an 80 get ice cube at 10 degrees Celsius is put into a flask

an 8.0 get ice cube at -10 degrees Celsius is put into a flask containing 100 cm^3 of water at 20 degrees celsius. By how much has the entropy of the cube water system changed when the equilibrium Is reached? the specific hear of ice is 2220 j/kg.K

Solution

The change in entropy in this case is

       S1 = integral ( dQ /T)

              = m C_ice ln (Tf / Ti)

              = (8 x 10^-3) (2220) ln (273 / 263)

              = 0.662 J/ K

Now the ice melts at 0^oC , Then

       S2 = Q / T = mL /T   = 8 * 333 / 273 = 9.7882 J /K

Now the water warms to lake temperature

         S3 = m Cw ln(Tf / Ti)

                = (1000)(100*10^-4 )(4187) ln (293 / 273)

                = 29.6 J /K

Therefore total change in entropy = 0.662 + 9.7882 + 29.6

         S = 40.0502 J /K