The coefficient of linear expansion (α) quantifies how much a material expands per degree change in temperature. It is given by the formula:
\[ \alpha = \frac{\Delta L / L_0}{\Delta T} \]
Where:
- \(\Delta L\) is the change in length,
- \(L_0\) is the original length,
- \(\Delta T\) is the change in temperature.
In your case, you have two dimensions (0.34 m and 0.28 m). Assuming you want to treat them individually, you can calculate the coefficient of linear expansion for each dimension by following these steps:
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Calculate the change in temperature (\(\Delta T\)): \[ \Delta T = T_1 - T_2 = 100°C - 60°C = 40°C \]
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Choose one dimension to calculate the change in length. Let's consider one dimension (the longer one, for example, 0.34 m):
Assuming the length changes linearly with temperature, we can look at the fractional change in length caused by the temperature change.
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Assume the original length (\(L_0\)) is 0.34 m: We will calculate the coefficient for this dimension:
Let's denote the new length after temperature change as \(L_1\) (which we don't know yet), but we know \(\Delta L\) is defined as: \[ \Delta L = L_1 - L_0 \]
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However, we don't have \(L_1\) directly given for temperature 60 °C. Instead, we can assume a hypothetical expansion amount due to the temperature change, or if the physical context allows (e.g., given expansion amounts), use that to adjust.
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If we hypothetically assume a length change (this is a necessary assumption since no other information is given), let's say the length at 60 °C is \(L_1 = 0.34 m + \Delta L\), where \(\Delta L\) is what we assume based on the material properties or actual measurement, but let's take a simple example formula of change.
If we assume for a fictitious number, let's say \(L_1\) is for calculation purposes \(L_1 = 0.34 m + 0.00136 m\) (as a hypothetical or you might assess this):
\[ \Delta L = L_1 - L_0 = 0.00136 , m \]
Plug these into the formula for linear expansion:
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Calculating:
- Where \(L_0 = 0.34 m\),
- \(\Delta L = 0.00136 m\),
- \(\Delta T = 40°C\):
\[ \alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \]
Substituting values,
\[ \alpha = \frac{0.00136,m}{0.34,m \cdot 40°C} \]
\[ \alpha ≈ \frac{0.00136}{13.6} \]
\[ \alpha ≈ 0.0001 , °C^{-1} \]
This method demonstrates how to find the linear expansion, but actual measurements or additional context of real thermal expansion numbers might vary per material context used, and welded together dimensions would typically expand isotropically.
Additionally, make sure to confirm actual length changes from physical responses if it’s material-dependent or machinable results as mentioned above.
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