = 72.241 degrees C??

    if the -41,860 was positive it came to a plausable answer. Now, I've checked over and and over and I don't understand where my error is but... 87,696,580/ (41,860 + 1,255,800) = 67.5 degrees C. Is one of my signs backwards or did i setup the problem incorrectly. Thanks for any help.
    Last edited: Mar 4, 2008
  2. jcsd
  3. Mar 4, 2008 #2

    Halp...I'll <3 u long time
  4. Mar 4, 2008 #3
    mc(-3 - final) + mLfusion - mc(70-final)=0

    Seems like it should be mc(-3 + final).

    This is the heat gained by the ice. The final temperature will be positive, so as you have it now, heat gained by ice comes out negative. Change that minus to a plus, and heat gained by ice becomes positive, like it should be.
  5. Mar 4, 2008 #4
    This is an alternative way to do it I suppose: I cant promise that it will work, but it might give you the correct answer. Im not an expert or anything but my last chapter was dealing with Thermodynamics and I had a lot of problems like this.

    Q = CMT
    Heat = heat capacity * mass * (Tfinal - Tinitial)
    Waters heat capacity is 4.184 and ice is 2.092

    4.184J/cal * 300kg (x-70C) = 2.092J/cal * 10kg (x-(-3))

    Good luck, and also I agree with turning it positive.

    Edit: I ran my calcs and I came up with around 68-69C.
    Last edited: Mar 4, 2008
  6. Mar 5, 2008 #5
    I understand the error now... in the equation Qlost = Qgained it isn't actually Qlost = Qgained but rather the magnitude of difference of Qlost and Qgained is equal I.E.

    |Qlost| = |Qgained| which sounds dumb because obviously one of them is going to be negative(losing heat) and one positive(gaining heat) but it just never clicked.

    So, in bringing one of them to the other side I needed to set up so they both ended up positive (or take their abs value). So for lost it's (InitalT - finalT) but for gained it's (finalT - initialT). I knew there was an error but I just could not understand why in actuallity it is what it is...now I do, thanks!
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