# A steel section of the Alaskan pipeline had a length of 68.2 m and a temperature of 21.0^{\circ}C...

## Question:

A steel section of the Alaskan pipeline had a length of 68.2 m and a temperature of {eq}21.0^{\circ}C {/eq} when it was installed. What is its change in length when the temperature drops to a frigid {eq}-45.7^{\circ}C {/eq}?

## Thermal Contraction:

A phenomenon in which an object undergoes fractional changes in size, shape, area, or volume, due to a change in the temperature is called a thermal expansion (or contraction). These changes, which are particularly evident for solids and metals, are quantitatively determined using the material's initial length, change in temperature, and a coefficient that dictates the fractional change in size of the material. Temperature changes in a unidimensional material is referred as a linear thermal expansion, with the change in length, {eq}\Delta L {/eq}, obtained through the formula:

$$\Delta L = \alpha L_0 \Delta T$$

where:

• {eq}L_0 = \rm initial\ length\ of\ material {/eq}
• {eq}\alpha = \rm coefficient\ of\ linear\ expansion\ of\ specific\ material {/eq}
• {eq}\Delta T = {\rm change\ in\ temperature} = T_f - T_i {/eq}

The change in length of the steel section of the Alaskan pipeline is due to the phenomenon of linear expansion or compression. An equation used to represent the change in length is expressed as

{eq}\Delta L = \alpha L_0 \Delta T {/eq}

Given:

{eq}L_0 = \rm 68.2\ m {/eq}

{eq}\alpha_{steel} = \rm 1.1 \times 10^{-5}\ ^{\circ}C^{-1} {/eq}

{eq}\Delta T = T_f - T_i = \rm -45.7^{\circ}C - 21.0^{\circ}C = -66.7^{\circ}C {/eq}

Thus, the change in the length of the steel section is calculated to be

{eq}\begin{align} \Delta L &= \alpha_{steel} L_0 \Delta T\\ &= \rm (1.1 \times 10^{-5}\ ^{\circ}C^{-1})(68.2\ m)(-66.7^{\circ}C)\\ \Delta L &= \rm 0.0500\ m \end{align} {/eq}

The negative sign suggests a compression of steel due to a drastic decrease in temperature. Therefore, the change in length of the Alaskan pipeline's steel section due to a drop in temperature is {eq}\boxed{\rm \ 0.0500\ m\ } {/eq}. 