To solve these problems involving thermal expansion of the square metal plate and the hole, we'll use the concept of linear thermal expansion.
7.2.1: Temperature of the plate when sides are 2.53 m long
The formula for linear expansion is given by:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
Where:
- \( \Delta L \) is the change in length (or side of the square in this case),
- \( L_0 \) is the original length,
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta T \) is the change in temperature.
First, we need to find the change in length:
\[ \Delta L = L - L_0 = 2.53, \text{m} - 2.5, \text{m} = 0.03, \text{m} \]
We know:
- Original length \( L_0 = 2.5, \text{m} \),
- Coefficient of linear expansion \( \alpha = 13 \times 10^{-6}, \text{°C}^{-1} \).
Now, plugging in the values into the expansion formula to find \( \Delta T \):
\[ 0.03, \text{m} = 2.5, \text{m} \cdot (13 \times 10^{-6}, \text{°C}^{-1}) \cdot \Delta T \]
Rearranging for \( \Delta T \):
\[ \Delta T = \frac{0.03, \text{m}}{2.5, \text{m} \cdot (13 \times 10^{-6}, \text{°C}^{-1})} \]
Calculating \( \Delta T \):
\[ \Delta T = \frac{0.03}{2.5 \cdot 13 \times 10^{-6}} \approx \frac{0.03}{3.25 \times 10^{-5}} \approx 923.08, \text{°C} \]
Now, we find the new temperature:
\[ T = T_0 + \Delta T = 32, \text{°C} + 923.08, \text{°C} \approx 955.08, \text{°C} \]
7.2.2: Diameter of the hole at the temperature in Question 7.2.1
With the hole being circular, the expansion still applies in a similar manner. The diameter will also expand linearly with temperature.
The original diameter of the hole is given as 850 mm or 0.85 m.
Using the same formula for linear expansion:
\[ \Delta d = d_0 \cdot \alpha \cdot \Delta T \]
Where:
- \( d_0 = 0.85, \text{m} \)
Now substituting:
\[ \Delta d = 0.85, \text{m} \cdot (13 \times 10^{-6}, \text{°C}^{-1}) \cdot 923.08, \text{°C} \]
Calculating \( \Delta d \):
\[ \Delta d = 0.85, \text{m} \cdot 13 \times 10^{-6} \cdot 923.08 \approx 0.85 \cdot 0.000012 = 0.000010569, \text{m} \approx 0.00001057, \text{m} \]
Now we add this change to the original diameter:
\[ d = d_0 + \Delta d = 0.85, \text{m} + 0.00001057, \text{m} \approx 0.85001057, \text{m} \]
Converting back to mm for clarity:
\[ d \approx 850.01057, \text{mm} \]
Final Answers:
- 7.2.1: The temperature of the plate is approximately 955.08 °C.
- 7.2.2: The diameter of the hole at this temperature is approximately 850.01 mm (or rounded, 850.01 mm).