A brass ruler is 300 mm long at 21 °C. It is heated to 100 °C and expands by 0.45 mm. Determine the coefficient of expansion for brass (in K-1). Give your answer to two significant figures in scientific notation. (i.e. 2.8E-6)

Question

BlackTom AI · Accepted Answer

Linear thermal expansion: ΔL = α L0 ΔT, so α = ΔL / (L0 ΔT).

Relevant concepts:
- α is the linear coefficient of thermal expansion with units K^-1.
- ΔL is the change in length, L0 is the original length, and ΔT is the temperature change in kelvin (same as °C interval).

Step-by-step reasoning and calculations:
1. Identify given quantities:
   - L0 = 300 mm
   - ΔL = 0.45 mm
   - Initial temperature = 21 °C, final = 100 °C, so ΔT = 100 − 21 = 79 K

2. Substitute into the formula:
   α = ΔL / (L0 ΔT) = 0.45 mm / (300 mm × 79 K)

3. Compute the denominator:
   300 × 79 = 23,700 (mm·K)

4. Divide:
   α = 0.45 / 23,700 = 0.00001898734177... K^-1

5. Express in scientific notation and round to two significant figures:
   0.000018987... ≈ 1.9 × 10^-5 K^-1, written as 1.9E-5

(Alternate unit check: converting to metres gives the same result: ΔL = 0.00045 m, L0 = 0.300 m → α = 0.00045/(0.300×79) = 1.8987…×10^-5.)

Why this answer is correct:
- The calculation directly applies the linear expansion formula and uses the correct temperature change (79 K). Units cancel properly (mm/mm·K), leaving K^-1.

Why other plausible answers would be incorrect:
- An answer like 1.9E-6 would be too small by a factor of 10 (likely from a unit slip or dividing by 10 extra).
- Answers such as 2.8E-6 or similar arise from arithmetic or ΔT mistakes (e.g., using ΔT = 100 or wrong length conversion).

Conclusion:
The coefficient of linear expansion for the brass ruler is 1.9E-5 K^-1, which matches the provided answer.

A brass ruler is 300 mm long at 21 °C. It is heated to 100 °C and expands by 0.45 mm. Determine the coefficient of expansion for brass (in K-1). Give your answer to two significant figures in scientific notation. (i.e. 2.8E-6)Numerical

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