Respuesta :

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Answer:

6.78 × 10⁵ J

Explanation:

Step 1: Given and required data

  • Mass of water (m): 300 g
  • Latent heat of vaporization of water (ΔH°vap): 2260 J/g

Step 2: Calculate the heat (Q) required to vaporize 300 grams of water

We will use the following expression.

Q = ΔH°vap × m

Q = 2260 J/g × 300 g

Q = 6.78 × 10⁵ J

6.78 × 10⁵ Joule will be required to convert 300 g of water to steam at 100 °C.