How much heat (in kj) is required to warm 13.0 g of ice, initially at -10.0 ∘c, to steam at 111.0 ∘c? the heat capacity of ice is 2.09 j/g⋅∘c and that of steam is 2.01 j/g⋅∘c, the heat of fusion for water is 6.02 kj/mol, and the heat of vaporization for water is 40.7 kj/mol.

Approximately 39.7 kJ.

Assumptions: the specific heat capacity of water is , the melting point of water is , and that the boiling point of water is .

Explanation:

It takes five steps to convert 13.0 grams of ice to steam at .

The following equation gives the amount of energy required to raise the temperature of an object  by a :

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In this equation,

Assume that there's no mass loss in this whole process. The value of would stay the same at .

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The equations for the energy of fusion and energy of vaporization are quite similar:

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where is the number of moles of the substance.

Look up the relative atomic mass of oxygen and hydrogen from a modern periodic table:

Hence the molar mass of water:

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Number of moles of molecules in :

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c) should be the answer.