We have following relevant data for water: normal melting point = 0.0 degree C,
ID: 527140 • Letter: W
Question
We have following relevant data for water: normal melting point = 0.0 degree C, normal boiling point = 100.0 degree C, C_s (H_2O(1)) = 4.18 J/g degree K, C_s(icc) = 2.09 J/g. degree C, C_s(H2 O(g)) = 1.84 J/g degree K, delta H_vap = 40.67 kJ/mol, delta H_freezing = -6.01 kJ/mol. a) what is the total heat change to convert 67.3 grams of H_2O at 30.0 degree C to -45.0 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. 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
While converting water liquid from 30 deg.c to -45 deg.c, the following changes will have to take place,
The total heat to be removed is sum of the above three heats
The total heat to be removed is sum of the above three heats
And this heat= latent heat of fusion* moles of water ( since latent heat is givenin Kj/mole, mass of water needs to be converted to moles.moles of water = mass/molar mass= 67.3/18 =3.74
Hence heat required for phase change of 3.74 moles of water= 3.74*(-6.01)KJ=-22.5 Kj
3. Ice at 0 deg. c becomes ice at -35 deg.c by loosing sensible heat and this heat= mass* specific heat of ice * temperature difference= 67.3*2.09*( -35-0)= -4923 Joules =-4.923 KJ
The total heat to be removed= -9.7195 -22.5-4.923 =-37.1425 KJ. Since heat has to be removed , it is shown as negative.
2. heat of fusion of phase change of ice to water= 6.01 Kj/mole. for the present problem it is 22.5 Kj
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