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A question on the Principle of Floatation

  1. Sep 8, 2006 #1
    This is not actually a question from my tutorial but it stemed out from there.
    So, here it goes:
    Consider a block of ice partially submerged on water. It is floating, so by the principle of floatation and Archimedes' Principle, the weight of water displaced equals to the weight of the block of ice.
    When the ice melts, we know that there is no rise of water level because the vol. of water produced by the ice when it melts equals to the vol. of water displaced.
    Here's comes the part i don't understand: The ice is only partially submerged in the water, so the vol. of water displaced by the ice should only be the vol. of ice submerged in the water, am i correct? Then, the total vol. of ice will have to be greater than the vol. of water displaced and therefore when it melts, it should produce a vol. of water greater than the water displaced, but why it doesnt? Or is that, the vol. of ice is different from the vol. of water it produced when melted?

    However, if i use the mathematical eqn and assuming the densities of ice and water are the same, I am able to get the vol. of water displaced =vol. of ice, which leaves me even more confused.
    The eqn is :
    M(ice)*g=Vol(water displaced/part of ice that is underwater)*p(water)*g
    p(ice)*vol(ice)=Vol(water displaced)*p(water)
    p(ice)=p(water)
    Thus: Volume of ice=Vol(of water displaced/Vol.part of ice underwater)

    Thanks a million for helping with my problem
     
  2. jcsd
  3. Sep 8, 2006 #2
  4. Sep 9, 2006 #3

    andrevdh

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    The watermark will stay at the same level if the melted ice produces the same volume of water as that which the remaining ice raises up due to the lost.

    Vol. water gained:

    [tex]\Delta m_i = \Delta m_w[/tex]

    which comes to

    [tex]\Delta V_w = \frac{\rho _i}{\rho _w}\Delta V_i[/tex]

    Vol. ice raised:

    Water displaced by the weight of melted ice [tex]\Delta m_i g[/tex]

    [tex]\Delta V_w = \frac{\Delta m_i}{\rho _w}[/tex]

    This translates to the same volume of water as that which the melted ice produces.
     
    Last edited: Sep 9, 2006
  5. Sep 10, 2006 #4
    I still don quite get it....When the ice is only partially submerged on the water, is e total volume of ice=the volume of water displaced by the ice?
    But, if we try to calculate the vol. of water displaced, we will use the vol. of ice which is submerged in the water as the vol. of water displaced right?

    What will happen if there are some impurities inside the ice? For example some grains of sand which has a higher density than water? Will the water level rise or drop when the ice melts?
     
  6. Sep 10, 2006 #5

    andrevdh

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    I considered what will happen if a little bit of ice, [tex]\Delta m_i[/tex], melted. The melted ice will add to the volume of existing water, causing the water level to rise. But since the ice cube is now lighter it will rise causing the water level to fall. What I was trying to indicate is that these two volumes will be the same. The first part of my previous post (#3) calculates the amount of water that will be added to the existing water due to the ice that melted.
     
  7. Sep 10, 2006 #6
    ok, so the symbol [tex]\Delta m_i[/tex] u have used in the previous post is actually the vol. of water that will be produced by the little bit of ice when it melts and not the vol. of water displaced by the ice?
    Then, can i say that the vol. of ice before it melts is different from the vol. of water if produces after it melts since the densities of ice and water are different?
     
  8. Sep 10, 2006 #7

    andrevdh

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    The term [tex]\Delta m_i[/tex] refers to the mass of the bit of ice that melts. And yes the volume of the initial bit of ice and the final bit of water will differ because of the difference in densities as you remark. This can be seen from

    [tex]\Delta m_i = \Delta m_w[/tex]

    which comes to

    [tex]\rho _i \Delta V_i = \rho _w \Delta V_w[/tex]

    so that the volume of the water created by the melted ice will be

    [tex]\Delta V_w = \frac{\rho _i}{\rho _w} \Delta V_i[/tex]

    so that the volume of water will be smaller since the density of ice is smaller than that of water.
     
  9. Sep 10, 2006 #8
    Ok, i think i got it. Thanks for your help! Will there be a case where both densities are the same? In that case, will there be a rise of water level?
     
  10. Sep 11, 2006 #9

    andrevdh

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    The density of water do decrease with temperature, but even at 100 degrees celsius it is still quite a bit denser (3.63 kg/gal) than ice (3.47 kg/gal). In answer to your second question: The analysis of the situation in post #4 is general, that is I did not use specific values, which means it will be true under all conditions.
     
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