Answer:
109.779BTU/lbm and 220.179BTU/lbm
Explanation:
We have the values of Pressure and Temperature,
[tex]P_1 = 10psi\\T_1 = 500\°F\\P_2 = 80psi\\T_2 = 800\°F[/tex]
At the tables, this conditions are for enthalpy and specific energy,
[tex]s_1=1.9693BTU/lbm.R\\s_2=1.8704BTU/lbm.R\\u_1=1182.2BTU/lbm\\u_2=1292.6BTU/lbm[/tex]
We calculate now the heat transfer for the system,
[tex]q_{1-2}=\frac{T_1+T_2}{2}(s_1-s_2)[/tex]
[tex]q_{1-2}=\frac{(500+460)(800+460)}{2}(1.9693-1.8704)[/tex]
[tex]q_{1-2}=109.779BTU/lbm[/tex]
We can calculate now the work transfer given by,
[tex]w=q-\Delta u[/tex]
[tex]w= 109.779-(1182.2-1292.6)[/tex]
[tex]w=220.179BTU/lbm[/tex]