Problem Statement
A circular aluminum plate has a radius of 10 cm at 20°C. Find the increase in area when heated to 100°C. ($\alpha_{Al} = 2.3 \times 10^{-5}$ °C$^{-1}$)
Given Information
- All quantities, constants, and constraints stated in the problem above
- Physical constants used as needed (see Concepts section)
Physical Concepts & Formulas
This problem draws on fundamental physical principles. The key is to identify which conservation law or field equation governs the system, then apply it systematically. Dimensional analysis can always be used to verify that the final answer has the correct units. Working from first principles — rather than memorising formulas — builds deeper understanding and allows tackling novel problems.
- Identify the relevant physical law (Newton’s laws, conservation of energy/momentum, Maxwell’s equations, etc.)
- State the mathematical form of that law as it applies here
- Check dimensions at every step: both sides of an equation must have the same units
Step-by-Step Solution
Problem Statement
A circular aluminum plate has a radius of 10 cm at 20°C. Find the increase in area when heated to 100°C. ($\alpha_{Al} = 2.3 \times 10^{-5}$ °C$^{-1}$)
Given Information
- Radius at 20°C: $r = 10$ cm $= 0.1$ m
- $\Delta T = 100 – 20 = 80°C$
- $\alpha = 2.3 \times 10^{-5}$ °C$^{-1}$
Physical Concepts & Formulas
For area expansion, the coefficient $\beta = 2\alpha$:
$$\Delta A = A_0 \beta \Delta T = A_0 (2\alpha) \Delta T$$
Original area: $A_0 = \pi r^2$
Step-by-Step Solution
Step 1: Compute original area.
$$A_0 = \pi (0.1)^2 = 0.01\pi \approx 3.1416 \times 10^{-2} \text{ m}^2$$
Step 2: Compute $\beta$.
$$\beta = 2\alpha = 2 \times 2.3 \times 10^{-5} = 4.6 \times 10^{-5} \text{ °C}^{-1}$$
Step 3: Compute $\Delta A$.
$$\Delta A = 0.01\pi \times 4.6 \times 10^{-5} \times 80$$
$$= 0.01\pi \times 36.8 \times 10^{-4}$$
$$= 3.1416 \times 10^{-2} \times 3.68 \times 10^{-3}$$
$$\approx 1.156 \times 10^{-4} \text{ m}^2$$
Worked Calculation
$$\Delta A = \pi (0.1)^2 \times 2 \times 2.3 \times 10^{-5} \times 80 \approx 1.16 \times 10^{-4} \text{ m}^2 \approx 1.16 \text{ cm}^2$$
Answer
$\Delta A \approx 1.16 \times 10^{-4}$ m² $\approx 1.16$ cm²
Physical Interpretation
The area expansion coefficient is twice the linear coefficient because expansion occurs in both dimensions simultaneously. A hole in a plate also expands — the material surrounding the hole pushes outward, making the hole larger, not smaller.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\Delta A = \pi (0.1)^2 \times 2 \times 2.3 \times 10^{-5} \times 80 \approx 1.16 \times 10^{-4} \text{ m}^2 \approx 1.16 \text{ cm}^2}$$
Physical Interpretation
The answer should be checked for dimensional consistency and physical reasonableness: is the magnitude in the expected range for this type of problem? Does the answer change in the correct direction when parameters are varied (e.g., increasing mass should increase momentum, increasing distance should decrease field strength)? These sanity checks are as important as the calculation itself.
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