Question

Take the heat capacity of ice as 2.108 J/(g*C) and the heat capacity of water as 4.184 J/(g*C) for this problem. For water, DH_fus = 6.001 kJ/mol at 0 C.

a. A 10.0 gram ice cube at -10.0 C is placed into 45.0 mL of water at 40 C. Determine the final physical state and temperature of the result. Calculate DS for the ice cube and the surrounding water, and

Use the value DS_tot to explain why this process is spontaneous.

b. use data above and from http://bilbo.chm.uri.edu/CHM112/tables/thermtable.htm to find the standard entropy of ice at zero C.

Answer 1

1. from -10 deg,c to 0 deg.c ice gains sensible heat and this is given by = mass of ice* specific heat* temperature difference = 10*2.108*10 = 210.8 joules

2. at 0 deg.c ice gains latent heat of fusion and this heat = moles of ice* heat of fusion of ice = (10/18)*6.001 Kj/mole = 3.3333 Kj= 3333.3 j

Before doing further calculations, we have to see whether water at 40 deg.c will be able to supply the heat required to add sensible heat to ice as well as heat fusion which= 3333.3+210.8 = 3544.1 joules

Volume of water given =45ml, assuming the density to be 1 g/ml, mass of water= 45 *1= 45 gm, sensible heat lost by water during reduction in temperature from 40 deg.c to 0 deg.c= mass of water* specific heat* temperature difference= 45*4.184*(40-0)= 7531.2 joules. This heat is more than the combined heat required for sensible heat of ice and latent heat of fusion. Hence liquid water at 0 deg.c undergoes further heating and if t= final equilibrium temperature

3544.1+10*4.184*(t-0)= 45*4.184*(40-t)=7531.2-188.3t

3544.1+41.84t-41.84= 7531.2-188.3t

Hence t*(41.84+188.3)= 7531.2-3544.1+41.84

Temperature ,t = 17.51 deg.c

Final state is liquid at 17.1 deg.c

Entropy change of ice= mass* specific heat* ln{(0+273/(-10+273)}=0.78 J/K

Entropy change of melting = heat of fusion/ Temperature = 6.001* 1000*(10/18)/273 =12.21 J/K

Entropy change of heating of liquid water from 0 deg.c to 17.1 deg.c

= 10*4.184*ln{(273+17.51)/273}=4. J/K

Total entropy change of ice during phase change= 0.78+12.21+4 =16.99 J/K

Entropy change of mixing of liquid water at 40 deg.c = mass of water*specific heat of water*ln{(17.51+273)/(40+273)}= -14.09 J/K

Overall entropy change= 16.99-14.09= 2.9 J/K

Entropy change is since +ve, the process is spontaneous.

Entropy of liquid water at 25 deg.c= 69.91 J/molK

Entropy of liquid water at 25 deg.c= 69.91*10/18 J/K ( 10 gm of water)=38.84 J//K

Entropy of liquid water at 0 deg.c= 38.84+ 10*4.184*ln(273/298) =35.2 J/K

Entropy change at 0 deg.c = 16.99 J/K

Entropy of liquid water at 0 deg.c- entropy of solid ice at 0 deg.c

=16.99

Entropy of solid ice at 0 deg.c= 35.2-16.99= 18.21 J/K