Final answer:
To vaporize 42.0 grams of water at 100°C, the heat required is calculated by multiplying the mass of the water by the heat of vaporization, which is 2,250 J per gram, resulting in 94,500 joules.
Step-by-step explanation:
To calculate the amount of heat required to vaporize 42.0 grams of water at 100°C, we need to use the heat of vaporization. The heat of vaporization of water at 100°C is approximately 2,250 J per gram. Therefore, we can find the total amount of heat by multiplying the mass of the water in grams by the heat of vaporization per gram.
The calculation is as follows:
Heat required (Q) = Mass of water (m) × Heat of vaporization (ΔHvap)
Q = 42.0 g × 2,250 J/g
Q = 94,500 J
So, to vaporize 42.0 grams of water at 100°C, 94,500 joules of heat is required.