An aluminum wire has a resistance of 5.50 Q at 10.0°C. Determine its resistance (in Q) at 410°C. The temperature coefficient of resistivity for aluminum wire is 3.90 x 10¬3 (°C)-'. (Assume that the temperature coefficient of resistivity was measured using the reference temperature 20°C.) 8.58
An aluminum wire has a resistance of 5.50 Q at 10.0°C. Determine its resistance (in Q) at 410°C. The temperature coefficient of resistivity for aluminum wire is 3.90 x 10¬3 (°C)-'. (Assume that the temperature coefficient of resistivity was measured using the reference temperature 20°C.) 8.58
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![**Problem:**
An aluminum wire has a resistance of 5.50 Ω at 10.0°C. Determine its resistance (in Ω) at 410°C. The temperature coefficient of resistivity for aluminum wire is 3.90 × 10⁻³ (°C)⁻¹. (Assume that the temperature coefficient of resistivity was measured using the reference temperature 20°C.)
**Solution:**
Resistance calculation box:
\[ \text{8.58} \, \Omega \]
**Explanation:**
This problem involves calculating the change in resistance of an aluminum wire as temperature changes. The temperature coefficient of resistivity (\(\alpha\)) allows us to determine how much the resistance of a material will increase per degree change in temperature.
**Formula for calculation:**
\[ R_t = R_0 \cdot (1 + \alpha \cdot (T - T_0)) \]
Where:
- \(R_t\) is the resistance at temperature \(T\).
- \(R_0\) is the original resistance (5.50 Ω at 10.0°C).
- \(\alpha\) is the temperature coefficient of resistivity for aluminum (3.90 × 10⁻³ (°C)⁻¹).
- \(T\) is the new temperature (410°C).
- \(T_0\) is the initial temperature (10.0°C).
**Calculation:**
\[ R_t = 5.50 \, \Omega \times (1 + 3.90 \times 10^{-3} \times (410 - 10)) \]
\[ R_t = 5.50 \, \Omega \times (1 + 3.90 \times 10^{-3} \times 400) \]
\[ R_t = 5.50 \, \Omega \times (1 + 1.56) \]
\[ R_t = 5.50 \, \Omega \times 2.56 \]
\[ R_t \approx 14.08 \, \Omega \]
Therefore, the resistance at 410°C is approximately 14.08 Ω. The incorrect value shown in the problem is 8.58 Ω.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F411fdfa3-98d6-4c96-a283-c3e756a59776%2Fb95866f8-8c89-4925-9c97-392baa7ae83e%2Ftkxaw2t_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem:**
An aluminum wire has a resistance of 5.50 Ω at 10.0°C. Determine its resistance (in Ω) at 410°C. The temperature coefficient of resistivity for aluminum wire is 3.90 × 10⁻³ (°C)⁻¹. (Assume that the temperature coefficient of resistivity was measured using the reference temperature 20°C.)
**Solution:**
Resistance calculation box:
\[ \text{8.58} \, \Omega \]
**Explanation:**
This problem involves calculating the change in resistance of an aluminum wire as temperature changes. The temperature coefficient of resistivity (\(\alpha\)) allows us to determine how much the resistance of a material will increase per degree change in temperature.
**Formula for calculation:**
\[ R_t = R_0 \cdot (1 + \alpha \cdot (T - T_0)) \]
Where:
- \(R_t\) is the resistance at temperature \(T\).
- \(R_0\) is the original resistance (5.50 Ω at 10.0°C).
- \(\alpha\) is the temperature coefficient of resistivity for aluminum (3.90 × 10⁻³ (°C)⁻¹).
- \(T\) is the new temperature (410°C).
- \(T_0\) is the initial temperature (10.0°C).
**Calculation:**
\[ R_t = 5.50 \, \Omega \times (1 + 3.90 \times 10^{-3} \times (410 - 10)) \]
\[ R_t = 5.50 \, \Omega \times (1 + 3.90 \times 10^{-3} \times 400) \]
\[ R_t = 5.50 \, \Omega \times (1 + 1.56) \]
\[ R_t = 5.50 \, \Omega \times 2.56 \]
\[ R_t \approx 14.08 \, \Omega \]
Therefore, the resistance at 410°C is approximately 14.08 Ω. The incorrect value shown in the problem is 8.58 Ω.
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