Answer:
work done = 117 kJ
Step-by-step explanation:
given data
mass m = 3 kg
constant pressure P = 200 kPa
temperature T = 200°C
solution
we know that work done by steam is express as
work done = pressure × ΔVolume ....................1
and here ΔVolume = final volume - initial volume
we use here steam table and get at pressure 200 kPa
final specific volume = 1.08052 m³/kg
and
initial specific volume = 0.885735 m³/kg
so here
ΔV = 3 × (1.08052 - 0.885735)
ΔV = 0.584 m³
so put value in equation 1 we get
work done by steam = 200 × 0.584
work done = 117 kJ