6. A piece of wire, original length 120 mm, is heated from 20 °C to 95 °C. If the final length of wire is 120,324mm . Calculate the linear expansion coefficient of the wire

1 answer

To calculate the linear expansion coefficient (α) of the wire, we can use the formula for linear expansion:

\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]

Where:

  • \( \Delta L \) = change in length
  • \( L_0 \) = original length
  • \( \alpha \) = linear expansion coefficient
  • \( \Delta T \) = change in temperature

Given:

  • Original length (\( L_0 \)) = 120 mm
  • Final length = 120.324 mm
  • Initial temperature = 20 °C
  • Final temperature = 95 °C

Step 1: Calculate the change in length (\( \Delta L \))

\[ \Delta L = \text{Final length} - L_0 = 120.324 , \text{mm} - 120 , \text{mm} = 0.324 , \text{mm} \]

Step 2: Calculate the change in temperature (\( \Delta T \))

\[ \Delta T = \text{Final temperature} - \text{Initial temperature} = 95 , \text{°C} - 20 , \text{°C} = 75 , \text{°C} \]

Step 3: Rearranging the linear expansion formula to find \( \alpha \)

We can rearrange the formula to solve for \( \alpha \):

\[ \alpha = \frac{\Delta L}{L_0 \cdot \Delta T} \]

Step 4: Substitute the values into the equation

\[ \alpha = \frac{0.324 , \text{mm}}{120 , \text{mm} \cdot 75 , \text{°C}} \]

Step 5: Calculate \( \alpha \)

\[ \alpha = \frac{0.324}{120 \cdot 75} = \frac{0.324}{9000} \approx 3.6 \times 10^{-5} , \text{°C}^{-1} \]

Conclusion

The linear expansion coefficient of the wire is approximately:

\[ \alpha \approx 3.6 \times 10^{-5} , \text{°C}^{-1} \]