We have following relevant data for water: normal melting point =0.0 degree C, n

Question

We have following relevant data for water: normal melting point =0.0 degree C, normal boiling point = 100.0 degree C, C_3 (H_2 O(1)) = 4.18 J/g degree K, C_s(ice) = 2.09 J/g. degree C, C_5(H_2 O(g)) = 1.84 J/g degree K, delta H_vap = 40.67 kJ/mol, delta H_freeling = -6.01 kJ/mol. a) What is the total heat change to convert 85.0 grams of H_20 at 130.0 degree C to 25.0 degree C? (Draw and find out how many physical states and phase change(s) should be there and what heat calculation equation to use in order to calculate heat for each step. b. if delta H_freezing = -6.01 kJ/mol, what will be the phase change heat of fusion (i.e., delta H_fus) of water?

Explanation / Answer

a) The following processes take place:

i) H2O (g, 130C) ------> H2O (g, 100C)

Heat involved, q1 = (mass of H2O,g)*Cs(H2O, g)*(change in temperature) = (85.0 g)*(1.84 J/g.K)*[(130 + 273) – (100 + 273)]K = 4692 J = (4692 J)*(1 kJ/1000 J) = 4.692 kJ.

ii) H2O (g, 100C) ------> H2O (l, 100C)

Moles of water = (85.0 g)/(18.0 g/mol) = 4.7222 mole.

Heat involved. q2 = (moles of water)*Hvap = (4.7222 mole)*(40.67 kJ/mol) = 192.0519 kJ.

iii) H2O (l, 100C) -----> H2O (l, 25C)

Heat involved, q3 = (mass of water)*Cs(H2O,l)*(change in temperature) = (85.0 g)*(4.18 J/g.K)*[(100 + 273) – (25 + 273)]K = 26647.5 J = (26647.5 J)*(1 kJ/1000 J) = 26.6475 kJ.

Total heat involved, q = q1 + q2 + q3 = (4.692 + 192.0519 + 26.6475) kJ = 223.3914 kJ 223.391 kJ (ans).

b) Given Hfreezing = -6.01 kJ/mol and the process is

H2O (s, 0C) ------> H2O (l, 0C)

However, we want the reverse process:

H2O (l, 0C) ------> H2O (s, 0C)

Hfus will be the reverse of Hfreezing, i.e, Hfus = -Hfreezing = -(-6.01 kJ/mol) = +6.01 kJ/mol (ans).

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