Answer to Question #105316 in Molecular Physics | Thermodynamics for Kule

Question

During a non-flow process 120 Btu are removed as heat from each pound of working substance while the internal energy decreases 85.5 Btu /lb. Calculate the work involved in the process in ft·lbf/lb

Answer 1

By the law of thermodynamics

$\Delta U=Q-W$

$W=Q-\Delta U$

As 120 Btu of heat is removed from the system which means $Q=-120\ \\$

and $\Delta U=-85.5 Btu$

$W=-120-(-85.5)=-120+85.5=-34.5Btu$