Answer
$c_{p}=1.045\text{ kJ/kgK}$
$c_{v}=0.748\text{ kJ/kg⋅K}$
Work Step by Step
The $c_p$ and $c_v$ of ideal gases depends on temperature only, and are expressed as $c_p(T)=d h(T) / d T$ and $c_v(T)=$ $d u(T) / d T$. Approximating the differentials as differences about $400 \mathrm{~K}$, the $c_p$ and $c_v$ values are determined to be $$
\begin{aligned}
c_p(400 \mathrm{~K}) & =\left(\frac{d h(T)}{d T}\right)_{T=400 \mathrm{~K}} \cong\left(\frac{\Delta h(T)}{\Delta T}\right)_{T=400 \mathrm{~K}} \\
& =\frac{h(410 \mathrm{~K})-h(390 \mathrm{~K})}{(410-390) \mathrm{K}} \\
& =\frac{(11,932-11,347) / 28.0 \mathrm{~kJ} / \mathrm{kg}}{(410-390) \mathrm{K}} \\
& =\mathbf{1 . 0 4 5} \mathbf{k J} / \mathbf{k g} \mathbf{K}
\end{aligned}
$$ (Compare: Table A-2b at $400 \mathrm{~K} \rightarrow c_\rho=1.044 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}$ ) $$
\begin{aligned}
c_v(400 \mathrm{~K}) & =\left(\frac{d u(T)}{d T}\right)_{T=400 \mathrm{~K}} \cong\left(\frac{\Delta u(T)}{\Delta T}\right)_{T=400 \mathrm{~K}} \\
& =\frac{u(410 \mathrm{~K})-u(390 \mathrm{~K})}{(410-390) \mathrm{K}} \\
& =\frac{(8,523-8,104) / 28.0 \mathrm{~kJ} / \mathrm{kg}}{(410-390) \mathrm{K}}=\mathbf{0 . 7 4 8} \mathbf{k J} / \mathbf{k g} \cdot \mathbf{K}
\end{aligned}
$$
