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Delvig [45]
3 years ago
7

If the change in enthalpy is -5074.3 kj, how much work is done during the combustion?

Physics
1 answer:
Anton [14]3 years ago
6 0
<span>Answer: I guess there is is a typo in your question both change in enthalpy and change in internal energy for the combustion are negative: â†H = -5074 kJ â†U = - 5084.2 kJ Work done to surrounding in a process is given by the integral: W = â« P dV from initial to final volume for a constant pressure process the integral simplifies to: W = P â™ â« dV = Pâ™â†V Enthalpy is defined as: H = U + Pâ™V So the change in enthalpy and the change in internal energy are related as: â†H = â†U + â†(Pâ™V) for a constant pressure process: â†H = â†U + Pâ™â†V Hence, â†H = â†U + W => W = â†H - â†U = -5074 kJ - (-5084.2) kJ = 10.2 kJ</span>
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What is the circle of least confusion?
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Answer and Explanation:

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A straight, vertical wire carries a current of 2.15 A downward in a region between the poles of a large superconducting electrom
enot [183]

Answer:

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b) F = 0.01234N to the north direction

c) F = 0.01234N

d) North of West

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Current flowing through the wire, I = 2.15 A

a) If the magnetic field direction is east

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b) If the magnetic field direction is south

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c) If the magnetic field direction is 30 degrees south

The angle between the magnetic field and the length of the wire still remains 90 degrees

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8 0
3 years ago
2.
gizmo_the_mogwai [7]

Answer:

a) P1=100kpa

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V2=?

P2=50kpa

rearranging mathematically the expression for Boyle's law

V2=(P1V1)/P2=(100×6)/50=12m³

b) same apartment as in (a) but only the value of P2 changes

=> V2=(100×6)/40=15m³

Explanation:

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4 0
3 years ago
A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch, and then resumes driving for the next 2.0 h through a di
Oksi-84 [34.3K]
<h2>82.353 km/hr</h2>

Explanation:

       The driver travels 135 km towards East in 1.5 hr. He stops for 45 min. He again travels 215 km towards East in 2.0 hr.

       The total displacement of the driver in the given time is ths sum of individual displacements, because all the displacements are in the same directon.

       Total displacement = 135\text{ }km\text{ }East\text{ }+\text{ }0\text{ }km\text{ }+\text{ }215\text{ }km\text{ }East\text{ }=\text{ }350\text{ }km\text{ }East

       Total time travelled = 1.5\text{ }hr\text{ }+\text{ }45\text{ }min\text{ }+\text{ }2.0\text{ }hr\text{ }=\text{ }90\text{ }min\text{ }+\text{ }45\text{ }min\text{ }+\text{ }120\text{ }min\text{ }=\text{ }255\text{ }min\text{ }=\text{ }4\text{ }hr\text{ }15\text{ }min\text{ }=\text{ }4.25\text{ }hr

      \text{Average velocity = }\dfrac{\text{Total displacement}}{\text{Total time taken}}=\dfrac{350\text{ }km}{4.25\text{ }hr}=82.353\text{ }\frac{km}{hr}

∴ Driver's average velocity = 82.353\text{ }\frac{km}{hr}

3 0
3 years ago
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