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ivann1987 [24]
3 years ago
9

A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the

temperature of this object is raised from 22 to 320°c. how much work is done by the expanding aluminum if the air pressure is 1.01 ✕ 105 pa? j
Physics
1 answer:
Alex17521 [72]3 years ago
4 0

Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5

So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

W = 3.12 J

Hope this helps!

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The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

<h3>What is pressure?</h3>

The physical force used to apply pressure to an object is defined as such. Per square inch of an object, a force is applied perpendicularly to its surface. For pressure, the fundamental formula is F/A. (Force per unit area). The Pascal is the unit of pressure (Pa).

The four different types of pressure are absolute, atmospheric, differential, and gauge pressure. Have you ever noticed that when you use a straw to drink something, the air actually gets suked out? In reality, you're applying "Pressure" as you drink the beverage.

A box has the dimensions of 50 cm × 30 cm × 15 cm

Let each face be A, B and C

The weight of the box = 150 N

Formula for pressure is

P = F/A

To find out which face of the box can withstand a pressure of 0.25 N/cm

we need find the area of each face and find its pleasure

Face A = 50 cm × 30 cm

Area A = l × b

            = 50 × 30

            = 1500 cm²

Pressure A = 150/1500

                  = 0.1 N/cm³

                    0.25 > 0.1

The surface can definitively withstand the pressure of Face A

Face B = 30 cm × 15 cm

Area A = l × b

            = 30 × 15

            = 450 cm²

Pressure A = 150/450

                  = 0.3 N/cm³

                    0.25 < 0.3

The surface could not withstand the pressure of Face B

Face C = 50 cm × 15 cm

Area A = l × b

            = 50 × 15

            = 750 cm²

Pressure A = 150/750

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                    0.25 > 0.2

The surface can definitively withstand the pressure of Face C

Thus, The surface can only withstand a pressure Face A (50cm × 30cm) and Face C (50cm × 15cm) as the surface can only withstand a pressure of 0.25 N/cm³.

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Six new refrigerator prototypes are tested in the laboratory. For each refrigerator, the electrical power P needed for it to ope
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Answer:

performance coefficient from largest to the smallest

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) P= 400 W, Qc,max/deltaT= 1200 J/s

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2.1.P= 1000 W, Qc,max/deltaT= 3000 J/s

P= 500 W, Qc,max/deltaT= 2000 J/s

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A refrigerator is a device that uses work to remove heat energy from a cold reservoir and deposit it into a hot reservoir. .A good refrigerator (with a large performance coefficient) will remove a large amount of heat energy from the cold reservoir for a small amount of work input

The performance coefficient  of a refrigerator is defined as the ratio of the heat energy removed from the cold reservoir  to the work  input to the refrigerator:

k=QC/W

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3.2000/500=4

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6.3000/1000=3

performance coefficient from largest to the smallest

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P= 250 W, Qc,max/deltaT= 1000 J/s

P= 750 W, Qc,max/deltaT= 1500 J/s

) P= 400 W, Qc,max/deltaT= 1200 J/s

P= 500 W, Qc,max/deltaT= 1500 J/s

P= 1000 W, Qc,max/deltaT= 3000 J/s

2, Rate at which they raise the temperature of the room.

rate at which temperature rises in the inner chamber of the refrigerator is proportional to the rate of energy used to dispel heat from the refrigerator

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P= 500 W, Qc,max/deltaT= 2000 J/s

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P= 500 W, Qc,max/deltaT= 1500 J/s

P= 400 W, Qc,max/deltaT= 1200 J/s

P= 250 W, Qc,max/deltaT= 1000 J/s

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