The change in entropy in this case is
      ΔS1 = integral ( dQ /T)
              =
m Cice ln (Tf / Ti)
              =
(8.5 x 10-3) (2220) ln (273 / 263)
              =
0.704 J/ K
Now the ice melts at 0oC , Then
      ΔS2 = Q / T = mL
/TÂ Â = 8.5 * 333 / 273 = 10.4 J /K
Now the water warms to lake temperature
        ΔS3 = m Cw ln(Tf /
Ti)
                =
(1000)(1.5*10-4 )(4187) ln (293 / 273)
                =
44.4 J /K
Therefore total change in entropy
        ΔS = 56.1 J /K
