Answer: 3.45 * 10^3 kJ
Step-by-step explanation:
With the information provided, we can see that this is a specific heat capacity problem that asks how much heat (i.e. energy) is required to boil a mass of water (m = 1530 g). We can use the following equation to solve for the heat, Q, needed to vaporize the water (given its specific enthalpy of vaporization, ΔHvap):
Q = m*ΔHvap, where m represents the amount of the substance
Since the units of ΔHvap is provided in kJ/mol, we need to convert the mass of water to moles. Given that the molar mass of water (H2O) is 18.01528 g/mol, we have 84.927 (3 sig. figs) mol H2O. We can now solve Q:
Q = (84.927 mol H2O) * (40.66 kJ/mol) = 3453.168 kJ (3 sig. figs) = 3.45 * 10^3 kJ