Well, this is quite a steamy situation! Let's break it down. We have steam condensing into water, so heat is being released. We also have an insulated container, which means no heat is lost to the surroundings. So, we can say that the heat lost by the steam is equal to the heat gained by the water.
First, let's calculate the heat lost by the steam. We know the mass of the steam is 0.524g, so we need to convert that to moles. Using the molar mass of water (18.015 g/mol), we find that the number of moles of steam is 0.524g / 18.015 g/mol ≈ 0.029 moles.
Now, let's calculate the heat lost by the steam. We can use the equation Q = ΔH∘vap * n, where Q is the heat lost, ΔH∘vap is the molar heat of vaporization (40.7 kJ/mol), and n is the number of moles of steam.
Q = 40.7 kJ/mol * 0.029 mol ≈ 1.1823 kJ
Since the heat lost by the steam is equal to the heat gained by the water, we can set up the equation Q = m * C * ΔT, where Q is the heat gained, m is the mass of the water (4.20g), C is the specific heat capacity of water (4.18 J/g⋅∘C), and ΔT is the change in temperature.
1.1823 kJ = 4.20g * 4.18 J/g⋅∘C * ΔT
Now, let's solve for ΔT:
ΔT = 1.1823 kJ / (4.20g * 4.18 J/g⋅∘C) ≈ 0.06 ∘C
So, the final temperature of the mixture is approximately 0.06 ∘C. That's pretty chilling!