Answer
$c_{p}-c_{v}=0.3192\text{ kJ/kg⋅K}$
Work Step by Step
The specific heat difference $\mathrm{c}_p-\mathrm{c}_\nu$ is given as $$
c_p-c_v=-T\left(\frac{\partial v}{\partial T}\right)_P^2\left(\frac{\partial P}{\partial v}\right)_T
$$ Approximating differentials by differences about the specified state,
$$
\begin{aligned}
c_p-c_v & \cong-T\left(\frac{\Delta v}{\Delta T}\right)_{P=15 \mathrm{MPa}}^2\left(\frac{\Delta P}{\Delta v}\right)_{T=80^{\circ} \mathrm{C}} \\
& =-(80+273.15 \mathrm{~K})\left(\frac{\boldsymbol{v}_{100 \mathrm{C}}-\boldsymbol{v}_{600^{\circ} \mathrm{C}}}{(100-60)^{\circ} \mathrm{C}}\right)_{P=15 \mathrm{MPa}}^2\left(\frac{(20-10) \mathrm{MPa}}{\boldsymbol{V}_{20 \mathrm{MPa}}-\boldsymbol{v}_{10 \mathrm{MPa}}}\right)_{T=80^{\circ} \mathrm{C}} \\
& =-(353.15 \mathrm{~K})\left(\frac{(0.0010361-0.0010105) \mathrm{m}^3 / \mathrm{kg}}{40 \mathrm{~K}}\right)^2\left(\frac{10,000 \mathrm{kPa}}{(0.0010199-0.0010244) \mathrm{m}^3 / \mathrm{kg}}\right) \\
& =0.3192\ \mathrm{kPa} \cdot \mathrm{m}^3 / \mathrm{kg} \cdot \mathrm{K}=\mathbf{0 . 3 1 9 2} \mathbf{~} \mathbf{k J} / \mathbf{k g} \cdot \mathbf{K}
\end{aligned}
$$
