A4.45 kg block of ice at 0.00 ? c falls into the ocean and melts. the average temperature of the ocean is 3.70 ? c, including all the deep water.

The ocean does not change temperature but it does lose some entropy ( he gives heat to melt the ice and to warm it to 3.70° C ).
I ) For the ice:
1 ) For melting the ice:
Q = m · Lf = 4.45 kg · 334 · 10³ J/kg = 1,486,300 J
Δ S = Q / T = 1,486,300 J / 273.15 K = 5.441 · 10³ J/K.
2 ) To warm the melted ice to 3.70° C:
Q = m c Δ T = 4.45 kg · 4,190 J / kgK · 3.70 K = 68,988.35 J
Δ S = m c ln( T2/ T1 ) = 4.45 kg · 4,190 J/kgK · ln ( 276.85 / 273.15 ) =
= 18,645.5 · ln ( 1.01354 ) = 18.645.5 · 0.013454 = 250.8692 J/K
II ) For the ocean:
Δ S = Q / T = ( - 68,988.35  + 1,486,300 ) / 276.85 = - 5,617.8 J/K
The net entropy change:
Δ S = ΔS ice + ΔS ocean = 5,441.1 + 250.8692 - 5,617.8 = 74.1692 J/K
74 J/K.