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Difficulty: MediumThermal Expansion of Solids (Linear, Area, and Volume Expansivity)

An aluminum electric cable suspended between two transmission poles has a length of 100 m100\text{ m} at an initial morning temperature of 20∘C20^\circ\text{C}. By afternoon, the cable temperature increases to 50∘C50^\circ\text{C}. Given that the linear expansivity of aluminum is 2.3×10−5 K−12.3 \times 10^{-5}\text{ K}^{-1}, what is the increase in the length of the cable in centimeters?

Answer: 6.9 cm

Answer

The increase in the length of the cable is 6.9 cm6.9\text{ cm}.
Applying the formula for linear expansion ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T, where L0=100 mL_0 = 100\text{ m}, α=2.3×10−5 K−1\alpha = 2.3 \times 10^{-5}\text{ K}^{-1}, and ΔT=30 K\Delta T = 30\text{ K}, gives ΔL=0.069 m\Delta L = 0.069\text{ m}. Converting this to centimeters yields 6.9 cm6.9\text{ cm}.

Step-by-Step Solution

1
Determine the temperature change
\Delta T = 30\text{ K}
Temperature change is the difference between final and initial temperatures: 50∘C−20∘C=30 K50^\circ\text{C} - 20^\circ\text{C} = 30\text{ K}.
2
Calculate expansion in meters using the linear expansivity formula
\Delta L = 0.069\text{ m}
\Delta L = L_0 \alpha \Delta T = 100 \times (2.3 \times 10^{-5}) \times 30 = 0.069\text{ m}.
3
Convert the change in length to centimeters
\Delta L = 6.9\text{ cm}
Since 1 m=100 cm1\text{ m} = 100\text{ cm}, multiply 0.069 m0.069\text{ m} by 100100 to obtain 6.9 cm6.9\text{ cm}.

Key Concept

Linear expansivity of solid conductors
Estimated Time:1m 30s
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Topics

Thermal Expansion of Solids (Linear, Area, and Volume Expansivity)

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