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Anton [14]
3 years ago
5

A person jogs for 4.0 km in 32 mins,then 2.0km in 22mins,and finallu 1.0 km in 16 mins.whqt is the joggers average speed in km p

er minute?
Physics
1 answer:
blondinia [14]3 years ago
5 0

Average speed = (total distance covered) / (time to cover the distance)

total distance covered = (4km + 2km + 1km) = 7 km

time to cover the distance = (32min + 22min + 16min) = 70 min

Average speed = (7 km) / (70 min)

Average speed = 0.1 km/minute

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The Entropy generated by the steam = 2.821 kJ/K

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Total volume of container = 5m³

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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

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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,  U_{f2} = 340.49 kJ/kg

(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}

From the steam table,

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

Therefore the entropy generated will be :

Entropy = mass* (S₂ - S₁)

Entropy = 3.33* (4.102 - 3.2548)

Entropy = 2.821 kJ/K

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