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Volume of melted liquid water- physics

  1. Oct 16, 2013 #1
    1. The problem statement, all variables and given/known data

    A cube of ice, 30 cm on each side, is melted into a measuring cup. What is the volume of the liquid water?

    2. Relevant equations

    I tried doing 30 x 30 x 30 to get 27000 cm^3 but this answer was incorrect.

    3. The attempt at a solution
    Not sure what I am doing wron! Any help would be very nice thank you!
     
  2. jcsd
  3. Oct 16, 2013 #2

    SteamKing

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    What happens when water freezes? (besides turning solid. Hint: it's why ice cubes and ice bergs float)
     
  4. Oct 16, 2013 #3
    I have to find the density?
     
  5. Oct 16, 2013 #4
    When a material melts its mass remains same. In this case density of ice and water are different (SteamKing gave you a hint about that). If density is different but mass is same so, volume must be different, right???.....Now try to solve it, I have given pretty much every information you need to solve this question.
     
  6. Oct 17, 2013 #5
    I have D=mv. V= 30^3 or 27000 cm^3 = Volume of the ice. Now I need to find volume of water. How do I find all this though if I don't know the mass
     
  7. Oct 17, 2013 #6
    Water density = 1 gram per every cubic centimeter. So .001 x 27000 will give you D. So I have 27=m(27000), solving for m I get .001. What do I do from here?.
     
  8. Oct 17, 2013 #7
    Check your equation.
    $$ρ=\frac{m}{V}$$
    ##ρ## stands for density.
    Relate mass of both liquid and ice by their respective Density and Volume. But you must know density of both, it will be given in your textbook/assignment.
     
  9. Oct 17, 2013 #8
    D=m/v. Density of ice= 27/27000 or .001.
     
  10. Oct 17, 2013 #9
    Sorry forget that. I have D= m / v. 27= m/ 27000. M= 729000. What do I do from here?
     
  11. Oct 17, 2013 #10
    If, and only if, your textbook/question does NOT provide two different values for ρ (density), assume the following ones:

    Water (liquid, at 4°C or 277.15K) = 1.0g/cm³
    Ice (solid, at 0°C or 273.15K) = 0.917g/cm³

    As mentioned, your density equation is wrong.

    As density is given by [itex]m\cdot l^{-3}[/itex], you should divide mass by volume (and not multiply them together).
     
  12. Oct 17, 2013 #11
    Okay so D = m/v. So for my initial block of ice, my volume is 27000 cm^3 so my Denisty would have to be 1 g/cm^3 x 27000? So my final mass is of my ice cube is 729000 correct? If I convert the grams to kilograms.
     
  13. Oct 17, 2013 #12
    Write the units (ALWAYS).
    my volume is 27000 cm^3
    Right.

    my Density would have to be 1 g/cm^3 x 27000
    You are assuming that the ICE density IS 1g/cm^3 which is incorrect.

    my final mass is of my ice cube is 729000 correct?
    No. The final mass is the volume times the density, as your density is wrong, the final mass is wrong.

    If your book does not provide you the two densities (solid and liquid), use the ones I posted.
    As we assume that liquid water has a density of 1g/cm^3, it's fine if you only have the ice density.
    But it is NOT the same as liquid water.
     
  14. Oct 17, 2013 #13
    If you have 27000 cm3 of ice, and its density is 0.917 g/cm3, how much mass do you have (in grams)?
    If you have this same mass of water in grams, and its density is 1.0 g/cm3, what volume of water do you have (in cm3)?

    Chet
     
  15. Oct 17, 2013 #14
    My apologies to besulzbach. I didn't mean to suggest in any way that he was someone who needed help in solving this problem. I just wanted to make use of the data on densities he provided. In retrospect, I should have quoted astru025 instead. Besulzbach, please accept my apologies.

    Chet
     
  16. Oct 17, 2013 #15
    You do not need to apologize, as I told you. I just found you quotation unexpected, nothing else.

    I guess that his textbook does NOT provide the constants that I got from Wolfram|Alpha, so I posted them. If that is really the case (something only the author of this topic can say), he should, and hopefully will, use mine or other similar values found on the web.

    Again, you do not need to apologize to me at all, Chestermiller.
     
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