Question

Calculate the amount of energy (in kJ) necessary to convert 397 g of liquid water from 0oC to water vapor at 172oC. The molar heat of vaporization (Hvap) of water is 40.97 kJ/mol. The specific heat for water is 4.184 J/g *oC, and for steam is 1.99 J/g *oC. (Assume that the specific heat values do not change over the range of temperatures in the problem.)

Answer 1

@Photon336

Answer 2

@ganeshie8

Answer 3

@Cuanchi I've attempted this problem and get 1137 kJ as my answer but I don't think it's right.

Answer 4

can you explain what did you do?

Answer 5

1. you have to calculate how many moles of water you have in 397g of water.
2. Then calculate the amount of energy to heat the water from 0C to 100C Q=mCe(water)ΔT.
3. calculate the amount of energy for the transition from liq to gas with the molar heat of vaporization
4 calculate the amount of energy to heat the vapor from 100 to 172C Q=mCe(vapor)ΔT

Answer 6

ms delta T = 397 g * 4.184 J/ goC *(172oC - 0oC)
397* 4.184J *(172)
J=285700.256

Answer 7

n delta H = 20.87 mol *40.79 kJ/mol
= 851.273 kJ

Answer 8

851 kJ + 286 kJ = 1137 kJ

Answer 9

@Photon336

Answer 10

q1= 397g x 4.18 J/gC x (100-0)= 165946 J
q2= (397g/18g) x 40.97 kJ/mol = 903616 J
q3= 397g x 1.99 J/gC x (172-100) = 56882 J

q total= 165946 J + 903616 J + 56882 J = ?????