Answer
$\overline{C_v}=5.434$ and $T\approx 396.45^{\circ} C$
Work Step by Step
The average value of $C_v$ is given as follows:
$\overline{C_v}=\dfrac{1}{675-20} \int_{20}^{675} [8.27 +10^{-5} (26 T-1.87 T^2)] \space dT$
Applying Simpson's Rule, we will plug in the above equation
$a=20; b=675$
Now, set $n=2$
So, $\overline{C_v}=5.434$
Now, the temperature is: $T= 8.27 +10^{-5} (26 T-1.87 T^2)=5.434 $
and $ T \approx 396.45^{\circ} C$
