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Dimas [21]
3 years ago
9

What is the quantity of heat energy required to convert 10g cube of ice at -30oC to steam at 120oC. also draw a graph of tempera

ture against energy to show this process.

Physics
1 answer:
julia-pushkina [17]3 years ago
7 0

30,869.2 J

Explanation:

Given parameters:

Mass of ice cube = 10g

Initial temperature = -30°C

Final temperature = 120°C

Specific heat capacity of water = 4.2J/g°C

Specific heat capacity of ice = 2.1J/g°C

Specific heat capacity of steam  = 1.996J/g°C

Latent heat of fusion of water(l) = 334J/g

Latent heat of vaporization = 2230J/g

Unknown:

Quantity of heat required = ?

Temperature-energy graph = ?

Solution:

The temperature energy profile is attached to this solution.

q = mc∅ₓ + mlₓ + mc∅ₙ + mlₙ + mc∅ₐ

qₓ = mc∅ₓ in converting ice from -30°C to ice at 0°C

qₓ is the amount of heat supplied to the ice that changes its temperature from -30°C to that at freezing point:

qₓ  = 10 x 4.2 x (0-(-30)) = 10 x 2.1 x 30 = 630J

qₓ = mlₓ in converting ice to water

qₓ here is the latent heat used to break the ice bonds without a change in temperature:

qₓ = ml = 10 x 334 = 3340J

qₙ = mc∅ₙ is the heat from water at 0°C to boiling point.

This is the heat required to take water from freezing temperature to its boiling point

qₙ = mc∅ₙ  = 10 x 4.2 x (100 - 0) = 4200J

qₙ  = mlₙ is the heat in vaporizing water

In vaporizing water, there is no temperature change when the hydrogen bonds are broken. The heat supplied is not used to raise temperature. We use the latent heat of vaporization:

qₙ   = 10 x 2230 = 22300J

qₐ = mc∅ₐ from vapor at 100°C to 120°C

This is the heat used to raise the temperature of vapor:

qₐ = 10 x 1.996 x (120-100) = 399.2J

The total heat:

  q = qₓ + qₙ + qₐ = 630J + 3340J + 4200J + 22300J + 399.2J =30,869.2 J

Learn more:

Specific heat brainly.com/question/7210400

#learnwithBrainly

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Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

a

 \alpha = 6261 \  rev/minutes^2

b

 \theta  = 613 \ revolutions

Explanation:

From the question we are told that

   The initial  angular speed is w_i =  1120 \ rev/minutes

    The angular speed after t = 13.8 s = \frac{13.8}{60 }  = 0.23 \ minutes  is w_f = 2560 \ rev/minutes

    The time for revolution considered is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  

 Generally the angular acceleration is mathematically represented as

         \alpha = \frac{w_f - w_i }{t}

=>      \alpha = \frac{2560  - 1120 }{0.23}  

=>      \alpha = 6261 \  rev/minutes^2

Generally the number of revolution made is t_r =  20 \ s  =  \frac{20}{60} = 0.333 \  minutes  is mathematically represented as

           \theta  =  \frac{1}{2}  * (w_i + w_f)*  t

=>      \theta  =  \frac{1}{2}  * (1120+ 2560 )*  0.333

=>      \theta  = 613 \ revolutions

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Answer:

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Explanation:

In quantum mechanics the moment operator is given by

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We apply this equation to the given wave functions

a)  φ = e^{ikx}

        .d φ dx = i k e^{ikx}

We replace

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The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero

            p = 0

b) φ = cos kx

           p = h’ k sen kx

The average sine function is zero,

          p = 0

c) φ = e^{-ax^{2} }

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Answer:

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This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

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E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.

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