Answer
$\Delta S_{water}=30.3\frac{J}{K}$
Work Step by Step
We know that:
$\Delta S_{water}=\int_{283}^{314}\frac{cm dT}{T}$
where $40.9^{\circ}C=314K$ and $10.0^{\circ}C=283K$
The above equation can be written as:
$\Delta S_{water}=cm\ln(\frac{T_f}{T_i})$
We plug in the known values to obtain:
$\Delta S_{water}=4190(0.0700)\ln(\frac{314}{283})=30.3\frac{J}{K}$
