An aluminum rod is 23.5 cm long at 20°C and has a mass of 350 g. If 10,500 J of energy is added to the rod by heat, what is the change in length of the rod? (The average coefficient of linear expansion for aluminum is 24 x 10-6 (°C)¯¹.) 0.20586 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. mm
An aluminum rod is 23.5 cm long at 20°C and has a mass of 350 g. If 10,500 J of energy is added to the rod by heat, what is the change in length of the rod? (The average coefficient of linear expansion for aluminum is 24 x 10-6 (°C)¯¹.) 0.20586 X Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error. mm
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Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
![**Thermal Expansion of an Aluminum Rod**
**Scenario:**
An aluminum rod is 23.5 cm long at 20°C and has a mass of 350 g. If 10,500 J of energy is added to the rod by heat, what is the change in length of the rod? (The average coefficient of linear expansion for aluminum is \( 24 \times 10^{-6} \, (°C)^{-1}.)
**User Input:**
The user provided the following calculation result:
\[ 0.20586 \, \text{mm} \]
**Feedback:**
The response provided was within 10% of the correct value. This discrepancy might be due to a round-off error or a mistake in the user’s calculation. Users are advised to carry out all intermediate results with at least four-digit accuracy to minimize round-off error.
**Detailed Explanation of the Process:**
1. **Calculate the temperature change (\(\Delta T\)):**
\[
\Delta T = \frac{\text{energy added}}{\text{mass} \times \text{specific heat capacity of aluminum}}
\]
The specific heat capacity of aluminum is approximately \( 900 \, \text{J/kg}^\circ\text{C} \).
2. **Determine the increase in temperature (\(\Delta T\)):**
\[
\Delta T = \frac{10,500 \, \text{J}}{0.35 \, \text{kg} \times 900 \, \text{J/kg}^\circ\text{C}}
\]
\[
\Delta T \approx 33.33^\circ\text{C}
\]
3. **Calculate the change in length (\(\Delta L\)):**
\[
\Delta L = L_0 \times \alpha \times \Delta T
\]
Where:
- \(L_0\) is the initial length (23.5 cm or 0.235 m)
- \(\alpha\) is the coefficient of linear expansion \((24 \times 10^{-6} \, (°C)^{-1})\)
- \(\Delta T\) is the temperature change (33.33°C)
4. **Finding \(\Delta L\):**
\[
\Delta L = 0.235 \, \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa5d0ed69-4925-4403-a796-f6681910a505%2F17dd874a-c1c5-4cd1-a10f-8f083d827e48%2F4jumu4h_processed.png&w=3840&q=75)
Transcribed Image Text:**Thermal Expansion of an Aluminum Rod**
**Scenario:**
An aluminum rod is 23.5 cm long at 20°C and has a mass of 350 g. If 10,500 J of energy is added to the rod by heat, what is the change in length of the rod? (The average coefficient of linear expansion for aluminum is \( 24 \times 10^{-6} \, (°C)^{-1}.)
**User Input:**
The user provided the following calculation result:
\[ 0.20586 \, \text{mm} \]
**Feedback:**
The response provided was within 10% of the correct value. This discrepancy might be due to a round-off error or a mistake in the user’s calculation. Users are advised to carry out all intermediate results with at least four-digit accuracy to minimize round-off error.
**Detailed Explanation of the Process:**
1. **Calculate the temperature change (\(\Delta T\)):**
\[
\Delta T = \frac{\text{energy added}}{\text{mass} \times \text{specific heat capacity of aluminum}}
\]
The specific heat capacity of aluminum is approximately \( 900 \, \text{J/kg}^\circ\text{C} \).
2. **Determine the increase in temperature (\(\Delta T\)):**
\[
\Delta T = \frac{10,500 \, \text{J}}{0.35 \, \text{kg} \times 900 \, \text{J/kg}^\circ\text{C}}
\]
\[
\Delta T \approx 33.33^\circ\text{C}
\]
3. **Calculate the change in length (\(\Delta L\)):**
\[
\Delta L = L_0 \times \alpha \times \Delta T
\]
Where:
- \(L_0\) is the initial length (23.5 cm or 0.235 m)
- \(\alpha\) is the coefficient of linear expansion \((24 \times 10^{-6} \, (°C)^{-1})\)
- \(\Delta T\) is the temperature change (33.33°C)
4. **Finding \(\Delta L\):**
\[
\Delta L = 0.235 \, \
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