A 1.0-kg block and a 2.0-kg block are pressed together on a horizontal frictionless surface with a compressed very light spring
1 answer:
Answer:
4. both blocks will both have the same amount of kinetic energy.
Explanation:
When the blocks are released free from the compression force, the spring exerts equal and opposite force on each block but the block with heavier (double) mass will attain slower ( half ) speed as compared to the lighter block according to the law of inertia. This works in synchronization to energy conservation.
Spring force is given as:
where: length of compression in the spring
<u>We know kinetic energy is given by:</u>
Hence the kinetic energy of both the blocks is equal when they are released to move free.
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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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