10) An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 x 10-6 K-1. A) 17.435 cm B) 0.348 cm C) 17.365 cm D) 0.0348 cm
10) An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length? Aluminum has a linear expansion coefficient of 25 x 10-6 K-1. A) 17.435 cm B) 0.348 cm C) 17.365 cm D) 0.0348 cm
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Chapter1: Units, Trigonometry. And Vectors
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![**Question 10: Thermal Expansion of Aluminum Rod**
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length?
**Given Data:**
- Initial length of the rod (\(L_0\)) = 17.400 cm
- Initial temperature (\(T_0\)) = 20°C
- Final temperature (\(T_f\)) = 100°C
- Linear expansion coefficient of aluminum (\(\alpha\)) = 25 × 10^-6 K^-1
**Formula:**
The formula for linear expansion is given by:
\[ \Delta L = L_0 \alpha \Delta T \]
Where:
- \(\Delta L\) = Change in length
- \(L_0\) = Initial length
- \(\alpha\) = Linear expansion coefficient
- \(\Delta T\) = Change in temperature
**Calculation:**
1. Calculate the change in temperature (\(\Delta T\)):
\[ \Delta T = T_f - T_0 \]
\[ \Delta T = 100°C - 20°C \]
\[ \Delta T = 80°C \]
2. Calculate the change in length (\(\Delta L\)):
\[ \Delta L = 17.400 \, \text{cm} \times 25 \times 10^{-6} \, \text{K}^{-1} \times 80 \, \text{K} \]
\[ \Delta L = 17.400 \times 0.000025 \times 80 \]
\[ \Delta L = 0.0348 \, \text{cm} \]
3. Calculate the new length (\(L_f\)):
\[ L_f = L_0 + \Delta L \]
\[ L_f = 17.400 \, \text{cm} + 0.0348 \, \text{cm} \]
\[ L_f = 17.4348 \, \text{cm} \]
**Answer Choices:**
A) 17.435 cm
B) 0.348 cm
C) 17.365 cm
D) 0.0348 cm
The correct answer is: **A) 17.435 cm**](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3997526e-7cda-40d8-9b92-5ffe558ca7d8%2F3c50b980-33fd-48a0-8c8c-88a070e9a73a%2F0c768mo_processed.png&w=3840&q=75)
Transcribed Image Text:**Question 10: Thermal Expansion of Aluminum Rod**
An aluminum rod 17.400 cm long at 20°C is heated to 100°C. What is its new length?
**Given Data:**
- Initial length of the rod (\(L_0\)) = 17.400 cm
- Initial temperature (\(T_0\)) = 20°C
- Final temperature (\(T_f\)) = 100°C
- Linear expansion coefficient of aluminum (\(\alpha\)) = 25 × 10^-6 K^-1
**Formula:**
The formula for linear expansion is given by:
\[ \Delta L = L_0 \alpha \Delta T \]
Where:
- \(\Delta L\) = Change in length
- \(L_0\) = Initial length
- \(\alpha\) = Linear expansion coefficient
- \(\Delta T\) = Change in temperature
**Calculation:**
1. Calculate the change in temperature (\(\Delta T\)):
\[ \Delta T = T_f - T_0 \]
\[ \Delta T = 100°C - 20°C \]
\[ \Delta T = 80°C \]
2. Calculate the change in length (\(\Delta L\)):
\[ \Delta L = 17.400 \, \text{cm} \times 25 \times 10^{-6} \, \text{K}^{-1} \times 80 \, \text{K} \]
\[ \Delta L = 17.400 \times 0.000025 \times 80 \]
\[ \Delta L = 0.0348 \, \text{cm} \]
3. Calculate the new length (\(L_f\)):
\[ L_f = L_0 + \Delta L \]
\[ L_f = 17.400 \, \text{cm} + 0.0348 \, \text{cm} \]
\[ L_f = 17.4348 \, \text{cm} \]
**Answer Choices:**
A) 17.435 cm
B) 0.348 cm
C) 17.365 cm
D) 0.0348 cm
The correct answer is: **A) 17.435 cm**
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