To solve these problems, we can use the concept of thermal expansion, which states that materials will expand or contract as temperature changes. For a linear expansion, the formula is given by:
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \]
where:
- \(\Delta L\) is the change in length,
- \(L_0\) is the original length,
- \(\alpha\) is the linear coefficient of expansion,
- \(\Delta T\) is the change in temperature.
1. Temperature of the metal plate if it is heated until the sides are 1.005m long.
Given:
- Original side length \(L_0 = 1.0 , \text{m}\)
- New side length \(L = 1.005 , \text{m}\)
- Linear coefficient of expansion \(\alpha = 12.5 \times 10^{-6} , \text{K}^{-1}\)
First, calculate \(\Delta L\):
\[ \Delta L = L - L_0 = 1.005 , \text{m} - 1.0 , \text{m} = 0.005 , \text{m} \]
Now, use the linear expansion formula to find \(\Delta T\):
\[ \Delta L = L_0 \cdot \alpha \cdot \Delta T \Rightarrow 0.005 = 1.0 \cdot (12.5 \times 10^{-6}) \cdot \Delta T \]
Rearranging this gives us:
\[ \Delta T = \frac{0.005}{12.5 \times 10^{-6}} = 400 , \text{K} \]
Now, to find the final temperature \(T\):
\[ T = T_0 + \Delta T = 283 , \text{K} + 400 , \text{K} = 683 , \text{K} \]
2. Diameter of the hole at this temperature.
The hole expands as well, so we use the same linear expansion formula for the diameter. The original diameter \(D_0\) of the hole is 600 mm or 0.6 m. The change in diameter \(\Delta D\) can be calculated as follows:
\[ \Delta D = D_0 \cdot \alpha \cdot \Delta T \]
Substituting the known values:
\[ \Delta D = 0.6 \cdot (12.5 \times 10^{-6}) \cdot 400 = 0.6 \cdot 0.005 = 0.003 , \text{m} \]
Now, calculate the new diameter \(D\):
\[ D = D_0 + \Delta D = 0.6 , \text{m} + 0.003 , \text{m} = 0.603 , \text{m} \] or 603 mm.
3. Increase in area of the hole at this temperature.
To find the increase in area, we first need to find the area of the hole before and after expansion.
Original Area \(A_0\):
\[ A_0 = \pi \left( \frac{D_0}{2} \right)^2 = \pi \left( \frac{0.6}{2} \right)^2 = \pi \left( 0.3 \right)^2 = 0.09\pi , \text{m}^2 \]
New Area \(A\) with the expanded diameter \(D = 0.603 , \text{m}\):
\[ A = \pi \left( \frac{D}{2} \right)^2 = \pi \left( \frac{0.603}{2} \right)^2 = \pi \left( 0.3015 \right)^2 \approx 0.0909\pi , \text{m}^2 \]
Increase in Area \(\Delta A\):
\[ \Delta A = A - A_0 = (0.0909\pi - 0.09\pi) = 0.0009\pi , \text{m}^2 \approx 0.00283 , \text{m}^2 , (\text{approximately}) \]
Summary of Results
- Final temperature of the metal plate: 683 K
- Diameter of the hole at this temperature: 603 mm
- Increase in area of the hole at this temperature: approximately 0.00283 m².