Answer:
To calculate the minimum amount of heat necessary to vaporize 90 g of water starting at 20 c
Explanation:
We will use heat balance equation
Q = q1+q2 = mCΔt + nL
where
m = mass of water = 90g
C = specifi heat of water =4.18J/gK
L = Latent heat of Vaporization = 40.7kJ/mol
Δt = temperature changes = 100 - 20 = 80 degree
water at 20 degrees to water at 100 degrees
then water at 100 degrees to vapor at 100 degrees
q1 = (90g)(4.18J/gK)(80K)
q1 = 30096J=30.1kJ
But
m =90 g water, gives n = 5mol water (divide by his molar mass which is 18g)
q2=nL= (40.7kJ/mol)(5mol)
q2=203.5kJ
the minimum amount of heat necessary to vaporize 90 g of water starting at 20 c = q1+q2 =30.1+203.5=233.6kJ
