To solve the problem of finding the increment in the length of a rod when its temperature increases from 27°C to 327°C, we will follow these steps:
### Step 1: Understand the Coefficient of Linear Expansion
The coefficient of linear expansion, denoted as \( \alpha \), is given by the formula:
\[
\alpha = \frac{2}{T}
\]
where \( T \) is the temperature in Kelvin. The temperature range is given as \( 300 \, \text{K} \leq T \leq 600 \, \text{K} \).
### Step 2: Convert Celsius to Kelvin
We need to convert the initial and final temperatures from Celsius to Kelvin:
- Initial temperature \( T_i = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
- Final temperature \( T_f = 327^\circ C = 327 + 273 = 600 \, \text{K} \)
### Step 3: Set Up the Length Change Formula
The change in length \( \Delta L \) can be calculated using the formula:
\[
\Delta L = L \cdot \alpha \cdot \Delta T
\]
where \( L \) is the original length of the rod, \( \alpha \) is the coefficient of linear expansion, and \( \Delta T \) is the change in temperature.
### Step 4: Calculate the Change in Temperature
The change in temperature \( \Delta T \) is:
\[
\Delta T = T_f - T_i = 600 \, \text{K} - 300 \, \text{K} = 300 \, \text{K}
\]
### Step 5: Integrate the Coefficient of Linear Expansion
Since \( \alpha \) varies with temperature, we need to integrate:
\[
\Delta L = \int_{T_i}^{T_f} L \cdot \alpha \, dT
\]
Substituting \( \alpha = \frac{2}{T} \):
\[
\Delta L = L \int_{300}^{600} \frac{2}{T} \, dT
\]
### Step 6: Perform the Integration
The integral of \( \frac{2}{T} \) is:
\[
\int \frac{2}{T} \, dT = 2 \ln T
\]
Now, we evaluate this from 300 K to 600 K:
\[
\Delta L = L \left[ 2 \ln T \right]_{300}^{600} = L \left( 2 \ln 600 - 2 \ln 300 \right)
\]
This simplifies to:
\[
\Delta L = L \cdot 2 \ln \left( \frac{600}{300} \right) = L \cdot 2 \ln(2)
\]
### Step 7: Substitute the Length of the Rod
Given that the length \( L = 1 \, \text{m} \):
\[
\Delta L = 1 \cdot 2 \ln(2) = 2 \ln(2) \, \text{m}
\]
### Step 8: Calculate the Final Length Change
Using the approximate value \( \ln(2) \approx 0.693 \):
\[
\Delta L \approx 2 \cdot 0.693 = 1.386 \, \text{m}
\]
### Final Answer
The increment in the length of the rod when its temperature increases from 27°C to 327°C is approximately:
\[
\Delta L \approx 1.386 \, \text{m}
\]
