https://r2.erweima.ai/mathbot/image/1725536198273dldh3acv.jpg

The image contains six thermal expansion problems. Let's solve them one by one.

### 1. Copper Wire Expansion
A copper wire measures 5 m at a temperature of 10°C. What is its length at 60°C?

The formula for linear expansion is:
\[
\Delta L = L_0 \cdot \alpha \cdot \Delta T
\]
where:
- \(L_0 = 5 \, \text{m}\) (initial length),
- \(\alpha_{\text{cobre}} = 17 \times 10^{-6} \, \text{°C}^{-1}\) (coefficient of linear expansion for copper),
- \(\Delta T = 60 - 10 = 50 \, \text{°C}\) (temperature change).

Solution:
\[
\Delta L = 5 \cdot 17 \times 10^{-6} \cdot 50 = 0.00425 \, \text{m}
\]
Thus, the final length is:
\[
L_f = L_0 + \Delta L = 5 + 0.00425 = 5.00425 \, \text{m}.
\]

### 2. Aluminum Wire Expansion
An aluminum wire measures 2 m at 20°C. What is its length at 60°C?

Using the same formula, but for aluminum:
- \(L_0 = 2 \, \text{m}\),
- \(\alpha_{\text{aluminio}} = 23 \times 10^{-6} \, \text{°C}^{-1}\),
- \(\Delta T = 60 - 20 = 40 \, \text{°C}\).

Solution:
\[
\Delta L = 2 \cdot 23 \times 10^{-6} \cdot 40 = 0.00184 \, \text{m}
\]
Thus, the final length is:
\[
L_f = L_0 + \Delta L = 2 + 0.00184 = 2.00184 \, \text{m}.
\]

### 3. Aluminum Rod Expansion
An aluminum rod is 40 cm (0.4 m) long. It is found at 80°C in a furnace. How much must the temperature be reduced for the rod to shorten by 80 cm?

Here, the change in length is negative:
- \(\Delta L = -80 \, \text{cm} = -0.08 \, \text{m}\),
- \(\alpha_{\text{aluminio}} = 23 \times 10^{-6} \, \text{°C}^{-1}\),
- \(L_0 = 0.4 \, \text{m}\).

Rearrange the linear expansion formula to find \(\Delta T\):
\[
\Delta T = \frac{\Delta L}{L_0 \cdot \alpha} = \frac{-0.08}{0.4 \cdot 23 \times 10^{-6}} = -869.57 \, \text{°C}.
\]
Thus, the temperature must be reduced by about 870°C.

### 4. Steel Wire Expansion
A steel wire is 1 m long at 20°C. What is its length at 80°C if the coefficient of linear expansion for steel is \(12 \times 10^{-6} \, \text{°C}^{-1}\)?

- \(L_0 = 1 \, \text{m}\),
- \(\alpha_{\text{acero}} = 12 \times 10^{-6} \, \text{°C}^{-1}\),
- \(\Delta T = 80 - 20 = 60 \, \text{°C}\).

Solution:
\[
\Delta L = 1 \cdot 12 \times 10^{-6} \cdot 60 = 0.00072 \, \text{m}.
\]
Thus, the final length is:
\[
L_f = 1 + 0.00072 = 1.00072 \, \text{m}.
\]

### 5. Bar Expansion with Steel and Copper
Two bars, one steel and one copper, each 2 m long, are heated to the same temperature. The steel bar lengthens by 1.2 mm more than the copper bar. Determine the temperature change.

Let:
- \(\alpha_{\text{cobre}} = 17 \times 10^{-6} \, \text{°C}^{-1}\),
- \(\alpha_{\text{acero}} = 12 \times 10^{-6} \, \text{°C}^{-1}\),
- \(L_0 = 2 \, \text{m}\),
- \(\Delta L_{\text{acero}} = \Delta L_{\text{cobre}} + 1.2 \, \text{mm} = 1.2 \times 10^{-3} \, \text{m}\).

Now, using the expansion formula for both materials:
\[
\Delta L_{\text{acero}} = 2 \cdot 12 \times 10^{-6} \cdot \Delta T,
\]
\[
\Delta L_{\text{cobre}} = 2 \cdot 17 \times 10^{-6} \cdot \Delta T.
\]
The difference in expansions is 1.2 mm:
\[
2 \cdot 12 \times 10^{-6} \cdot \Delta T - 2 \cdot 17 \times 10^{-6} \cdot \Delta T = 1.2 \times 10^{-3}.
\]
Simplifying:
\[
2 \cdot (12 \times 10^{-6} - 17 \times 10^{-6}) \cdot \Delta T = 1.2 \times 10^{-3},
\]
\[
-10 \times 10^{-6} \cdot \Delta T = 1.2 \times 10^{-3},
\]
\[
\Delta T = \frac{1.2 \times 10^{-3}}{5 \times 10^{-6}} = 240 \, \text{°C}.
\]

### 6. Steel Cable Expansion
A steel cable must touch the ground at noon due to expansion from morning to noon. What temperature increase is required? The length is 100 m, and the temperature difference between morning and noon is 30°C.

Let:
- \(L_0 = 100 \, \text{m}\),
- \(\alpha_{\text{acero}} = 12 \times 10^{-6} \, \text{°C}^{-1}\),
- \(\Delta T = 30 \, \text{°C}\).

Solution:
\[
\Delta L = 100 \cdot 12 \times 10^{-6} \cdot 30 = 0.036 \, \text{m} = 3.6 \, \text{cm}.
\]
The steel cable expands by 3.6 cm at noon.

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Would you like more details on any of the steps, or have any questions?

### Five Related Questions:
1. How does the coefficient of linear expansion vary between materials?
2. What is the relationship between the temperature change and the change in length for different materials?
3. How is thermal expansion important in construction and engineering?
4. How do we calculate the area or volume expansion for materials?
5. What happens if a material contracts instead of expands?

### Tip:
Always use consistent units when calculating thermal expansion to avoid errors!

The image contains six thermal expansion problems involving the linear expansion of copper, aluminum, and steel wires or rods with varying initial lengths and temperature changes.

Learn how to solve thermal expansion problems for materials such as copper, aluminum, and steel. These problems include calculating changes in length based on initial length, temperature change, and material properties. Suitable for high school students (Grades 10-12).

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