Temperature Needed For Equal Lengths Of Two Different Metal Bars

Hey everyone! Let's dive into an interesting physics problem involving thermal expansion. We have two bars made of different materials, and we want to figure out the temperature at which they'll be the same length. Sounds like fun, right? Let's break it down!

Understanding Thermal Expansion

Before we jump into the calculations, let's quickly recap what thermal expansion is all about. Thermal expansion is the tendency of matter to change in volume in response to changes in temperature. When a substance is heated, its particles move more, thus maintaining a greater average separation. Because of this, solids, liquids, and gases tend to expand in volume when heated. Conversely, they contract when cooled. This property is described by a coefficient of thermal expansion, which varies depending on the material.

For solids, we often talk about linear expansion, which is the change in length. The formula for linear expansion is:

ΔL = L₀ * α * ΔT

Where:

• ΔL is the change in length,
• L₀ is the original length,
• α is the coefficient of linear expansion, and
• ΔT is the change in temperature.

Knowing this, we can tackle our problem!

The Problem: Two Bars, Different Materials

Here’s the scenario: We have two bars. Bar A starts with a length of 75 cm, and Bar B starts at 75.3 cm, both at 0°C. The materials have different coefficients of linear expansion: 5.4 x 10⁻⁵ ºC⁻¹ for Bar A and 2.4 x 10⁻⁵ ºC⁻¹ for Bar B. Our mission is to find the temperature at which both bars will have the same length.

Breaking Down the Approach

To solve this, we'll use the concept of thermal expansion to determine how each bar's length changes with temperature. The key is to set up equations for the final lengths of both bars and find the temperature at which these lengths are equal. Let's walk through the steps:

1. Define the variables: Let's define our variables clearly. We have the initial lengths (L₀) and the coefficients of linear expansion (α) for both bars. We need to find the change in temperature (ΔT) that makes their final lengths the same.
2. Write the equations for final lengths: The final length (L) of each bar can be expressed as the initial length plus the change in length due to thermal expansion. So, we’ll write separate equations for Bar A and Bar B.
3. Set the final lengths equal: Since we want to find the temperature at which the lengths are the same, we'll set the two equations equal to each other.
4. Solve for ΔT: This will involve some algebraic manipulation to isolate ΔT, giving us the temperature change needed.
5. Calculate the final temperature: Remember, ΔT is the change in temperature from the initial temperature (0°C). So, we’ll add ΔT to 0°C to get the final temperature.

Step-by-Step Solution

Let's get our hands dirty with the calculations!

1. Define the Variables

• For Bar A:
  • L₀A = 75 cm
  • αA = 5.4 x 10⁻⁵ ºC⁻¹
• For Bar B:
  • L₀B = 75.3 cm
  • αB = 2.4 x 10⁻⁵ ºC⁻¹
• ΔT = Change in temperature (what we want to find)

2. Write the Equations for Final Lengths

The final length (LA) of Bar A at temperature T can be expressed as:

LA = L₀A + ΔLA = L₀A + L₀A * αA * ΔT

Substituting the values for Bar A:

LA = 75 + 75 * (5.4 x 10⁻⁵) * ΔT

Similarly, the final length (LB) of Bar B is:

LB = L₀B + ΔLB = L₀B + L₀B * αB * ΔT

Substituting the values for Bar B:

LB = 75.3 + 75.3 * (2.4 x 10⁻⁵) * ΔT

3. Set the Final Lengths Equal

To find the temperature at which the lengths are equal, we set LA = LB:

75 + 75 * (5.4 x 10⁻⁵) * ΔT = 75.3 + 75.3 * (2.4 x 10⁻⁵) * ΔT

4. Solve for ΔT

Now, let's solve for ΔT. First, we'll expand the terms and rearrange the equation:

75 + 0.00405 * ΔT = 75.3 + 0.0018072 * ΔT

Move the constants to one side and the ΔT terms to the other:

0. 00405 * ΔT - 0.0018072 * ΔT = 75.3 - 75

Simplify:

0. 0022428 * ΔT = 0.3

Now, divide by 0.0022428 to isolate ΔT:

ΔT = 0.3 / 0.0022428

ΔT ≈ 133.76 °C

5. Calculate the Final Temperature

Since the initial temperature was 0°C, the final temperature (T) is:

T = 0 + ΔT

T ≈ 133.76 °C

Final Answer

So, the temperature at which the two bars will have the same length is approximately 133.76°C. That's pretty cool (or hot, in this case!), right? We've successfully used the principles of thermal expansion to solve a real-world problem. Understanding how materials respond to temperature changes is crucial in many engineering applications, from bridge design to electronics manufacturing.

Additional Insights and Considerations

It’s worth noting a few additional points that can provide a deeper understanding of thermal expansion and its applications.

Material Properties Matter: The coefficient of linear expansion is a material-specific property. Different materials expand at different rates when subjected to the same temperature change. For instance, aluminum expands almost twice as much as steel for the same temperature increase. This is why engineers need to carefully consider the materials they use in structures and devices that will experience temperature variations.

Practical Applications: Thermal expansion isn't just a theoretical concept; it has numerous practical applications and implications. Some examples include:

• Bimetallic Strips: These are used in thermostats and other temperature-sensitive devices. They consist of two different metals bonded together. Because the metals have different coefficients of expansion, the strip bends when heated, which can be used to open or close an electrical circuit.
• Expansion Joints in Bridges: Bridges are designed with expansion joints to allow for thermal expansion and contraction of the materials. Without these joints, the bridge could buckle or crack under extreme temperature changes.
• Shrink Fitting: This technique involves heating a metal part to expand it, fitting it over another part, and then allowing it to cool. As the heated part cools, it contracts, creating a tight fit. This is commonly used in manufacturing and engineering to join components securely.

Limitations of the Linear Expansion Formula: The formula ΔL = L₀ * α * ΔT works well for small temperature changes. However, for very large temperature changes, the relationship between temperature and expansion might not be perfectly linear, and more complex models may be needed.

Volume Expansion: While we focused on linear expansion, it’s important to remember that materials also undergo volume expansion. The coefficient of volume expansion is typically about three times the coefficient of linear expansion for solids.

By considering these additional insights, we can appreciate the broader context and implications of thermal expansion in various fields.

Wrapping Up

I hope you found this exploration of thermal expansion and our problem-solving journey insightful. Physics is all about understanding the world around us, and thermal expansion is a fascinating example of how temperature affects materials. Keep exploring, keep questioning, and keep learning! If you have any questions or want to dive deeper into this topic, feel free to ask. Happy calculating!