If 1.80 m3 of water at 0°C is frozen and cooled to -10°C by being in contact with a great

The mass of the water to freeze is:
m = ρ·V = 1000kg/m³ ·2m³ = 2000kg

The heat exchanged by the freezing water
Q = m·( - L + c·ΔT)
with latent heat of fusion of water
L = 333.55kJ/kg
heat capacity of ice
c = 2.05kJ/kgK
=>
Q1 = 2000kg · ( - 333.55kJ/kg + 2.05kJ/kgK·(-25K)
= 2000kg · ( - 384.8kJ/kg)
= -769.600kJ

The same amount is absorbed by the surrounding excess of ice.
Q2 = -Q1 = 769.600kJ

To get the change of entropy of the freezing ice and the surrounding ice, use relation:
dS = dQ/T

For the freezing ice portion:
dS1= m·( - L/Tm + c·dT/T)
(Tm is the absolute melting/freezing temperature,i.e. Tm = 273.15K)
=>
ΔS1 = m·( - L/Tm + ? Tm?Tf c/T dT)= m·( - L/Tm + c·ln(Tf/Tm) )
(Tf is the final absolute temperature, i.e. Tf =(273.165 -25)K=248.15K )
So
ΔS1 = 2000kg · ( - 333.55kJ/kg/273.15K + 2.05kJ/kgK·ln(248.15/273.15) )
= -2836kJ/K

Because surrounding ice is in large absorbs the heat at (almost) constant temperature. Therefore
ΔS2 = Q2/T2
= 769.600kJ/248.15K
= 3101kJ/K

The total change of entropy is the sum of the partial changes:
ΔS = ΔS1 + Δs2
= -2836kJ/K + 3101kJ/K
= 265kJ/K