Chapter 15, Problem 99A

(a)

Interpretation Introduction

Interpretation: The mass of water in grams in the cylindrical vessel needs to be calculated using the given information in question.

Concept Introduction: The volume of the cylinder is calculated as follows:

  $V=\pi r^{2}h$

Here, r is the radius and h is the height of the cylinder.

The density is related to mass and volume as follows:

  $d=\frac{m}{V}$

Here, m is mass and V is volume.

Explanation of Solution

The height of the cylindrical vessel is 28.0 cm and its diameter is 3.00 cm. It is filled with water at $50{}^{\text{o}}\text{C}$ . The density of water is 0.988 ${\text{g/cm}}^{\text{3}}$ . From the given height and diameter, the volume of the cylindrical vessel can be calculated as follows:

  $V=\pi r^{2}h$

Substitute the values,

  $\begin{array}{c}V=\left(3.14\right){\left(\frac{3.00\text{ cm}}{2}\right)}^2\left(28.0\text{ cm}\right)\\ =197.82{\text{ cm}}^3\end{array}$

Now, the volume of the cylinder is equal to the volume of water filled in it.

Now, density is related to mass and volume as follows:

  $d=\frac{m}{V}$

Or,

  $m=d\times V$

Now, the density of water is 0.988 ${\text{g/cm}}^{\text{3}}$ .

Substitute the values,

  $\begin{array}{c}m=\left(0.988{\text{ g/cm}}^3\right)\left(197.82{\text{ cm}}^3\right)\\ =195.45\text{ g}\end{array}$

Therefore, the mass of water is 195.45 g.

(b)

Interpretation Introduction

Interpretation: The mass of water in milligrams in the cylindrical vessel needs to be calculated using the given information in question.

Concept Introduction: The volume of the cylinder is calculated as follows:

  $V=\pi r^{2}h$

Here, r is the radius and h is the height of the cylinder.

The density is related to mass and volume as follows:

  $d=\frac{m}{V}$

Here, m is mass and V is volume.

Explanation of Solution

The calculated mass of water can be converted into milligrams using the following conversion factor:

  $\frac{1000\text{ mg}}{1\text{ g}}$

Thus,

  $\begin{array}{c}m=\left(195.45\text{ g}\right)\left(\frac{1000\text{ mg}}{1\text{ g}}\right)\\ =1.95\times {10}^5\text{ mg}\end{array}$

Thus, the mass of water in mg is $1.95\times {10}^5\text{ mg}$ .

(c)

Interpretation Introduction

Interpretation: The mass of water in kilogram in the cylindrical vessel needs to be calculated using the given information in question.

Concept Introduction: The volume of the cylinder is calculated as follows:

  $V=\pi r^{2}h$

Here, r is the radius and h is the height of the cylinder.

The density is related to mass and volume as follows:

  $d=\frac{m}{V}$

Here, m is mass and V is volume.

Explanation of Solution

The calculated mass of water can be converted into kilograms as follows:

  $\frac{{10}^{-3}\text{ kg}}{1\text{ g}}$

Thus,

  $\begin{array}{c}m=\left(195.45\text{ g}\right)\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)\\ =0.95\text{ kg}\end{array}$

Thus, the mass of water in kg is $0.95\text{ kg}$ .