Answer:
Δs = 0.133 KJ/.K
Step-by-step explanation:
Given that
V= 20 L = 0.02 m³
At P₁= 200KPa and T₁=150 C :
v₁=0.93 m³/kg
s₁=7.2 KJ/kg.K
At T₂=40 C :
v₁ = v₂
vf= 0.001 m³/kg
vg = 19.51 m³/kg
v₂ = vf + x(vg-vf)
0.93 = 0.001 + x(19.51 - 0.001)
x=0.047
s₂ = sf + x(sg-sf)
s₂ = 0.57 + 0.047(7.638) KJ/kg.K
s₂ = 0.93 KJ/kg.K
Δs= s₁ - s₂
Δs= 7.2 - 0.93 =6.2 KJ/kg.K
m=V/v₁
m=0.02/0.93 = 0.021 kg
Δs = 0.021 x 6.2 = 0.133 KJ/.K
Δs = 0.133 KJ/.K