Water has a specific heat of 4.18 j/gk and a molar heat of vaporization of 40.7 kj. calculate the minimum amount of heat necessary to vaporize 90 g of water starting at 20 c

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

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