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 .

Step one: heat the 13.0 gram of ice from to . The change in temperature would be .Step two: supply the heat of fusion to convert that 13.0 gram of ice to water.Step three: heat the 13.0 gram of water from to . The change in temperature would be .Step four: supply the heat of vaporization to convert that 13.0 gram of water to steam.Step five: heat the 13.0 gram of steam from to . The change in temperature would be .Energy required for step one, three, and five

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

.

In this equation,

is the specific heat of this substance, is the mass of the substance, and is the change in the temperature of the object.

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

.

Energy required for step two and four

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

.

.

where is the number of moles of the substance.

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

H: 1.008,O: 15.999.

Hence the molar mass of water:

.

Number of moles of molecules in :

.

Energy required for all five steps, combined

.

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