Video: Calculating Linear Expansion

An aluminum beam is 20 meters long at 25°C. What is the change in length of the beam if its temperature is raised to 55°C? (The coefficient of linear expansion of aluminum is 24 × 10⁻⁶/°C. [A] 1.4 × 10² m [B] 1.4 × 10³ m [C] 1.4 × 10⁻² m [D] 2.4 × 10⁻⁶ m [E] 4.8 × 10⁻⁵ m

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Video Transcript

An aluminum beam is 20 meters long at 25 degrees Celsius. What is the change in length of the beam if its temperature is raised to 55 degrees Celsius? The coefficient of linear expansion of aluminum is 24 times 10 to the negative sixth per degree Celsius.

Let’s begin by underlining the important information in our problem and writing down our known variables. Our aluminum beam is initially 20 meters long. We can use the variable 𝐿 subscript zero for the initial length. This measurement is taken when the initial temperature is 25 degrees Celsius. We can write down the variable 𝑇 subscript zero to represent the initial temperature. We are solving for the change in the length of the beam. The change in length can be represented by the variable Δ𝐿, where Δ means the difference. We are looking for this change when we have a final temperature of 55 degrees Celsius. The variable 𝑇 subscript f can be used to represent final temperature.

In the parentheses under the problem, we are given that the coefficient of linear expansion for aluminum is 24 times 10 to the negative sixth per degree Celsius. The variable used to represent coefficient of linear expansion is the Greek letter 𝛼. This coefficient is dependent on both the material and the state of the matter being expanded or contracted. Materials expand differently based on the differences in the force between atoms. Gases expand more easily because atoms are not held as strongly together. We can draw a diagram to visualize what’s happening in the problem. We have our aluminum beam which is solid at an initial length of 20 meters and temperature 25 degrees Celsius. The yellow dots represent some of the atoms that make up the aluminum beam.

As we add heat to the aluminum beam to raise its temperature, the kinetic energy of these atoms will begin to increase. This means that the atoms will move around more as represented by the pink arrows which are exaggerated to help us visualize how the atoms motion changes when we add heat. Each atom will take up more space than before and the material will begin to expand. When the beam reaches its final temperature of 55 degrees Celsius, it will be at a new length that we will call 𝐿. We are solving for the change in length from the initial 20 meters to the new unknown length 𝐿. To solve for this change in length, we must use the equation for thermal linear expansion.

This equation states that the change in length, Δ𝐿, equals the initial length, 𝐿 subscript zero, times the coefficient of linear expansion, 𝛼, times the change in temperature, Δ𝑇. Based on the values given in the problem, we can begin to substitute in for our variables and solve our equation. Δ𝐿 remains in the equation as that is the variable we are solving for. We replace 𝐿 subscript zero for the initial length of 20 meters. 𝛼 was given as 24 times 10 to the negative sixth per degree Celsius. Our Δ𝑇, change in temperature, is given as our final temperature, 55 degrees Celsius, minus our initial temperature, 25 degrees Celsius. We simplify the equation by subtracting our temperatures, 25 from 55, to get 30. When we multiply out our values, we get 14400 times 10 to the negative sixth meters.

We are left with meters because our temperature change units of degrees Celsius cancel out with our coefficient of linear expansion units of per degree Celsius, leaving meters. Our answer right now is given to three significant figures. But all of the variables in the problem were given to only two significant figures. So, we must round our value to be two significant figures. By looking at our answers, we can see that they’re in scientific notation. This means that you can only have one digit in front of the decimal place. We move the decimal four places to the left which is equivalent to dividing by 10 to the fourth. So, we must add four to our negative six exponent to keep our value the same as before. This gives us a final answer of 1.4 times 10 to the negative second meters. This matches answer choice c.

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