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:
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 required amount of heat will be "5830 KJ".
Given:
Amount of water,
As we know,
Latent heat of vaporization,
Now,
The amount of heat required will be:
= [tex]m\times L_{vap}[/tex]
= [tex]2.58\times 2260[/tex]
= [tex]5830 \ KJ[/tex]
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