Answer to Question #105397 in Molecular Physics | Thermodynamics for Kule

Question

If the above problems is carried out as indicated, but with the additional provision that the process is also performed at a constant temperature of 327.9 F, calculate the change of entropy for the steam during the process, Btu/b.°R.

Answer 1

So, $\Delta Q=889.2$ $Btu/lb$

The change of entropy ($T=327.9 F=787.5 R=const$ )

$\Delta S=\int{\frac{\delta Q}{T}}=\frac{889.2}{787.5}=1.129$ $\frac{Btu}{lb\cdot °R}$ Answer