Final answer:
To find the heat absorbed when 70 grams of water are vaporized, multiply the mass by the heat of vaporization, 2,250 J/g, which yields 157,500 J or 157.5 kJ.
Step-by-step explanation:
The student asked how many kilojoules of heat are absorbed when 70 grams of water is completely vaporized at its boiling point. The heat of vaporization of water is about 2,250 J per gram, meaning that every gram of water that transitions from liquid to gas (vapor) requires an energy input of approximately 2,250 J. Given that 1 kJ equals 1,000 J, we can convert joules to kilojoules.
To calculate the total heat absorbed, we multiply the number of grams of water by the amount of energy needed per gram:
Heat absorbed = Mass of water (g) × Heat of vaporization (J/g)
Heat absorbed = 70 g × 2,250 J/g = 157,500 J
To convert joules to kilojoules:
Heat absorbed = 157,500 J ÷ 1,000 = 157.5 kJ