Problem Statement
A steel rod of length 1.0 m is heated from 20°C to 100°C. Find the increase in length. (Coefficient of linear expansion of steel $\alpha = 1.2 \times 10^{-5}$ per °C)
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 steel rod of length 1.0 m is heated from 20°C to 100°C. Find the increase in length. (Coefficient of linear expansion of steel $\alpha = 1.2 \times 10^{-5}$ per °C)
Given Information
- Original length: $L_0 = 1.0$ m
- Initial temperature: $T_1 = 20°C$
- Final temperature: $T_2 = 100°C$
- $\alpha_{steel} = 1.2 \times 10^{-5}$ °C$^{-1}$
Physical Concepts & Formulas
When a solid is heated, atoms vibrate more vigorously and the average inter-atomic spacing increases. For small temperature changes, the fractional increase in length is proportional to $\Delta T$:
$$\Delta L = L_0 \alpha \Delta T$$
Step-by-Step Solution
Step 1: Find the temperature change.
$$\Delta T = T_2 – T_1 = 100 – 20 = 80°C$$
Step 2: Apply the linear expansion formula.
$$\Delta L = L_0 \alpha \Delta T = 1.0 \times 1.2 \times 10^{-5} \times 80$$
Step 3: Calculate.
$$\Delta L = 1.2 \times 10^{-5} \times 80 = 96 \times 10^{-5} = 9.6 \times 10^{-4} \text{ m}$$
Worked Calculation
$$\Delta L = 1.0 \times 1.2 \times 10^{-5} \times 80 = 9.6 \times 10^{-4} \text{ m} = 0.96 \text{ mm}$$
Answer
$\Delta L = 9.6 \times 10^{-4}$ m $= 0.96$ mm
Physical Interpretation
A 1-meter steel rod expands by less than 1 mm over an 80°C temperature rise. This small but significant expansion is why railway tracks have expansion gaps, bridge decks have expansion joints, and engineers must account for thermal stress in large structures.
Worked Calculation
Substituting all given numerical values with their units into the derived formula:
$$\text{Numerical result} = \text{given expression substituted with values}$$
Answer
$$\boxed{\Delta L = 1.0 \times 1.2 \times 10^{-5} \times 80 = 9.6 \times 10^{-4} \text{ m} = 0.96 \text{ mm}}$$
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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