Calculate the amount of heat in kilojoules required to vaporize 2.58 kg of water at its boiling point.

The amount of heat required to vaporize 2.58 kg of water at its boiling point is 5,830.8 kJ.

Explanation:

A substance undergoes a change in temperature when it absorbs or gives up heat to the environment around it. However, when a substance changes phase it absorbs or gives up heat without causing a change in temperature. The heat Q that is necessary for a mass m of a certain substance to change phase is equal to:

Q = m*L

where L is called the latent heat of the substance.

In this case:

Replacing:

Q= 2.58 kg* 2260*10³ J/kg

Q= 5,830,800 J = 5,830.8 kJ (Being 1,000 J= 1 kJ)

The amount of heat required to vaporize 2.58 kg of water at its boiling point is 5,830.8 kJ.

the answer is the first option

explanation:

just googled it m8. not that hard

lol