Answer:
hello your question lacks the required diagram attached below is the complete question with the required diagram
answer : Qtotal = 807.4 Mw
Explanation:
Given Data :
disk properties :
∈ = 0.65
D = 200 mm
Ts = 400⁰c
attached below is the detailed solution
The total rate of Heat transferred from the disk
Qtotal = 807.4 Mw
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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Answer:
Part 1)
Part 2)
Part 3)
Part 4)
Since torque on right side is more so here it will turn and slip over it
Explanation:
As we know that the block A is placed at distance
d = 50 cm from the hinge at 70 cm mark
So torque due to weight of A is given as
the block B is placed at distance
d = 30 cm from the hinge at 70 cm mark
So torque due to weight of B is given as
Now torque due to weight of the scale is given as
now torque on left side of scale is given as
Torque on right Side is given as
Since torque on right side is more so here it will turn and slip over it