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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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Answer:

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(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

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If the tension in the string is increased by 15%, we will have;

f_{(1  \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}}  }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L}  = 1.3225 \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1

f_{(1  \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz  = 740.6 \  Hz

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The value of the fundamental frequency will then be given as follows;

f_{(1  \, new)} =  \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \times \dfrac{2 \times L}{3} }  =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}}  }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz =  840 \ Hz

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

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