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2 years ago

Which formula can be used to solve problems related to the first law of thermodynamics? q = w u q = u – deltaw q = u w q = u – w

Physics

1 answer:

Orlov [11]2 years ago

3 0

"Q = ΔU + W" is the equation is used to solve the questions related to "First law of thermodynamics".

<h3> What is the first law of thermodyanamics?</h3>

"First law of thermodynamics" states that "energy" neither created nor destroyed, but it can transfer from "one form of energy" to "another form of energy".

This "First law of thermodynamics" is also called as "law of conversation of energy". The formula for "First law of thermodynamics" of a system is that "change in internal energy of a system" is same as the difference of "heat energy" flows across the " boundaries of a system" and the "work done" on the system.

ΔU = Q - W

Q = ΔU + W

Where, "ΔU" is "change in internal energy", "Q" is "heat transferred and "W" is "work done.

Hence "Q = ΔU + W" is the equation is used to solve the questions related to "First law of thermodynamics".

To know more about the First law of thermodynamics follow

brainly.com/question/15071682

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Gravity is all ways pulling down and the normal force acting on top of the object and for it to have to push or pull to the object

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3 years ago

Urgent need help 100 points

Sati [7]

Answer: The velocity with which the sand throw is 24.2 m/s.

Explanation:

Explanation:

acceleration due to gravity, a = 3.9 m/s2

height, h = 75 m

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Use third equation of motion

The velocity with which the sand throw is 24.2 m/s.

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2 years ago

Percent Yield Lab Report

Vlada [557]

Based on the data obtained from the reaction, the following conclusions can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

<h3>What is the percent yield of the reaction?</h3>

Equation of the reaction is given below:

2 Mg + O₂ ----> 2 MgO

Trial 1

Mass of empty crucible with lid = 26.698 (g)

Mass of Mg metal, crucible, and lid = 27.040 (g)

Mass of MgO, crucible, and lid = 27.198 (g)

Mass of metal = 27.040 - 26.698 = 0.342

Mass of MgO = 27.198 - 26.698 = 0.500

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.342/24 = 0.01425

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.01425

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.500/40 = 0.0125

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percentage yield = 0.0125/0.01425 * 100%

Percent yield of MgO = 87.7%

Trial 2

Mass of empty crucible with lid = 26.691 (g)

Mass of Mg metal, crucible, and lid = 27.099 (g)

Mass of MgO, crucible, and lid = 27.361 (g)

Mass of metal = 27.099 - 26.691 = 0.408

Mass of MgO = 27.361 - 26.691 = 0.670

<h3>Moles of Mg used</h3>

moles of Mg = mass/molar mass

molar mass of Mg = 24 g

moles of Mg = 0.408/24 = 0.0170

<h3>Moles of MgO expected</h3>

Based on the equation of reaction;

moles of MgO expected = 0.0170

moles of MgO produced = mass/molar mass

molar mass of MgO = 40 g/mol

moles of MgO produced = 0.670/40 = 0.01675

<h3>Percent yield</h3>

Percent yield = (moles of MgO produced/moles of MgO expected) * 100%

Percent yield = 0.01675/0.0170 * 100%

Percent yield of MgO = 98.5%

Average percent yield = (87.7 + 98.5)% / 2

Average percent yield = 89.0%

Based on the data obtained from the reaction, the following conclusion can be made;

according to the law of conservation of mass, the mass of the product was higher than the reactant because of the mass of oxygen added during the combustion.
the percent yield is less than 100% because of loss in mass of either reactants or products.
the errors could have occurred during the weighing and transfer of reactants and products
repeated measurements are required in order to improve accuracy

Learn more about law of conservation of mass at: brainly.com/question/1824546

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2 years ago

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The answer would be a frog!

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3 years ago

Read 2 more answers

Betty weighs 400 N and she is sitting on a playground swing seat that hangs 0.21 m above the ground. Tom pulls the swing back an

disa [49]

Answer:

4.15 m/s

Explanation:

As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:

ΔE = ΔK + ΔU =0

If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:

h₀ = 1.09 m -0.21 m = 0.88 m

⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J

As Uf = 0, ΔU = Uf -U₀ = -352 J

As the swing starts from rest, K₀=0, so we can say:

ΔK = Kf = $\frac{1}{2} *m*vf^{2}$ (1)

As ΔK = -ΔU ⇒ ΔK = 352 J (2)

From (1) and (2) we can solve for vf, as follows:

$vf = \sqrt{\frac{2*352J}{40.8kg}} = 4.15 m/s$

So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.

5 0

3 years ago