Question

A steel is calibrated at ${20^ \circ }C$.On a cold day when the temperature is - ${15^ \circ }C$, percentage error in the tape will be [${\alpha _{steel}} = 1.2 \times {10^{ - 5}}^ \circ {C^{ - 1}}$] ?

Answer 1

Hint:Let us have some idea about thermal expansion. Thermal expansion, which does not include phase transitions, is the tendency of matter to change its form, area, volume, and density in response to a change in temperature.

Complete step by step answer:
The average molecular kinetic energy of a material is a monotonic feature of temperature. When a material is heated, the molecules tend to vibrate and move more, resulting in a greater distance between them. It's rare to find substances that contract as the temperature rises, and they only happen in a few temperature ranges.

The material's coefficient of linear thermal expansion is defined as the relative expansion (also known as strain) divided by the change in temperature, and it varies with temperature. Particles move faster as their energy increases, weakening the intermolecular forces between them and thereby expanding the material. Let us now come to the problem:
$\Delta l = l\alpha \Delta t$
Where, $\Delta l = $change in length, $\Delta t = $ Change in temperature and $\alpha = $Linear coefficient of expansion.
Here, $\Delta t = - 15 - 20 = - {35^ \circ }C$
So, $\Delta l = 1 \times 1.2 \times {10^{ - 5}} \times - 35$
And percentage error $ = \dfrac{{\Delta l}}{l} \times 100 = - 0.042$

Hence, the percentage error will be $- 0.042$.

Note: The term "linear expansion" refers to a change in one dimension (length) rather than a change in volume (volumetric expansion). The change in length measurements of an object due to thermal expansion is related to temperature change by a coefficient of linear thermal expansion, to a first approximation (CLTE).