Answer
$\Delta S = 0.03675 \space J/K$
Work Step by Step
$\Delta S = \int \frac{nC_vdT}{T}$
Integrate time from 5.00 to 10.00 and leave aside the constants
$\Delta S = nA \int^{10}_{5} T^2dT$
$\Delta S = nA [\frac{T^3}{3}]^{10}_5$
$\Delta S = \frac{nA}{3}[10^3 -5^3] $
Substitute $A = 3.15 \times 10^{-5} J/mol.K^4$
and $n = 4.0 mol$
$\Delta S = \frac{(4) (3.15 \times 10^{-5} J/mol.K^4)}{3}[10^3 -5^3] $
$\Delta S = 0.03675 \space J/K$
