Question

How much energy is required to melt 10.0 g of water at 0°C? (ΔH°fus = 6.01 kJ/mol; ΔH°vap = 40.76 kJ/mol; c = 75.3 J/mol • °C)

Answer #1

Energy is required to melt 10.0 g of water at 0°C

Given- ΔH°fus = 6.01 kJ/mol

Heat of fusion is the measure of the intramolecular energy which holds the molecules together and prevents them from a phase change without adding energy. In other word it's the amount of energy needed to change water from a solid to a liquid state.

In this case ΔH°fus = 6.01 kJ/mol, so convert 10g water into mol of water

Since a mole of water weighs 18 grams

10 grams of water = 10/18ths mol

q = (10.0 g / 18.0 g /mol) (6.01 kJ/mol) =3.34KJ

**3.34KJ energy is required to melt 10.0 g of water at
0°C.**

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How much ice at 0°C [Cp = 37.1 J/(mol ·°C), ΔH°fus = 6.01 kJ/mol]
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Melting point: 0 degrees C
Boiling Point: 100 degrees C
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Delta H(vap): 40.67 kJ/mol
cp(s): 37.1 J/mol x degrees C
cp(l): 75.3 J/mol x degrees C
cp(g): 33.6 J/mol x degrees C

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Do not enter units of measurement, do not enter the answer in
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Calculate the enthalpy change, ΔH, for the process in
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For water, ΔHvap = 44.0 kJ/mol at 25.0 ∘C and
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