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

An aluminium rod of initial length 2.0 m2.0\text{ m} at 20∘C20^\circ\text{C} expands by 0.96 mm0.96\text{ mm} when heated. If the linear expansivity of aluminium is 2.4×10−5 K−12.4 \times 10^{-5}\text{ K}^{-1}, what is the rise in temperature of the rod?

Answer: 20 K

Answer

The rise in temperature of the aluminium rod is 20 K20\text{ K}.
The fractional change in length depends on linear expansivity and temperature change through ΔL=L0αΔT\Delta L = L_0 \alpha \Delta T. Substituting the converted expansion ΔL=9.6×10−4 m\Delta L = 9.6 \times 10^{-4}\text{ m}, initial length L0=2.0 mL_0 = 2.0\text{ m}, and linear expansivity α=2.4×10−5 K−1\alpha = 2.4 \times 10^{-5}\text{ K}^{-1} gives ΔT=9.6×10−42.0×2.4×10−5=20 K\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times 2.4 \times 10^{-5}} = 20\text{ K}.

Step-by-Step Solution

1
Convert change in length from millimeters to meters
\Delta L = 9.6 \times 10^{-4}\text{ m}
Units must be consistent with initial length in meters.
2
Rearrange the linear thermal expansion formula \Delta L = L_0 \alpha \Delta T for temperature change \Delta T
\Delta T = \frac{\Delta L}{L_0 \alpha}
To isolate the unknown quantity \Delta T.
3
Substitute values into the rearranged formula and compute \Delta T
\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times (2.4 \times 10^{-5})} = 20\text{ K}
Evaluating the mathematical expression yields the required temperature rise.

Key Concept

Linear Expansivity and Thermal Expansion of Solids
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Topics

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

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