Q14PE - College Physics (Urone)

Question

Global warming will produce rising sea levels partly due to melting ice caps but also due to the expansion of water as average ocean temperatures rise. To get some idea of the size of this effect, calculate the change in length of a column of water  \(1.00{\rm{ km}}\) high for a temperature increase of  \(1.00^\circ {\rm{C}}\) . Note that this calculation is only approximate because ocean warming is not uniform with depth.

Step-by-Step Solution

Answer

The change in length of water level is $70.0\times {10}^{-6} \mathrm{km}$

Step 1: Describe how length and temperature are related.

Global warming produces a rise in the water level in the sea. So, a linear expansion occurred in the water level. The linear expansion caused by the variation in temperature is expressed as follows;

$\mathbf{∆}\mathbf{L}\mathbf{=}\mathbf{\alpha L}\mathbf{∆}\mathbf{T}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$

Here, $\mathbf{∆}\mathbf{L}$ is the change in length,  $\mathbf{\alpha }$ is the coefficient of linear expansion,  L is the initial length, and  $\mathbf{∆}\mathbf{T}$ is the temperature change.

Step 2: Calculate the change in length.

The initial length and change in temperature are given.

$\begin{array}{rcl}∆\mathrm{T}& =& 1.00°\mathrm{C}\\ \mathrm{L}& =& 1.00 \mathrm{KM}\end{array}$

The coefficient of linear expansion of water is unknown, but you can calculate it from the coefficient of volume expansion of water. It is denoted as $\beta$.

$\begin{array}{rcl}\mathbf{\beta }& \mathbf{=}& \mathbf{210}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathbf{1}\mathbf{/}\mathbf{°}\mathbf{C}\\ \mathbf{\alpha }& \mathbf{=}& \frac{\mathbf{\beta }}{\mathbf{3}}\\ \mathbf{\alpha }& \mathbf{=}& \frac{\mathbf{210}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathbf{1}\mathbf{/}\mathbf{°}\mathbf{C}}{\mathbf{3}}\\ \mathbf{\alpha }& \mathbf{=}& \mathbf{70}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathbf{1}\mathbf{/}\mathbf{°}\mathbf{C}\end{array}$

Substitute the values in the equation (1), and we get,

$\begin{array}{rcl}\mathbf{∆}\mathbf{L}& \mathbf{=}& \left(70\times {10}^{-6} 1/°C\right)\mathbf{\times }\left(1.00 \mathrm{km} \times 1.00°C\right)\\ & \mathbf{=}& \mathbf{70}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathbf{km}\end{array}$

Therefore, the change in length of water level is $\mathbf{70}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{ }\mathbf{km}$ .