A polymer rod 583mm long expands by 170mm when its temperature is raised from 0°C to 148°C. What is the Coefficient of thermal expansion? Answer to 4 decimal places and only put in the numerical value. Do not write the unit in your answer.

Question

BlackTom AI · Accepted Answer

We start by identifying the key quantities for linear thermal expansion. The change in length ΔL is 170 mm, the original length L0 is 583 mm, and the temperature change ΔT is 148°C. The linear coefficient of thermal expansion α is defined by α = ΔL / (L0 · ΔT).

Next, compute the denominator: L0 · ΔT = 583 mm × 148°C. Multiplying 583 by 148 gives 86284 (mm·°C).

Then form the ratio: α = 170 / 86284 ≈ 0.0019703 (approximately).

To report the value to four decimal places, we examine the fifth decimal place. The approximate value 0.0019703 has digits after the decimal as 0, 0, 1, 9, 7, ...; the fourth decimal place is 9 and the fifth is 7, so we round the fourth decimal place up. This yields α ≈ 0.0020 when rounded to four decimal places.

Summary of steps in order: (1) apply α = ΔL / (L0 ΔT), (2) compute L0 ΔT, (3) divide ΔL by that product, (4) round the result to four decimal places. The final numeric value to four decimals is 0.0020.

A polymer rod 583mm long expands by 170mm when its temperature is raised from 0°C to 148°C. What is the Coefficient of thermal expansion? Answer to 4 decimal places and only put in the numerical value. Do not write the unit in your answer.简答题

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