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):
[tex]dV = \beta V_{0} dT\\dV = (69*10^{-6})( C)^{-1} * (1.50*10^{-3})m^{3} * (298)C\\dV = 3.0843*10^-5[/tex]
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.
[tex]W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J[/tex]
W = 3.12 J
Hope this helps!
This question involves the concepts of volumetric thermal expansion and pressure-work formula.
The work done by expanding the aluminum is "3.11 J".
First, we will use the formula of volumetric thermal expansion, to find out the change in volume:
[tex]\Delta V = \beta V\Delta T[/tex]
where,
ΔV = change in volume = ?
β = coefficient of volumetric expansion of aluminum = 69 x 10⁻⁶ °C⁻¹
V = Original Volume = 1.5 x 10⁻³ m³
ΔT = Change in temperature = 320 °C - 22 °C = 298 °C
Therefore,
[tex]\Delta V = (69\ x\ 10^{-6}\ ^oC^{-1})(1.5\ x\ 10^{-3}\ m^3)(298\ ^oC)\\\Delta V = 3.084\ x\ 10^{-5}\ m^3\\\\[/tex]
Now, we will use the pressure-work formula to find out the work done:
[tex]W = P\Delta V\\W = (1.01\ x\ 10^5\ Pa)(3.084\ x\ 10^{-5}\ m^3)[/tex]
W = 3.11 J
Learn more about the volumetric thermal expansion here:
https://brainly.com/question/14042569?referrer=searchResults
The attached picture illustrates the phenomenon of volumetric thermal expansion.
