Answer to Question #105765 in Molecular Physics | Thermodynamics for Wapak

Question

During a non-flow process 120 Btu of heat is removed from each lbm of the working substance while the internal energy of the working substance decreases by 85.5 Btu/lbm. Determine how much work is involved in the process and indicate whether the work is done on or done by the working substance.

Answer 1

As per the question,

Heat removed during non-flow process=120 Btu

The decrease of the internal energy =85.5 Btu/lbm

work is involved in the process=?

Now, by the law of the thermodynamics,

$\Delta U=Q-W$

$W=Q-\Delta U$

Hence,

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