Hello,
This is a question about supercooled water that I am having trouble on:
Data: Cp(ice) = 76.1 J/K mol Cp(water) = 37.15 J/K mol dHfus(H2O) = 6.01 kJ/mol
Consider 28.5 g of supercooled (liquid) water at -12.0 oC and 1.00 bar. This metastable state suddenly freezes to ice at the same temperature and pressure.
(i) Treating the metastable state as an equilibirum state, calculate the heat released in this process. This I calculated to be 1.58 mol x 6.01 kJ/mol = 9.51 kJ
(ii) What is the change in entropy of the system? I am not sure here. I figured that the energy released from the sudden fusion of the water would be applied to the newly-formed ice to heat it up to 0
oC . Then, since dS = nC
pln(T
2/T
1) at constant pressure, dS would equal 5.4 J/K mol...?
(iii) Assuming the temperature of the surroundings is -12 oC, what is the change in entropy of the surroundings? Would this simply be dS
sur = q
p / T = -23 J/K mol ? However, this wouldn't make sense since then dS
system + dS
sur < 0