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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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C. They conducted experiments with uranium-containing minerals and pure uranium.

E. They discovered two new radioactive elements

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Convert the following measurements as indicated, show work. write the answer in scientific notation.
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Answer:

A. 0.95 m

B. 1,100 ml

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Explanation:

A. 95/100

B. 1.1 x 1,000

C. 17,000/1,000

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What best describes the angle between a changing electric field and the electromagnetic wave produced by it?
MrRa [10]

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3 years ago
The specific heat of a substance is 0.215 J/g°C. How much energy is required to raise the temperature of 20 g of the substance f
sergey [27]

Taking into account the definition of calorimetry and sensible heat, the amount of energy required is 68.8 J.

<h3>Calorimetry</h3>

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

<h3>Sensible heat</h3>

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

In this way, between heat and temperature there is a direct proportional relationship. The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body.

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where:

  • Q is the heat exchanged by a body of mass m.
  • c is the specific heat substance.
  • ΔT is the temperature variation.

<h3>Energy required in this case</h3>

In this case, you know:

  • Q= ?
  • c= 0.215 \frac{J}{gC}
  • m= 20 g
  • ΔT= Tfinal - Tinitial= 88 C - 72 C= 16 C

Replacing in the definition of sensible heat:

Q = 0.215 \frac{J}{gC}× 20 g× 16 C

Solving:

<u><em>Q=68.8 J</em></u>

Finally, the amount of energy required is 68.8 J.

Learn more about calorimetry:

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6 0
2 years ago
A blue supergiant star has surface temperature 27 kK and has total power output 100000 times that of the Sun.
Free_Kalibri [48]

Answer:

Wien peak ( λmax ) is 107.40 nm

radius of super giant is 1.086 ×10^{10} m

Explanation:

given data

temperature 27 kK

power = 100000 times of Sun

Sun radius =  6.96 × 10^8 m

to find out

Wien peak ( λmax ) and radius of supergiant (r)

solution

we will apply here first wien law to find Wien peak  that is

λmax = b / t

λmax = 2.9 × 10^{-3} / 27000 = 1.0740 × 10^{-7}

so Wien peak ( λmax ) is 107.40 nm

and

now we apply steafay law that is

P = σ × A × T^{4}   .........................1

and we know total power output 100000 time of Sun

so we say  

4πr²sT^{4}  = 100000 × 4πR²sTs^{4}

r² = 100000  × R²Ts^{4} / T^{4}

put here value

r² = 100000  × (6.96×6000^{8} )² × 6000^{4} / 27000^{4}

r² = 1.18132 ×10^{20}

r = 1.086 ×10^{10} m

so radius of super giant is 1.086 ×10^{10} m

3 0
3 years ago
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