A) Using the steam tables;
Steam at 600^{\circ} \mathrm{C} and 10 \,\mathrm{bar} \rightarrow 3698.6 \frac{\mathrm{kJ}}{\mathrm{kg}}
Steam at 400^{\circ} \mathrm{C} and 1 \,\mathrm{bar} \rightarrow 3278.6 \frac{\mathrm{kJ}}{\mathrm{kg}}
\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=3278.6 \frac{\mathrm{kJ}}{\mathrm{kg}}-3698.6 \frac{\mathrm{kJ}}{\mathrm{kg}}=-\mathbf{4 2 0 . 0} \frac{\mathbf{k J}}{\mathbf{k g}}
B)
\begin{aligned}& \mathrm{d} \underline{\mathrm{H}}=\mathrm{C}_{\mathrm{P}} \mathrm{dT}\\& \frac{C_{P}^{*}}{R}=A+B T+C T^{2}+D T^{3}+E T^{4}\\& \mathrm{d} \underline{\mathrm{H}}=\left(\mathrm{RA}+\mathrm{RBT}+\mathrm{RCT}^{2}+\mathrm{RDT}^{3}+\mathrm{RET}^{4}\right) \mathrm{dT}\\& \int_{\text {in }}^{\text {out }} \underline{\mathrm{H}}=\int_{\text {in }}^{\text {out }}\left(R A+R B T+R C T^{2}+R D T^{3}+R E T^{4}\right) d T\\& \left(\underline{\mathrm{H}}_{\text {out }}-\underline{\mathrm{H}}_{\text {in }}\right)= \mathrm{RA}\left(\mathrm{T}_{\text {out }}-\mathrm{T}_{\text {in }}\right)+\frac{\mathrm{RB}}{2}\left(\mathrm{~T}_{\text {out }}{ }^2-\mathrm{T}_{\text {in }}{ }^2\right)+\frac{\mathrm{RC}}{3}\left(\mathrm{~T}_{\text {out }}{ }^3-\mathrm{T}_{\text {in }}{ }^3\right) \\ & \qquad\qquad\qquad +\frac{\mathrm{RD}}{4}\left(\mathrm{~T}_{\text {out }}{ }^4-\mathrm{T}_{\text {in }}{ }^4\right)+\frac{\mathrm{RE}}{5}\left(\mathrm{~T}_{\text {out }}{ }^5-\mathrm{T}_{\text {in }}{ }^5\right)\\& \mathrm{T}_{\mathrm{out}}=673 \mathrm{~K}\\& \mathrm{T}_{\text {in }}=873 \mathrm{~K}\end{aligned}
\begin{array}{|c|c|c|c|c|c|c|} \hline\bf Name & \bf Formula & A & \bf B\times10^3 & \bf C\times10^5 & \bf D\times10^8 & \bf E\times10^{11} \\\hline \rm Water & \rm H_2O & 4.395 & -4.186 & 1.405 & -1.564 & 0.632 \\ \hline \end{array}
\begin{aligned}\left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)= & \left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol} \mathrm{K}}\right)\{(4.395)(673 \mathrm{~K}-873 \mathrm{~K}) \\& +\frac{1}{2}\left(-4.186 \times 10^{-3}\right)\left(673 \mathrm{~K}^{2}-873 \mathrm{~K}^{2}\right) \\& +\frac{1}{3}\left(1.405 \times 10^{-5}\right)\left(673 \mathrm{~K}^{3}-873 \mathrm{~K}^{3}\right) \\& +\frac{1}{4}\left(-1.564 \times 10^{-8}\right)\left(673 \mathrm{~K}^{4}-873 \mathrm{~K}^{4}\right) \\& \left.+\frac{1}{5}\left(6.32 \times 10^{-12}\right)\left(673 \mathrm{~K}^{5}-873 \mathrm{~K}^{5}\right)\right\}\\\left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)=&-7702 \frac{\mathrm{kJ}}{\mathrm{kmol}}\left(\frac{1 \mathrm{kmol}}{18.02 \mathrm{~kg}}\right)=-\mathbf{4 2 7 . 4 1} \frac{\mathbf{k J}}{\mathbf{k g}}\end{aligned}
C) Based on Figure 2-3:
Nitrogen at 300 \mathrm{~K} and 10 \,\mathrm{bar} \rightarrow 459 \frac{\mathrm{kJ}}{\mathrm{kg}}
Nitrogen at 200 \mathrm{~K} and 1 \,\mathrm{bar} \rightarrow 357 \frac{\mathrm{kJ}}{\mathrm{kg}}
