Answer
a) $\Delta u =6193.71\ kJ/kg$
b) $\Delta u =6233.40\ kJ/kg$
c) $\Delta u =6109.80\ kJ/kg$
Work Step by Step
a) Assuming ideal gas behaviour:
$c_v = \frac{1}{M}\bar{c}_v=\frac{1}{M}(\bar{c}_p-R_u)$
$c_v = \frac{1}{M}\bar{c}_v=\frac{1}{M}(a+bT+cT^2+dT^3-R_u)$
$\Delta u = \int c_v dT$
$\Delta u = \frac{1}{M}((a-R_u)T+b/2.T^2+c/3.T^3+d/4.T^4)|_{T_1}^{T_2}$
With $T_2=800K, T_1=200K$, and the constant values from table A-2c:
$\Delta u =6193.71\ kJ/kg$
b) Since $T_{avg} = 500K$ and from table A-2b: $c_{v,\ avg} \approx c_{v,\ 500K}=10.389\ kJ/kg.K$
$\Delta u = c_{v,\ avg} \Delta T$
$\Delta u =6233.40\ kJ/kg$
c)At room temperature: $c_v=10.183\ kJ/kg.K$
$\Delta u =6109.80\ kJ/kg$
