Determine the angular velocity ω of the telescope as it orbits around the Sun.
1 answer:
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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
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