A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga
1 answer:
Answer:
Total work done in expansion will be
Explanation:
We have given pressure P = 2.10 atm
We know that 1 atm
So 2.10 atm
Volume is increases from 3370 liter to 5.40 liter
So initial volume
And final volume
So change in volume
For isobaric process work done is equal to
So total work done in expansion will be
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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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Answer:
Explanation:
Since fluid is pumping in and out at the same rate (5L/min), the total fluid volume in the tank stays constant at 350L. Only the amount of salt and its concentration changed overtime.
Let A(t) be the amount of salt (g) at time t and C(t) (g/L) be the concentration at time t
A(0) = 10 g
Brine with concentration of 1g/L is pouring in at the rate of 5L/min so the salt income rate is 5 g/min
The well-mixed solution is pouring out at the rate of 5L/min at concentration C(t) so the salt outcome rate is 5C g/min
But the concentration is total amount of salt over 350L constant volume
C = A / 350
Therefore our rate of change for salt A' is
A' = 5 - 5A/350 = 5 - A/70
This is a first-order linear ordinary differential equation and it has the form of y' = a + by. The solution of this is
So
with A(0) = 10
c + 350 = 10
c = 10 - 350 = -340