Question

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An aluminum beam is 10.0 m long at a temperature of 25.0 °C. The temperature of the beam is raised to 75.0 °C. What is
the change in the length of the beam due to thermal expansion, if aluminum has a coefficient of linear expansion of 2.40 E 5
*C-1?
O0.500 cm
O 1.20 cm
1.80 cm
02.50 cm

Answer 1

Answer:

${\Delta L=0.012 \ m$

Explanation:

Given:

$L_0=10.0 \ m\\\Delta T_0=25.0 \ \textdegree C\\\Delta T_f=75.0 \ \textdegree C\\\alpha_{Al}=2.40 \times 10^{-5} \ \textdegree C^{-1}$

Find:

$\Delta L= \ ?? \ m$

Using the formula for linear expansion.

$\boxed{\left\begin{array}{ccc}\text{\underline{Formula for Linear Expansion:}}\\\\ \Delta L=\alpha L_0 \Delta T\end{array}\right}$

Where...

• "ΔL" represents the change in length
• "α" represents the coefficient of linear expansion
• "L_0" represents the initial length of the object
• "ΔT" represents the change in temperature

$\hrulefill$

Plug the known values into the formula for linear expansion.  

$\Delta L=\alpha L_0 \Delta T\\\\\Longrightarrow \Delta L=(2.40 \times 10^{-5})(10.0)(75.0-25.0)\\\\\therefore \boxed{\boxed{\Delta L=0.012 \ m}}$

Thus, the change in length is found.

Answer 2

The change in length of the aluminum beam is 0.012 m.

Length of the aluminum beam, L = 10 m

Initial temperature of the beam, T₁ = 25°C

Final temperature of the beam, T₂ = 75°C

The coefficient of linear expansion of aluminum, α = 2.4 x 10⁻⁵⁻⁵⁻⁻C⁻¹

When a material's temperature increases, a phenomenon known as linear expansion occurs, which results in an increase in the material's length.

The length of a material that is one unit long changes as the temperature rises by ten degrees Celsius, which is how the coefficient of linear expansion is stated.

The equation for the change in length of the aluminum beam is given by,

ΔL = αLΔT

ΔL = 2.4 x 10⁻⁵x 10 x(75 - 25)

ΔL = 2.4 x 10⁻⁴x 50

ΔL = 0.012 m

To learn more about linear expansion, click:

https://brainly.com/question/14780533

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