Final answer:
The amount of heat required to vaporize 78 g of liquid water at 100°C to vapor at 100°C is 42120 calories.
Step-by-step explanation:
To calculate the heat required to turn 78 g of liquid water at 100°C to vapor at 100°C, we need to use the equation: q = m × ΔHvap, where q is the heat, m is the mass of water, and ΔHvap is the heat of vaporization. Given that the heat of vaporization of water is 540 cal/g, we can substitute the values into the equation:
q = 78 g × 540 cal/g = 42120 cal
Therefore, the amount of heat required to vaporize 78 g of liquid water at 100°C to vapor at 100°C is 42120 calories.