Answer
Please see the work below.
Work Step by Step
As We know that
$\Delta S=\int_{25}^{10}n C\frac{dT}{T}$
We plug in the known values to obtain:
$\Delta S=40\times \frac{31}{(343)^3}\int_{25}^{10}T^2dT$
$=\frac{3\times 10^{-5}}{3}((10)^3-(25)^3)=-.148\frac{J}{K}$
The negative sign shows that copper is cooled.
