Question

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

Answer 1

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)

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