Answer:
6.78 × 10⁵ J
Explanation:
Step 1: Given and required data
- 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.
