Question

Four pounds of certain gas receive 50 BTU of heat during a constant temperature reversible non-flow process at 165 0F. Find a.) the change of specific entropy, BTU/lbm.0R, and b.) the work done, BTU. Show the process on the p-V and T-s diagram.

Answer 1

According to the First Law of Thermodynamics

$Q=\Delta U+W$ or $\frac{Q}{m}=\frac{\Delta U}{m}+\frac{W}{m}\to q=u+w$

If $T=165°F=const$ then $\Delta U =0 \to u=0$

(a) the change of specific entropy

$q=T(s_2-s_1)=T\cdot \Delta s \to \Delta s=\frac{q}{T}=\frac{Q}{m\cdot T}=\frac{50}{4\cdot 165}=0.076 \frac{BTU}{lb\cdot°F}$

(b) the work done

$W=Q=50$ $BTU$