Answer:
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:
- m=2.58 kg
- The heat of vaporization of water is L=2260*10³ J/kg
Replacing:
Q= 2.58 kg* 2260*10³ J/kg
Q= 5,830,800 J = 5,830.8 kJ (Being 1,000 J= 1 kJ)
<u><em>The amount of heat required to vaporize 2.58 kg of water at its boiling point is 5,830.8 kJ.</em></u>
