Answer
$\Delta S = 71 J/K$
Work Step by Step
The energy when the ice melt is
$Q = mL_F$
$Q = (4.50 kg)(334 × 10^3 J/kg)$
$Q = 1.503 \times 10^6 J$
The entropy is
$\Delta S_{ice} = \frac{Q}{T} $
$\Delta S_{ice} = \frac{1.503 \times 10^6 J}{273.15 K} $
$\Delta S_{ice} = 5.502 \times 10^3 J/K$
Warm the melted ice to $3.50 ^oC$
$Q = mc \Delta T$
$Q = (4.50 kg)(4190 J/kg.K) (3.50K)$
$Q = 6.599 \times 10^4 J$
$\Delta S_{ice} = mc ln(\frac{T_2}{T_1})$
$\Delta S_{ice} =(4.50 kg)(4190 J/kg.K) ln(\frac{276.65 K}{273.15 K})$
$\Delta S_{ice} = 240.1 J/K$
For the ocean,
$\Delta S = \frac{-6.599 \times 10^4 J - 1.503 \times 10^6 J}{276.65 K} $
$\Delta S = -5.671 \times 10^3 J/K $
The entropy increases which is expected as a natural phenomena of the world.
The net entropy change is
$ΔS = ΔS_{ice} + ΔS_{ocean} $
$\Delta S = (5.502 \times 10^3 J/K + 240.1 J/K) + (-5.671 \times 10^3 J/K)$
$\Delta S = 71 J/K$
