During which phase does the arrangement of water molecules become more orderly?
1 answer:
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Answer:
The Entropy generated by the steam = 2.821 kJ/K
Explanation:
Total volume of container = 5m³
Heat transfer does not exist between system and surrounding, dQ = 0
At the first chamber, temperature of water at saturated liquid is 300°C
From the steam table:
Specific enthalpy of saturated liquid at 300°C ,
Specific internal energy of saturated liquid at 300°C, 1332.7 kJ/kg
For closed system, the first law of thermodynamics state that:
dQ = dw + dU..................(1)
work done for free expansion, dw =0
0 = 0 + dU
dU = 0 , i.e. U₁ = U₂
At the second chamber,
The final pressure, P₂ = 50 kPa
From the steam table, at P₂ = 50 kPa, = 340.49 kJ/kg
Let the dryness fraction at the second chamber = x
Since U₁ = U₂
1332.7 = 340.49 + x2140.7
Dryness fraction, x = 0.463
From steam table, the specific volume is,
V₂ = 5 m³
1.5 = 5/m₂
m₂ = 3.33 kg
At 300°C
From the steam table,
Therefore the entropy generated will be :
Entropy = mass* (S₂ - S₁)
Entropy = 3.33* (4.102 - 3.2548)
Entropy = 2.821 kJ/K
Answer:
48.7 J
Explanation:
For a mass-spring system, there is a continuous conversion of energy between elastic potential energy and kinetic energy.
In particular:
- The elastic potential energy is maximum when the system is at its maximum displacement
- The kinetic energy is maximum when the system passes through the equilibrium position
Therefore, the maximum kinetic energy of the system is given by:
where
m is the mass
v is the speed at equilibrium position
In this problem:
m = 3.6 kg
v = 5.2 m/s
Therefore, the maximum kinetic energy is:
Answer:
The potential difference between the plates increases
Explanation:
As we know that the capacitance of the capacitor is given by:
(1)
where
q = charge
C = capacitance
V = Voltage or Potential Difference
Also, the capacitance of a parallel plate capacitor is given as:
(2)
where
A = Area of the plates
D = Separation distance between the plates
Now, from eqn (1) and (2):
Now, from the above eqn we can say that:
Potential difference depends directly on the separation distance between the plates of the capacitor and is inversely dependent on the area of the plates of the capacitor.
Therefore, after disconnecting, if the separation between the plates is increased the potential difference across it also increases.