To find the coefficient of thermal expansion for the rod, we need to understand how materials expand when they are heated. The coefficient of linear thermal expansion (\(\alpha\)) quantifies how much a material expands per degree of temperature change. The formula to calculate this coefficient is:
Formula for Coefficient of Thermal Expansion
The coefficient of linear thermal expansion is given by:
\(\alpha = \frac{\Delta L}{L_0 \cdot \Delta T}\)
Where:
- \(\Delta L\) = Change in length of the rod
- L0 = Original length of the rod
- \(\Delta T\) = Change in temperature
Step-by-Step Calculation
Now, let's break down the calculation step by step:
1. Determine the Change in Length (\(\Delta L\))
The original length of the rod at 20°C is 20.05 cm, and its length at 270°C is 20.11 cm. The change in length can be calculated as:
\(\Delta L = L_{final} - L_{initial} = 20.11 \, \text{cm} - 20.05 \, \text{cm} = 0.06 \, \text{cm}\)
2. Calculate the Original Length (L0)
The original length of the rod is:
L0 = 20.05 \, \text{cm}
3. Find the Change in Temperature (\(\Delta T\))
The initial temperature is 20°C, and the final temperature is 270°C. Thus, the change in temperature is:
\(\Delta T = T_{final} - T_{initial} = 270°C - 20°C = 250°C\)
4. Plug Values into the Formula
Now we can substitute these values into the formula for the coefficient of thermal expansion:
\(\alpha = \frac{0.06 \, \text{cm}}{20.05 \, \text{cm} \cdot 250°C}\)
5. Perform the Calculation
Calculating the denominator:
20.05 cm × 250°C = 5012.5 cm·°C
Now, substituting back into the equation:
\(\alpha = \frac{0.06}{5012.5} \approx 1.195 \times 10^{-5} \, \text{°C}^{-1}\)
Final Result
The coefficient of thermal expansion for the material of the rod is approximately:
\(\alpha \approx 1.20 \times 10^{-5} \, \text{°C}^{-1}\)
This value indicates how much the rod expands per degree Celsius increase in temperature. Understanding this concept is crucial in fields like engineering and materials science, where temperature changes can significantly affect the performance and integrity of materials.