Answer
$W_b/m=151.10\ kJ/kg$
$Q/m = -75.50\ kJ/kg$
Work Step by Step
For a polytropic process:
$P_1^{1-n}T_1^n=P_2^{1-n}T_2^n$
With $n=1.2, P_1=120\ kPa, T_1=10°C=283.15K, P_2=800\ kPa$, we get to:
$T_2=388.45\ K$
The boundary work is given by($R=0.2870\ kJ/kg.K$):
$W_b=\frac{-1}{n-1}(mR(T_2-T_1))$
$W_b/m=151.10\ kJ/kg$
The energy balance for this system($c_v=0.718\ kJ/kg.K$):
$Q-W_b=\Delta U=mc_v\Delta T$
$Q/m = -75.50\ kJ/kg$
