Question

A metal plate has a surface area of 3 m² at 30° c the increase in area when it is heated to 80° c is what ?linear expansivity of the metal is 4.0 * 10 ^ - 5 k^-1

Answer 1

To find the increase in surface area of a metal plate when heated, we can use the formula for the change in area due to temperature change:

\[
\Delta A = A_0 \cdot 2\alpha \cdot \Delta T
\]

Where:
- \( \Delta A \) is the change in area,
- \( A_0 \) is the initial area,
- \( \alpha \) is the linear expansivity,
- \( \Delta T \) is the change in temperature.

Given:
- \( A_0 = 3 \, \text{m}^2 \)
- \( \alpha = 4.0 \times 10^{-5} \, \text{K}^{-1} \)
- Initial temperature \( T_1 = 30^\circ C \) and final temperature \( T_2 = 80^\circ C \), so \( \Delta T = 80^\circ C - 30^\circ C = 50 \, \text{K} \)

Now we can calculate the change in area:

\[
\Delta A = 3 \, \text{m}^2 \cdot 2 \cdot (4.0 \times 10^{-5} \, \text{K}^{-1}) \cdot 50 \, \text{K}
\]

Calculating this step by step:

1. Calculate \( 2\alpha \):
\[
2\alpha = 2 \cdot 4.0 \times 10^{-5} = 8.0 \times 10^{-5} \, \text{K}^{-1}
\]

2. Multiply \( 2\alpha \) by \( \Delta T \):
\[
8.0 \times 10^{-5} \, \text{K}^{-1} \cdot 50 \, \text{K} = 4.0 \times 10^{-3}
\]

3. Now multiply by the initial area \( A_0 \):
\[
\Delta A = 3 \, \text{m}^2 \cdot 4.0 \times 10^{-3} = 0.012 \, \text{m}^2
\]

Therefore, the increase in area when the metal plate is heated from 30°C to 80°C is:

\[
\Delta A = 0.012 \, \text{m}^2
\]

Thus, the area increases by **0.012 m²**.