11.2 Density  (Page 2/4)

Calculating the mass of a reservoir from its volume

A reservoir has a surface area of $\text{50}\text{.}0{\text{km}}^2$ and an average depth of 40.0 m. What mass of water is held behind the dam? (See [link] for a view of a large reservoir—the Three Gorges Dam site on the Yangtze River in central China.)

Strategy

We can calculate the volume $V$ of the reservoir from its dimensions, and find the density of water $\rho$ in [link] . Then the mass $m$ can be found from the definition of density

Solution

Solving equation $\rho =m/V$ for $m$ gives $m=\rho V$ .

The volume $V$ of the reservoir is its surface area $A$ times its average depth $h$ :

The density of water $\rho$ from [link] is $1\text{.}\text{000}\times {\text{10}}^3{\text{kg/m}}^3$ . Substituting $V$ and $\rho$ into the expression for mass gives

Discussion

A large reservoir contains a very large mass of water. In this example, the weight of the water in the reservoir is $\text{mg}=1\text{.}\text{96}\times {\text{10}}^{\text{13}}\text{N}$ , where $g$ is the acceleration due to the Earth's gravity (about $9\text{.}\text{80}{\text{m/s}}^2$ ). It is reasonable to ask whether the dam must supply a force equal to this tremendous weight. The answer is no. As we shall see in the following sections, the force the dam must supply can be much smaller than the weight of the water it holds back.