Final answer:
To determine the mass of water that boils at 100ºC and absorbs 479,564.88J of energy, we use the equation Q = mLv and calculate the mass using the given latent heat of vaporization of water. The mass of water is approximately 211.858 grams.
Step-by-step explanation:
To calculate the mass of water that boils at 100ºC and absorbs 479,564.88J of energy, we can use the specific heat capacity equation:
Q = mcΔT
Given:
Q = 479,564.88J
ΔT = 100ºC
Since the water is boiling, we assume there is no change in temperature during the phase change. Therefore, the equation becomes:
Q = mLv
Where Lv is the latent heat of vaporization of water. From the given information, we can find that Lv = 40.7 kJ/mol.
We can convert the given energy to kJ by dividing by 1000:
479,564.88J ÷ 1000 = 479.56488 kJ
Now, we can calculate the mass of water by rearranging the equation:
m = Q ÷ Lv
m = 479.56488 kJ ÷ 40.7 kJ/mol = 11.781 mol
Finally, we can convert the number of moles to grams by multiplying by the molar mass of water:
m = (11.781 mol) × (18 g/mol)
= 211.858 g
Therefore, the mass of water that boils at 100ºC and absorbs 479,564.88J of energy is approximately 211.858 grams.