Chapter 10, Problem 76GP

To determine

To estimate: The amount of water delivered.

Answer to Problem 76GP

The amount of water delivered was $2.31\times {10}^7\text{L}$ .

Explanation of Solution

Given:

The horizontal cross-sectional area of the Caribbean $A=2650{\text{m}}^{\text{2}}$

The height of the ship from the sea $h=8.50\text{m}$

Formula used:

  $F_{\text{buoyant}}=V_{\text{displaced}}{\rho }_{\text{sea}\text{water}}g=m_{\text{fresh}}g$

Where $F_{\text{buoyant}}$ is the buoyant force,

  $V_{\text{displaced}}$ displaced water volume and

  $m_{\text{fresh}}$ is mass of fresh water.

Calculation:

The buoyant force must be equal to the weight of the displaced seawater. But buoyant force on the ship is also equal to the weight of the displaced seawater. It is given as,

  $m_{fresh}=\left(2650{\text{m}}^{\text{2}}\right)\times \left(8.50\text{m}\right)\times \left(1025\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.\right)=2.31\times {10}^7\text{kg}$

The volume is given as,

  $V_{\text{fresh}}=\frac{m_{\text{fresh}}}{{\rho }_{\text{fresh}}}=\frac{2.31\times {10}^7\text{kg}}{1.00\times {10}^3\raisebox{1ex}{$\text{kg}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{3}}$}\right.}=2.31\times {10}^7\text{L}$

Conclusion:

The amount of water delivered was $2.31\times {10}^7\text{L}$ .