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lisabon 2012 [21]
2 years ago
10

A vessel with an unknown volume is filled with 10 kg of water at 90oC. Inspection of the vessel at equilibrium shows that 8 kg o

f the wateris in the liquid state. What is the pressure in the vessel, and what is the volume of the vessel
Physics
1 answer:
damaskus [11]2 years ago
8 0

Hey there!

In this case, it is possible to solve this problem by using the widely-known steam tables which show that at 90 °C, the pressure that produces a vapor-liquid mixture at equilibrium is about 70.183 kPa (Cengel, Thermodynamics 5th edition).

Moreover, for the calculation of the volume, it is necessary to calculate the volume of the vapor-liquid mixture, given the quality (x) it has:

x=\frac{m_{steam}}{m_{total}}

Thus, since 8 kg correspond to liquid water, 2 kg must correspond to steam, so that the quality turns out:

x=\frac{2kg}{10kg} =0.20

Now, at this temperature and pressure, the volume of a saturated vapor is  2.3593 m³/kg whereas that of the saturated liquid is 0.001036 m³/kg and therefore, the volume of the mixture is:

v=0.001036m^3/kg+0.2(2.3593-0.001036 )m^3/kg=0.4727m^3/kg

This means that the volume of the container will be:

V=10kg*0.4727m^3/kg\\\\V=4.73m^3

Regards!

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The charge present determines a force to be attractive or repulsive.

The charges acquired by two bodies determines the Force as Attractive Or Repulsive.

Electric Force applied due to Electrical charges is same in magnitude but opposite in direction. This corresponds this phenomenon equivalent to the Newton's Third Law.

Examples of the experiments and observations:

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This occurs due to the Electric charges present in the comb that induces charge in paper pieces leading to their attraction.

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If both the objects are charged the same i.e. either positive or negative then the Force is Repulsive and if the charges are Oppositely charged then the force is attractive.

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A rocket in deep space has an empty mass of 150 kg and exhausts the hot gases of burned fuel at 2500 m/s. It is loaded with 600
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Answer:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

Explanation:

The speed of the rocket is given the Tsiolkovsky's differential equation, whose solution is:

v (t) = v_{o} - v_{ex}\cdot \ln \frac{m}{m_{o}}

Where:

v_{o} - Initial speed of the rocket, in m/s.

v_{ex} - Exhaust gas speed, in m/s.

m_{o} - Initial total mass of the rocket, in kg.

m - Current total mass of the rocket, in kg.

Let assume that fuel is burned linearly. So that,

m(t) = m_{o} + r\cdot t

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m_{o} = 750\,kg

The fuel consumption rate is:

r = -\frac{600\,kg}{30\,s}

r = -20\,\frac{kg}{s}

The function for the current total mass of the rocket is:

m(t) = 750\,kg - (20\,\frac{kg}{s} )\cdot t

The speed function of the rocket is:

v(t) = - 2500\,\frac{m}{s}\cdot \ln \frac{750\,kg -(20\,\frac{kg}{s} )\cdot t}{750\,kg}

The speed of the rocket at given instants are:

v(10\,s) \approx 775.387\,\frac{m}{s}

v(20\,s)\approx 1905.350\,\frac{m}{s}

v(30\,s) \approx 4023.595\,\frac{m}{s}

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