Final answer:
11.3 kilojoules are required to vaporize 5.00 g of water at 100°C using the given heat of vaporization for water, 2260 J/g.
Step-by-step explanation:
The question asks how many kilojoules are required to convert 5.00 g of liquid water at 100°C to steam at 100°C. Given that the heat of vaporization for water is 2260 J/g, the total heat required (Q) can be calculated using the formula Q = m x Lv, where m is the mass in grams and Lv is the latent heat of vaporization in joules per gram. Since 1 kilojoule (kJ) is equal to 1000 joules (J), we can convert the result to kilojoules by dividing by 1000.
To find the answer, we multiply the mass of the water (5.00 g) by the latent heat of vaporization (2260 J/g), which gives:
Q = 5.00 g x 2260 J/g = 11300 J
Converting this to kilojoules:
Q = 11300 J / 1000 = 11.3 kJ
Therefore, it requires 11.3 kilojoules to vaporize 5.00 g of water at 100°C.