How would you estimate how much all the ice in a skating rink weighs? Use variables like D for depth, L for length, and such.

I am just stumped because if you get the volume, you turn it into mass, you will essentially have the mass of a cube, how can I change that to make it the mass of ice. With the desity or something like that?

  • $\begingroup$ What have you thought of already? Please indicate what you've done. People will be much more inclined to help you if you show that you are interested in your question. At the moment it just looks like you are too lazy to do your own homework. Why would we want to do it? $\endgroup$ – Ian Miller Apr 26 '16 at 4:10
  • $\begingroup$ @IanMiller Done $\endgroup$ – Dan Apr 26 '16 at 4:16
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    $\begingroup$ $mass=volume\times density$ The density of ice is $0.9167 g/cm^3$ at 0 °C. $\endgroup$ – Ian Miller Apr 26 '16 at 4:19

I would also approximate the ice in the rink as a volume of a rectangle. V=LWH According to this site:


30 m by 60 m is the size of the official Olympic games rink with an average of 2.5 cm for the thickness of the ice.

And as 1m is 100cm, 60m is 6000cm and 30m is 3000cm

Using the volume formula we get $V= 4.5\times10^7 cm^3$ for the ice.

Now plugging this into the density equation to get mass we get,

$M = 4.125\times10^4 kg$

  • $\begingroup$ Great explanation! Many thanks, Dan $\endgroup$ – Dan Apr 26 '16 at 5:58
  • $\begingroup$ Wait, you did you get M=4.125*10^4? You volume was V=4.5×10^7cm^3 , and the density of ice is 0.95. Could you explain this? $\endgroup$ – Dan Apr 28 '16 at 1:53
  • $\begingroup$ Mass is in kg not g. $\endgroup$ – Erock Brox Apr 30 '16 at 1:30

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