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I have known the reason why skate can slide over ice is that water's melting curve in terms of pressure and temperature has a negative slope. If the pressure due to our mass increases sufficiently high, the ice starts to melt.

But someone says, according to latest research, it is not true because the time for which we pressure the ice is too short so the ice can't has enough time to melt.

Which is more reasonable between the two arguments?

Rijul Gupta
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MS.Kim
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  • Do you happen to have a reference of the 'latest research' that claims the time is too short? – Michiel Oct 20 '13 at 08:54
  • @Michiel No, I have no reference of it yet. I have just told the claim from my acquaintance. I will ask him whether he has a reliable reference or not. –  Oct 20 '13 at 10:03
  • I think this will have a better chance on Physics, so I'm migrating it… – F'x Oct 20 '13 at 21:49
  • Some references: http://www.ncbi.nlm.nih.gov/pubmed/1517252 , http://scitation.aip.org/content/aapt/journal/ajp/63/10/10.1119/1.18028 . Unfortunately these are both paywalled. –  Oct 20 '13 at 23:33
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    Possible duplicate: http://physics.stackexchange.com/q/1720/2451 – Qmechanic Oct 20 '13 at 23:59

1 Answers1

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Decrease of the melting temperature with pressure increase is not enough to explain skating (I conducted the calculation myself, but please see http://scitation.aip.org/content/aapt/journal/ajp/63/10/10.1119/1.18028). AFAIK, skating can be explained by ice melting due to heating caused by friction.

EDIT(10/22/2013): OK, so let us use the Clausius-Clapeiron relation (http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation ): $\frac{dP}{dT}=\frac{L}{T\triangle v}$. Let us assume that $\triangle T=1K$, $L=334\frac{kJ}{kg}$ (http://en.wikipedia.org/wiki/Latent_heat#Specific_latent_heat ), $T=273K$, $\triangle v=10^{-4}\frac{m^3}{kg}$ (1 kg of water has a volume of $10^{-3}m^3$, and volume of ice is 10% greater than volume of water), so $\triangle P \approx1.22\cdot 10^7 Pa$. Let us also assume that the area of a skate is $S=1mm\times 30cm=3\cdot 10^{-4}m^2$ and the mass of the skater is $m=80kg$. Then the pressure exerted by the skater is $\frac{mg}{S}=\frac{80\cdot 9.8}{3\cdot 10^{-4}}\frac{N}{m^2}\approx 2.88\cdot 10^6 Pa$. Therefore, the skater exerts pressure on ice that is four times less than that required to decrease the melting point by 1 degree Celcius.

akhmeteli
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  • Sorry, I missed the earlier comment by Ben Crowell. – akhmeteli Oct 20 '13 at 23:46
  • wow that's really interesting! Can you put a summary of your calcs in your answer (don't bother if your calcs are handwritten - just thought you may have them LaTeXed somewhere as I am doing more and more these days). – Selene Routley Oct 20 '13 at 23:54
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    If you have access to the Colbeck paper, could you give us a sketch of what it says, and/or what your own calculations say? –  Oct 21 '13 at 00:15
  • I just used the Clausius-Clapeiron equation (http://en.wikipedia.org/wiki/Clausius%E2%80%93Clapeyron_relation ). Sorry, won't be able to give details earlier than tomorrow evening (Central time), in the best case. – akhmeteli Oct 21 '13 at 01:49