Heat of fusion (¦¤Hfus) is used for calculations involving a phase change between solid and liquid, with no temperature change. For H2O, ¦¤Hfus=6.02 kJ/mol.

Specific heat capacity (C) is used for calculations that involve a temperature change, but no phase change. For liquid water, C=4.184 J/(g⋅¡ãC).
Heat of vaporization (¦¤Hvap) is used for calculations involving a phase change between liquid and gas, with no temperature change. For H2O, ¦¤Hvap=40.7 kJ/mol.

How much heat is required to boil 85.5g of water at its boiling point?

I see a lot of information here but no question.

To find out how much heat is required to boil 85.5g of water at its boiling point, we need to use the formula:

Q = m * ¤Hvap

Where:
Q is the amount of heat required (in joules).
m is the mass of the substance (in grams).
¤Hvap is the heat of vaporization (in joules per gram).

First, let's convert the mass of water from grams to moles because the heat of vaporization is given in kJ/mol. We can use the molar mass of water (H2O) to make this conversion.

The molar mass of water (H2O) is:
2 * (1.008 g/mol) + 16.00 g/mol = 18.02 g/mol

Now, let's convert the mass of water from grams to moles:
85.5 g / 18.02 g/mol = 4.75 mol

Next, we can calculate the heat required using the formula mentioned earlier:

Q = m * ¤Hvap
Q = 4.75 mol * 40.7 kJ/mol

To cancel out the units and obtain the answer in joules, we need to convert kJ to J:
1 kJ = 1000 J, so
Q = 4.75 mol * 40.7 kJ/mol * 1000 J/1 kJ

Calculating this expression will give us the amount of heat required to boil 85.5g of water at its boiling point.