Problem Statement
A metal cube of side $L = 10\,\text{cm}$ has a coefficient of linear expansion $\alpha = 2\times10^{-5}\,\text{K}^{-1}$. Find the increase in (a) length of a side, (b) surface area, and (c) volume when heated from $30^{\circ}\text{C}$ to $80^{\circ}\text{C}$.
Given Information
- $L_0 = 10\,\text{cm} = 0.10\,\text{m}$
- $\alpha = 2\times10^{-5}\,\text{K}^{-1}$
- $\Delta T = 50\,\text{K}$
Physical Concepts & Formulas
For isotropic expansion: linear expansion DeltaL = alpha*L*DeltaT; area expansion DeltaA = 2*alpha*A*DeltaT; volume expansion DeltaV = 3*alpha*V*DeltaT.
- $\Delta L = \alpha L_0\Delta T$
- $\Delta A = 2\alpha A_0\Delta T$
- $\Delta V = 3\alpha V_0\Delta T$
Step-by-Step Solution
Step 1 — Linear: $$\Delta L = (2\times10^{-5})(0.10)(50) = 1\times10^{-4}\,\text{m} = 0.1\,\text{mm}$$
Step 2 — Area (A_0 = 6L^2 for cube): $$\Delta A = 2\alpha A_0\Delta T = 2(2\times10^{-5})(6\times0.01)(50) = 2(2\times10^{-5})(3) = 1.2\times10^{-4}\,\text{m}^2$$
Step 3 — Volume: $$\Delta V = 3\alpha V_0\Delta T = 3(2\times10^{-5})(10^{-3})(50) = 3\times10^{-6}\,\text{m}^3 = 3\,\text{cm}^3$$
Worked Calculation
$$\Delta L=0.1\,\text{mm};\;\Delta A=1.2\times10^{-4}\,\text{m}^2;\;\Delta V=3\,\text{cm}^3$$
Answer
$$\boxed{\Delta L=0.1\,\text{mm},\quad \Delta A=1.2\times10^{-4}\,\text{m}^2,\quad \Delta V=3\,\text{cm}^3}$$
Side increases by 0.1 mm; total surface area by 1.2e-4 m^2 (1.2 cm^2); volume by 3 cm^3.
Physical Interpretation
The coefficients 1:2:3 for linear:area:volume expansion are exact for small expansions (first-order in alpha). This 3-to-1 volume-to-linear ratio is a direct consequence of 3-dimensional geometry.
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