What is the change in entropy of the bullet during this process?

A bullet of mass m slams into a concrete wall and melts on impact. If the temperature of the speeding bullet was T₁ and if the specific heat of bullet's material is c and its latent heat of fusion is L, what is the change in entropy of the bullet during this process? The melting point of lead is T₂.
Heating:
dS₁ = dQ₁/T
dQ₁ = cmdT
dS₁ = cmdT/T = cmd(lnT)
ΔS₁ = ∫cmd(lnT) = cm∫d(lnT) = cm(lnT₂-lnT₁) = cm ln(T₂/T₁)
ΔS₁ = cm ln(T₂/T₁)
Melting:
dS₂ = dQ₂/T₂
ΔS₂ = ΔQ₂/T₂ = Lm/T₂
ΔS = ΔS₁+ΔS₂ = cm ln(T₂/T₁)+Lm/T₂
A 16-gram lead bullet slams into a concrete wall and melts on impact. If the temperature of the speeding bullet was 25°C and if the specific heat of lead is 0.13 kJ/(kgK) and its latent heat of fusion is 23 kJ/kg, what is the change in entropy of the bullet during this process? The melting point of lead is 601K.
m=16g
T₁=25°C=25°+273.15=298.15K
T₂=601K
c=0.13 kJ/(kgK)
L=23 kJ/kg
ΔS=cm ln(T₂/T₁)+Lm/T₂
Calculation by Google calculator:
0.13kJ/(kg*K) * 16g * ln(601 / 298.15) in J/K =
1.4580763 J / K
23kJ/kg * 16g / 601K in J/K =
0.612312812 J / K
1.4580763 J / K + 0.612312812 J / K = 2.07038911 J / K
ΔS≈2.1J/K