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

1. User Answers drader
Answer:
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

2. User Answers Anonym
Answer: The energy required to vaporize the given amount of water is 22600 J
Explanation:
The conversion involved in this process are:
[tex]H_2O(l)(100^oC)\rightarrow H_2O(l)(100^oC)[/tex]
To calculate the amount of heat required at constant temperature, we use the equation:
[tex]q=m\times L_{vap}[/tex]
where,
[tex]q[/tex] = amount of heat absorbed = ?
m = mass of water = 10 g
[tex]L_{vap}[/tex] = latent heat of vaporization = 2260 J/g
Putting all the values in above equation, we get:
[tex]q=10g\times 2260J/g=22600J[/tex]
Hence, the energy required to vaporize the given amount of water is 22600 J