Answer to Question #105411 in Molecular Physics | Thermodynamics for Kule

Question

One pound of saturated steam at 689.5 kPa is compressed in a reversible non-flow process from the specific volume of 0.2768 m?/kg to a specific volume of 0.001107 m3/kg. If the pressure remains constant at 689.5 kPa and the internal energy decreases 1878.2 kJ/kg during the process, determine how much heat will be transferred and whether it will be added to our abstracted from the working substance .

Answer 1

$1$ pound $=$ $0.45359237$ $kg$

$Q=\Delta U+p\Delta V\to \frac{Q}{m}=\frac{\Delta U}{m}+p\frac{\Delta V}{m}$

$q_m=-1878.2+689.5\cdot(0.001107-0.2768)\approx-2069\frac{kJ}{kg}$

$q=-2069\cdot0.45359237\approx-938.5kJ$

The heat will be transferred from the working substance.