Chapter 1, Problem 70QP

A human brain weighs about 1 kg and contains about ${\text{10}}^{\text{11}}\text{ cells}$. Assuming that each cell is completely filled with water $\left(\text{density = 1 g/mL}\right)$, calculate the length of one side of such a cell if it were a cube. If the cells are spread out into a thin layer that is a single cell thick, what is the surface area in square meters?

Interpretation Introduction

Interpretation:

The length of one side of each cell if it were a cube and the surface area if cells are spread out into a thin layer are to be calculated.

Concept introduction:

Dimensional analysis is used to set up and solve a unit conversion problem using conversion factors.

Dimensional analysis is a way to convert units of measurement. In order to convert one unit to another, the relationship between those units is to be predicted. These relationships are called conversion factors.

$\text{Volume = Surface area × height}$.

$\text{Surface area}=\frac{\text{Volume}}{\text{height}}$.

${\text{1mL =1cm}}^3$, to convert mL to cubic centimeter, the conversion factor is $\frac{1{\text{cm}}^3}{1\text{mL}}$.

Answer to Problem 70QP

Solution: $2\times {10}^{-3}\text{ cm}$, $50{\text{ m}}^{\text{2}}$

Explanation of Solution

Given information: The mass of the brain $=1\text{ kg}$.

The number of cells $={10}^{11}$.

The density of water $=1\text{ g/mL}$.

The mass of a brain cell is as follows:

$\begin{array}{c}\text{The mass of brain cell}=\frac{\text{1000 g}}{{\text{1×10}}^{\text{11}}\text{cells}}\\ =1\times {10}^{-8}\text{ g/cell}\end{array}$

Assume that each cell is completely filled with water.

The volume of each cell is as follows:

$\begin{array}{c}V=\left(\frac{{\text{1×10}}^{\text{-8}}\cancel{\text{g}}}{\text{1 cell}}\right)\left(\frac{\text{1 }\cancel{\text{mL}}}{\text{1 }\cancel{\text{g}}}\right)\left(\frac{{\text{1 cm}}^{\text{3}}}{\text{1 }\cancel{\text{mL}}}\right)\\ =1\times {10}^{-8}{\text{ cm}}^{\text{3}}\text{/cell}\end{array}$

Assume that the cell is cubic.

The volume of a cube is as:

$\text{Volume}=a^{\text{3}}$

Here, $a$ is the length of the side.

Rearrange the above equation to determine the length.

The length of a single cell is as:

$\begin{array}{c}a={\left(\text{V}\right)}^{\text{1/3}}\\ ={\left(1\times {10}^{-8}{\text{cm}}^{\text{3}}\right)}^{\text{1/3}}\\ =\text{ 0}\text{.002 cm}\\ =2\times {10}^{-3}\text{ cm}\end{array}$

So, the length of a single cell is $2\times {10}^{-3}\text{ cm}$.

There are ${10}^{11}$ cells that are spread in a thin layer.

A single cell thick surface area can be calculated as follows:

$\text{Volume = Surface area × height}$ (1)

First, calculate the volume for total cells and then calculate the surface area.

The volume of total cells is as follows:

$\begin{array}{c}{\text{Volume 10}}^{\text{11}}\text{of cells }=\left(1\times {10}^{-8}{\text{cm}}^{\text{3}}\text{/}\cancel{\text{cell}}\right)\left({\text{10}}^{\text{11}}\cancel{\text{cell}}\right)\\ =1\times {10}^3{\text{ cm}}^{\text{3}}\end{array}$

Substitute the values of height (equal to length) of a cell and volume of total cells in equation (1).

$\begin{array}{c}\text{Surface area}=\frac{\text{Volume}}{\text{height}}\\ =\left(\frac{{\text{1000 cm}}^{\text{3}}}{\text{0}\text{.002 cm}}\right)\\ =\left(5\times {10}^5{\text{cm}}^{\text{2}}\right){\left(\frac{1\times {10}^{-2}\text{m}}{\text{1cm}}\right)}^{\text{2}}\\ =\left(5\times {10}^5\cancel{{\text{cm}}^{\text{2}}}\right)\left(\frac{1\times {10}^{-4}{\text{m}}^{\text{2}}}{\text{1}\cancel{{\text{cm}}^{\text{2}}}}\right)\\ =50{\text{ m}}^{\text{2}}\end{array}$

Hence, the surface area is $50{\text{ m}}^{\text{2}}$.

Conclusion

The length of one side of each cell is $2\times {10}^{-3}\text{ cm}$ and the surface area is $50{\text{ m}}^{\text{2}}$.