Pure liquid water can be supercooled to temperatures below i

Pure liquid water can be supercooled to temperatures below its freezing point. Consider 10 kg of water cooled as a liquid to -10 Celsius. Let\'s now add a small crystal of ice (whose mass is negligible) as a \"seed\" the subcooled liquid causing a change of state to the ice at 0 Celsius. If the process occurs adiabatically and at constant (atmospheric) pressure (a) what is the mass of water that solidifies and (b) what is the change in entropy of the system?

Solution

since process os adiabatic no heat is given or taken

heat of fusion = m*c*dT

m1*s = m*c*dT

m1 = mass of water converted to ice

s=

dT = -10C

m = 10000g

c = 4.186 j/gK

s = 333j/g

m1 = 1257.06g = 1.257Kg................ANS

as theprocess is adiabatic change in entropy = 0