\widehat{\mathrm{H}}_{\mathrm{out}}-\widehat{\mathrm{H}}_{\mathrm{in}}=357 \frac{\mathrm{kJ}}{\mathrm{kg}}-459 \frac{\mathrm{kJ}}{\mathrm{kg}}=-\mathbf{1 0 2} \frac{\mathbf{k J}}{\mathbf{k g}}
D)
\begin{aligned}&\mathrm{d} \underline{\mathrm{H}}=\mathrm{C}_{\mathrm{P}} \mathrm{dT} \\& \frac{C_{P}^{*}}{R}=A+B T+C T^{2}+D T^{3}+E T^{4} \\& \mathrm{~d} \underline{\mathrm{H}}=\left(\mathrm{RA}+\mathrm{RBT}+\mathrm{RCT}^{2}+R D T^{3}+\mathrm{RET}^{4}\right) \mathrm{dT} \\& \int_{\text {in }}^{\text {out }} \mathrm{d} \underline{\mathrm{H}}=\int_{\text {in }}^{\text {out }}\left(R A+R B T+R C T^{2}+R D T^{3}+R E T^{4}\right) d T\\& \left(\underline{H}_{\text {out }}-\underline{H}_{\text {in }}\right)= \mathrm{RA}\left(\mathrm{T}_{\text {out }}-\mathrm{T}_{\text {in }}\right)+\frac{\mathrm{RB}}{2}\left(\mathrm{~T}_{\text {out }}{ }^2-\mathrm{T}_{\text {in }}{ }^2\right)+\frac{\mathrm{RC}}{3}\left(\mathrm{~T}_{\text {out }}{ }^3-\mathrm{T}_{\text {in }}{ }^3\right) \\ & \qquad\qquad\qquad +\frac{\mathrm{RD}}{4}\left(\mathrm{~T}_{\text {out }}{ }^4-\mathrm{T}_{\text {in }}{ }^4\right)+\frac{\mathrm{RE}}{5}\left(\mathrm{~T}_{\text {out }}{ }^5-\mathrm{T}_{\text {in }}{ }^5\right)\\& \mathrm{T}_{\text {out }}=200 \mathrm{~K}\\& \mathrm{T}_{\text {in }}=300 \mathrm{~K}\end{aligned}
\begin{array}{|c|c|c|c|c|c|c|}\hline \bf Name & \bf Formula & A & \bf B\times10^3 & \bf C\times10^5 & \bf D\times10^8 & \bf E\times10^{11} \\\hline \rm Nitrogen & \rm N_2 & 3.539 & -0.261 & 0.007 & 0.157 & -0.099 \\ \hline\end{array}
\begin{aligned}\left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)= & \left(8.314 \frac{\mathrm{kJ}}{\mathrm{kmol} \mathrm{K}}\right)\{(3.539)(200 \mathrm{~K}-300 \mathrm{~K}) \\& +\frac{1}{2}\left(-2.61 \times 10^{-4}\right)\left(200 \mathrm{~K}^{2}-300 \mathrm{~K}^{2}\right) \\& +\frac{1}{3}\left(7.00 \times 10^{-8}\right)\left(200 \mathrm{~K}^{3}-300 \mathrm{~K}^{3}\right) \\& +\frac{1}{4}\left(-1.57 \times 10^{-9}\right)\left(200 \mathrm{~K}^{4}-300 \mathrm{~K}^{4}\right) \\& \left.+\frac{1}{5}\left(9.9 \times 10^{-13}\right)\left(200 \mathrm{~K}^{5}-300 \mathrm{~K}^{5}\right)\right\}\\ \left(\underline{\mathrm{H}}_{\mathrm{out}}-\underline{\mathrm{H}}_{\mathrm{in}}\right)=&-2909 \frac{\mathrm{kJ}}{\mathrm{kmol}}\left(\frac{\mathrm{kmol}}{28.02 \mathrm{~kg}}\right)=-\mathbf{1 0 3 . 8} \frac{\mathbf{k J}}{\mathbf{k g}}\end{aligned}
E) The answers from parts a and b differ by ~7 kJ/kg, the answers from parts c and d by ~2 kJ/kg. In both cases the percent error is ~1.7%. Whether this is an acceptable level of error depends upon the application. The differences are attributable to the fact that the ideal gas model is an approximation that is best at low pressure, and these processes included pressures up to 10 bar.