Calculating the change in entropy, did I calculate correctly?

1. The problem statement, all variables and given/known data
A 3.5 kg block of cobber at 100 degrees celsius (373 K) is put in 0.8 kg water at 0 degrees celsius (273 K).
The equilibrium temperature is 30 degrees celsius (303 K).
Calculate the change of entropy for the system of cobber and water.

2. Relevant equations

3. The attempt at a solution
Change of entropy for the cobber block:
[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=3.5kg\cdot 390\frac{J}{kg K}\left ( \ln\frac{303 K}{373 K} \right )=-284 J/K[/itex]

Change of entropy for the water:
[itex]ΔS=S_2-S_1=\int^2_1\frac{dQ}{T}=\int^{T_2}_{T_1}mc\frac{dT}{T}=mc \ln\frac{T_2}{T_1}=0.8kg\cdot 4190\frac{J}{kg K}\left ( \ln\frac{303 K}{273 K} \right )=349 J/K[/itex]

Total change of entropy:
[itex]ΔS=-284 J/K+349 J/K=+65 J/K[/itex]

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