Answer:
ΔH = q = 8.02 x [tex]10^{4} J[/tex]
Explanation:
q = ΔH = nΔH vaporization + nCp,m steam . ΔT
= 1.65 mol x 40656 J [tex]mol^{-1}[/tex] + 1.65 mol x 33.58 J [tex]mol^{-1} K^{-1}[/tex] x (610K - 373K) = 8.02 x [tex]10^{4} J[/tex]
w = - P external ΔV = [tex]-10^{5} Pa[/tex] x (1.65 x 5.06 x [tex]10^{-2} m^{3}[/tex] - 1.65 x 1.89 x [tex]10^{-5} m^{3}[/tex]) = -8.34 x [tex]10^{3}[/tex] J
ΔU = w + q = -8.34 x [tex]10^{3}[/tex] J + 8.02 x [tex]10^{4} J[/tex] = 7.18 x [tex]10^{4} J[/tex]