Final answer:
Approximately 3.41 grams of liquid water would vaporize when 7700 joules of energy is absorbed at 100°C, using the latent heat of vaporization for water which is 2256 kJ/kg.
Step-by-step explanation:
We can determine the mass of water that vaporizes using the supplied energy and the latent heat of vaporization for water. The latent heat of vaporization for water is about 2256 kJ/kg, which implies that a kilogram of water requires 2256 kJ of energy to vaporize at 100°C.
First, convert the energy from joules to kilojoules:
Energy (kJ) = 7700 joules × (1 kJ / 1000 J)
Energy (kJ) = 7.7 kJ
Next, use the formula Q = m × Lv, where Lv is the latent heat of vaporization, Q is the amount of heat, and m is the mass, to find the mass (m):
m = Q / Lv
m = 7.7 kJ / 2256 kJ/kg
m ≈ 0.00341 kg
To convert to grams, as most laboratory measurements are in grams:
m ≈ 3.41 grams
Therefore, approximately 3.41 grams of liquid water would vaporize when 7700 joules of energy is absorbed at 100°C.