How much heat is required to boil 90.0g of water at its boiling point?

Do you mean to boil all of the water away? That's

q = mass H2O x heat vaporization

Or do you mean to boil the first molecule of water? If so that takes much less heat.

To calculate the heat required to boil water at its boiling point, we need to use the equation:

Q = m * ΔHvap

Where:
Q is the heat required (in joules)
m is the mass of water (in grams)
ΔHvap is the heat of vaporization of water (in joules/gram)

The heat of vaporization of water is approximately 40.7 kJ/mol, or 2260 J/g.

Now, let's calculate the heat required:

Q = 90.0g * 2260 J/g = 203,400 J

Therefore, approximately 203,400 joules of heat are required to boil 90.0 grams of water at its boiling point.

To calculate the heat required to boil water, we can use the formula:

Q = m * ΔH

Where:
Q is the heat energy (in joules),
m is the mass of the water (in grams), and
ΔH is the heat of vaporization for water (in joules per gram).

To find ΔH, we can use the value for the heat of vaporization for water, which is approximately 2260 joules per gram.

Now let's plug in the values into the equation:

Q = 90.0 g * 2260 J/g
= 203,400 J

Therefore, it would require 203,400 joules of heat to boil 90.0 grams of water at its boiling point.