Question

30 grams of steam at 130C are colled to the boiling point, condensed, then cooled to 0C and frozen into ice. What is the total change in entropy of the water? what is the total change in gibbs energy? state any assumptions you are making.

Answer 1

We will use the following data for our calculation,

specific heat of steam = 1.84 J/g.K

specific heat of water = 4.184 J/g.K

specific het of ice = 2.03 J/g.K

Heat of condensation of water = -2260 J/g

Given mass of steam = 30 g

Starting steam temperature = 130 C

Let us calculate heat Q of reaction.

Q1 = heat change from 130 C (steam) to 100 C (steam) cooling

      = m x Cp x dT

      = 30 g x 1.84 x (130 - 100)

      = 165.6 J

Q2 = heat change from 100 C (water) condensation to 100 C (water)

      = m x dHcond

      = 30 g x (-2260)

      = -67800 J

Q3 = heat change from 100 C (water) to 0 C (water) cooling

      = m x Cp x dT

      = 30 g x 4.184 x (100 - 0)

      = 12552 J

Q4 = heat change from 0 C (water) to 0 C (ice) freezing

      = m x Cp

      = 30 g x 2.03

      = 60.9 J

Total heat change for the reaction,

Q = Q1 + Q2 + Q3 + Q4

   = 165.6 - 67800 + 12552 + 60.9

   = -55021.5 J

Now, at constant pressure,

Q = dH (enthalpy)

thus,

entropy change for reacton dS = -(-55.022 x 103)/(130+273)

                                                 = 14.06 kJ/K

Free energy change for reaction,

dG = dH - TdS

      = -55.02 - 403 x 14.06

      = -5721.20 kJ