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

A copper rod has an initial length of 100 cm100\text{ cm} at 25∘C25^\circ\text{C}. When its temperature is increased to 75∘C75^\circ\text{C}, its length becomes 100.085 cm100.085\text{ cm}. What is the area expansivity of copper in 10−5 K−110^{-5}\text{ K}^{-1}?

Answer: 3.4 10^-5 K^-1

Answer

The area expansivity of copper is 3.4×10−5 K−13.4 \times 10^{-5}\text{ K}^{-1}, giving a numerical value of 3.43.4 in units of 10−5 K−110^{-5}\text{ K}^{-1}.
Linear expansivity is obtained as α=ΔLL0ΔT=0.085 cm100 cm×50 K=1.7×10−5 K−1\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085\text{ cm}}{100\text{ cm} \times 50\text{ K}} = 1.7 \times 10^{-5}\text{ K}^{-1}. Since area expansivity β\beta is related to linear expansivity by β=2α\beta = 2\alpha, multiplying by 22 yields 3.4×10−5 K−13.4 \times 10^{-5}\text{ K}^{-1}, or 3.43.4 in units of 10−5 K−110^{-5}\text{ K}^{-1}.

Step-by-Step Solution

1
Calculate the temperature change
ΔT=75∘C−25∘C=50 K\Delta T = 75^\circ\text{C} - 25^\circ\text{C} = 50\text{ K}
Thermal expansion depends directly on the change in temperature.
2
Find the change in length
ΔL=100.085 cm−100 cm=0.085 cm\Delta L = 100.085\text{ cm} - 100\text{ cm} = 0.085\text{ cm}
The expansion is the difference between the final length and original length.
3
Determine linear expansivity (α\alpha)
α=ΔLL0ΔT=0.085100×50=1.7×10−5 K−1\alpha = \frac{\Delta L}{L_0 \Delta T} = \frac{0.085}{100 \times 50} = 1.7 \times 10^{-5}\text{ K}^{-1}
Linear expansivity defines fractional change in length per degree temperature change.
4
Compute area expansivity (β\beta)
β=2α=2×(1.7×10−5)=3.4×10−5 K−1\beta = 2\alpha = 2 \times (1.7 \times 10^{-5}) = 3.4 \times 10^{-5}\text{ K}^{-1}
Area (superficial) expansivity is equal to twice the linear expansivity.

Key Concept

Relationship between linear expansivity (α\alpha) and area expansivity (β=2α\beta = 2\alpha).
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Topics

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

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