ok cool... here's the answer... i found it at
http://jchemed.chem.wisc.edu/JCESoft/CCA/CCA3/MAIN/ENTROPY/PAGE1.HTM
See... entropy decreases when ice freezes and energy decreases... true... but entropy concept just tells us that this phenomenon occurring spontaneously is highly improbable... its simply impossible for steam to condense to boiling water and then freeze to ice all by itself... Thus entropy is a quantity that reflects our empirical knowlegde about the direction of natural processes. Everything else (pressure,etc)
held constant, a given amount of heat added to a hot body increases its entropy less than when added to a cold body.
the following i got from http://www.maxwellian.demon.co.uk/art/eia/
Melting point, heat of melting
As an application of these conclusions, let's see if we can explain why there is something like a melting point. Refer to example no. 8; entropy is lowered upon freezing. But isn't freezing a natural process? And didn't we learn that for natural proceses, entropy increases?
It does. Keep in mind that we have excluded from consideration the heat of melting that is released when something freezes. Now we have to look at the entropy effects of this released heat.
Let us place the box of example no. 8 in a heat insulated vessel, containing, say, a very large amount of ethanol (it does not freeze itself at -2 °C). As soon as the water in the box freezes, a large amount of heat is set free; it warms up the freezing water to 0 °C (this is why temperature of freezing water is always 0 °C at normal pressure) and the excess is transferred to its surroundings, the ethanol in this case. Since we have very much ethanol, its temperature does not rise appreciably. So the ice at 0 °C eventually cools down to -2 °C; that is why we can avoid entropy effects of temperature change and focus on those of phase change. Thus total heat of melting is transferred to the ethanol, increasing its entropy.
The increase of entropy of ethanol must balance out the decrease of entropy when the ice is formed. Otherwise the process would not take place. In fact, if you prevent this heat from leaving the system, the substance in it would not freeze, at least not completely.
Now we have learned that the same amount of heat added to a cold body increases its entropy more than when added to a hot body. It follows that there must exist a temperature T(0), above which the entropy increase of the surroundings is not enough to balance out the decrease of the freezing water. Beneath T(0), entropy increase of surroundings is higher, and at T(0) entropy changes of freezing water and of surroundings are exactly equal. This is exactly how we defined entropy at the top of this article:
i got this question from...http://people.ouc.bc.ca/woodcock/121-notes/quiz6_answers.htm
4a. Calculate the entropy change when 1 mole of water freezes at 0º C. The heat of fusion for water is 6.01 kJ/mol.
Note: watch the units. The usual units for entropy change are J/mol.K, whilst those for enthalpy change are kJ/mol.
For a phase change: DS = DH / T = 6010 J / mol / 273 K = 22.0 J /mol.K.