Question

How much energy is required to vaporize 10 grams of water at its boiling point? In Joules.

Answer 1

On the other hand, the molecules in liquid water are held together by relatively strong hydrogen bonds, and its enthalpy of vaporization, 40.65 kJ/mol, is more than five times the energy required to heat the same quantity of water from 0 °C to 100 °C

Answer 2

Answer: The energy required to vaporize the given amount of water is 22600 J

Step-by-step explanation:

The conversion involved in this process are:

`H_2O(l)(100^oC)\rightarrow H_2O(l)(100^oC)`

To calculate the amount of heat required at constant temperature, we use the equation:

`q=m* L_(vap)`

where,

`q` = amount of heat absorbed = ?

m = mass of water = 10 g

`L_(vap)` = latent heat of vaporization = 2260 J/g

Putting all the values in above equation, we get:

`q=10g* 2260J/g=22600J`

Hence, the energy required to vaporize the given amount of water is 22600 J