Answer to Question #105398 in Molecular Physics | Thermodynamics for Kule

Question

One pound of saturated steam at 100 psia is expanded in a reversible non-flow process from a specific volume of 0.017736 #3/16 to a specific volume of 4.434 ftp/lb. If the pressure remains constant and the internal energy increases 807.5 Btu /lb during the process, determine how much heat is transferred, Btu /lb, and whether it will be added to our abstracted from the working substance.

Answer 1

We know that,

$U=Q+W$

$P_1 V_1=\dfrac{100\times 0.0177236\times 144}{778}=0.3280Btu/Ib$

$P_2V_2= \dfrac{100\times 144\times 4.434}{778}=82.06889Btu/Ib$

So, $W=82.06889-0.328=81.740Btu/Ib$

Hence, $Q=807.5+81.740=889.2Btu/Ib$