Given that the heat of fusion of water is -6.02 kJ/mol, that the heat capacity of H2O(l) is 75.2 kJ/mol*K and that the heat capacity of H2O(s) is 37.7 kJ/mol*K, calculate the heat of fusion of water at -11 K.

Answer 1

Read your book boy!

Answer 2

A substance's heat of fusion is the quantity of thermal energy involved in the solid/liquid phase change. When water freezes it releases 6.02kJ for each 18.0 grams (1 mole) that freezes. When it melts, each mole must gain 6.02 kJ of energy. Remember that there is no temperature change during phase changes - this is a change that, for pure water, occurs at 0 oC! Asking what water's heat of fusion is at -12 oC is kind of strange.....the heat of fusion of pure water is always the same but it's only important or is only involved when the substance is freezing or melting which doesn't occur at -12 oC.

But I guess for general learning purposes, this would be how to solve it.

The water is supercooled.
so it has a deficit of:
11° x (75.2 J/mol·K - 37.7 J/mol·K )
= 562.5 J/mol.
therefore you'll see
-6.02 kJ/mol + 0.562
= -5.458 kJ/mol

Answer 3

Why does the temperature does not change ? Shouldn't it in Kelvin because the heat capacity is J/mol K ?