A 4.45 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

Answer: 74 J/K.