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: $\overline{C_v}=\dfrac{1}{675-20} \int_{20}^{675} [8.27 +10^{-5} (26 T-1.87 T^2)] dT ....(1)$
Now, by using Simpson's Rule, we need to plug into equation (1) $a=20; b=675; $ and let us suppose that $n=2$
Thus, we have $\overline{C_v}=5.434$; this implies that for this average value of temperature, the value of $n=4$ is sufficient.
Now, the temperature is: $T= 8.27 +10^{-5} (26 T-1.87 T^2)=5.434 \implies T \approx 396.45^{\circ} C$
So, $\overline{C_v}=5.434$ and $T\approx 396.45^{\circ} C$
