Julie is making a capacitor with an area of 2.5 × 10–5 m2. The capacitance is 5.5 pF. What is the distance between the plates?
1 answer:
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Answer:
change in entropy is 1.44 kJ/ K
Explanation:
from steam tables
At 150 kPa
specific volume
Vf = 0.001053 m^3/kg
vg = 1.1594 m^3/kg
specific entropy values are
Sf = 1.4337 kJ/kg K
Sfg = 5.789 kJ/kg
initial specific volume is calculated as
FROM STEAM Table
at 200 kPa
specific volume
Vf = 0.001061 m^3/kg
vg = 0.88578 m^3/kg
specific entropy values are
Sf = 1.5302 kJ/kg K
Sfg = 5.5698 kJ/kg
constant volume so
Change in entropy
=3( 3.36035 - 2.88) = 1.44 kJ/kg
Mass = 0.201kg
Energy = 15J
temperature change = 10C
Energy(E) = mass(m) × specific heat capacity(c) × temperature change(θ)
we can rearrange this to make specific heat capacity the subject
c =
c =
c =7.46268657
Energy = 15J
temperature change = 10C
Energy(E) = mass(m) × specific heat capacity(c) × temperature change(θ)
we can rearrange this to make specific heat capacity the subject
c =
c =
c =7.46268657