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Adding Ice to Water Problem

  1. Apr 15, 2005 #1

    Working out:

    Q_b = m_beaker * c_water * change in temp. of beaker
    Q_b = (0.205)(4190)(26.7-79.9) = -45696.14J

    Q_1 = m_ice * c_ice * change in temp. of ice
    Q_1 = m_ice(2100)(26.7- -10) = m_ice(77070)

    Q_2 = m_ice * L_f
    Q_2 = m_ice(334000)

    Q_b + Q1 + Q2 = -45696.14 + m_ice(77070) + m_ice(334000) = 0

    Solving this for m_ice, I get: 0.111kg

    I'm told that the answer is wrong with a hint saying "Remember that the ice turns into liquid water above the melting point."

    I'm not sure how to go about this, can someone help me out, isn't that hint above the working out I did for Q_2 ?
  2. jcsd
  3. Apr 16, 2005 #2


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    The mistake is above I think... the ice first heats up... then turns to water then heats up again... so it should be
    Q_1 = m_ice(2100)(0--10) + m_ice(4190)(26.7-0)= 132873 m_ice

    Q_2 and Q_b look right.
  4. Apr 16, 2005 #3


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    You analysis is incorrect.

    1.The ice is will receive heat in the domain -10°--->0°.
    2.The ice at 0° will receive heat to pass into liquid water (no temperature & pressure variation).
    3.The initial mass of ice,now in the liquid state will receive heat to reach equilibrium temp.
    4.The initial mass of hot water will give away heat (all the heat added from points 1.-3.) till it reaches equilibrium temp.

    These hints are all u need...For now.

  5. Apr 16, 2005 #4
    dextercioby I understand the theory behind it, I have a nice diagram that explains this. But I don't know how to put this into numbers

    At the time I posted my problem I have thought of what learningphysics said and got
    Q_1 = m_ice(2100)(0--10) + m_ice(4190)(26.7-0)= 132873 m_ice

    But I'm not entirely sure that it is correct
  6. Apr 16, 2005 #5


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    If you replace your Q1 with the one above... I think you'll get the right answer...

    I'd organize it like this:

    Heat change of the water already in the beaker:
    Q_b = (0.205)(4190)(26.7-79.9) = -45696.14J (just as you did)

    Net het change of the ice as it turns to water at 26.7 (3 parts) :
    Heat change of the ice while it is still in ice form (it's temperature is rising from -10 to 0):
    1)Q_i = m_ice(2100)(0--10)

    Heat change of the ice as it turns to liquid (temperature is at 0):
    2)Q_c = m_ice(334000)

    Heat change of the liquid that was previously ice (temperature rises from 0 to 26.7):
    3)Q_d = m_ice(4190)(26.7 - 0)

    Q_ice = Q_i + Q_c + Q_d

    So total heat change = Q_b+ Q_ice =0
    Last edited: Apr 16, 2005
  7. Apr 16, 2005 #6
    Thanks for your help guys.

    I worked it out, what you said learningphysics, I done it before you posted just wasn't sure if it was right. Didn't think it was that simple. So didn't bother putting it in my 1st post :p

    Thanks again.
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