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Setler79 [48]
3 years ago
14

Kim holds a pinwheel in the air and says it can be used to model a source of energy. Identify the type of energy Kim's pinwheel

models. Explain the type of energy source used.
Physics
1 answer:
Natali5045456 [20]3 years ago
7 0

Answer:

it is kenetic

Explanation: its in motion:D

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If 50 identical light bulbs are connected in series across
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Consider a well-insulated rigid container with two chambers separated by a membrane. The total volume is 5.0 cubic meters. The f
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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 , h_{f} = 1344.8 kJ/kg

Specific internal energy of saturated liquid at 300°C, U_{f1} =  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

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The final pressure, P₂ = 50 kPa

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(U_{fg} )_{2} =  2142.7 kJ/kg

Let the dryness fraction at the second chamber = x

U_{2} = U_{f2} + U_{fg2}

U_{2} = 340.49 + x2140.7Since U₁ = U₂

1332.7 = 340.49 + x2140.7

Dryness fraction, x = 0.463

From steam table, the specific volume is, u_{f2} = 0.00103 m^{3} /kg\\

u_{2} = u_{f2} + xu_{fg2}

u_{2} = 0.00103 + 0.463(3.2393)\\u_{2} = 1.5 m^{3} /kg\\

u_{2} = \frac{v_{2} }{m_{2} }

V₂ = 5 m³

1.5 = 5/m₂

m₂ = 3.33 kg

At 300°C S_{1} = S_{f} = 3.2548 kJ/kg-k\\

S_{2} = S_{f2} + xS_{fg2}

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S_{f2} = 1.0912 kJ/kg-k\\S_{fg2} = 6.5019 kJ/kg-k\\S_{2} = 1.0912 + 0.463(6.5019)\\S_{2} = 4.102 kJ/kg-k

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Entropy = mass* (S₂ - S₁)

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