Chapter 1, Problem 1.37QP

(a)

Interpretation Introduction

Interpretation:

Height and weight of the given person has to be calculated in meter and kilogram unit.

Concept Introduction:

General unit conversions

$\begin{array}{l}\text{1 m}\text{=}\text{3}\text{.28}\text{ft}\\ \text{1 lb}\text{=}\text{453}\text{.6}\text{g}\\ \text{1000}\text{g}\text{=}\text{1}\text{kg}\end{array}$

Answer to Problem 1.37QP

• Height of the person in meter unit is $\text{1}\text{.8}\text{m}$
• Weight of the given person in kilogram unit is $\text{83}\text{.0}\text{kg}$

Explanation of Solution

Height of the person is $\text{6}\text{.0 ft}$

Weight of the given person is $\text{183}\text{lb}$

General unit conversions

$\begin{array}{l}\text{1 m}\text{=}\text{3}\text{.28}\text{ft}\\ \text{1 lb}\text{=}\text{453}\text{.6}\text{g}\\ \text{1000}\text{g}\text{=}\text{1}\text{kg}\end{array}$

Calculation of height in meter unit

$\begin{array}{l}\text{6}\text{.0}\text{ft}\text{×}\frac{\text{1}\text{m}}{\text{3}\text{.28}\text{ft}}\\ \text{=}\text{1}\text{.8}\text{m}\end{array}$

Calculation of weight in kilogram unit

$\begin{array}{l}\text{183}\text{lb}\text{×}\frac{\text{453}\text{.6}\text{g}}{\text{1}\text{lb}}\text{×}\frac{\text{1}\text{kg}}{\text{1000}\text{g}}\\ \text{=}\text{83}\text{.0088}\text{kg}\\ \text{=}\text{83}\text{kg}\end{array}$

• Height of the person in meter unit is $\text{1}\text{.8}\text{m}$
• Weight of the given person in kilogram unit is $\text{83}\text{.0}\text{kg}$

Height and weight of the given person was calculated in meter and kilogram unit by the use of given data and application of general unit conversions.

(b)

Interpretation Introduction

Interpretation:

Distance in miles that light would travel in 1 hour has to be calculated.

Concept Introduction:

By definition, $1milesperhour=1.60kilometerperhour$

Answer to Problem 1.37QP

Speed limit in kilometers per hour is $112kilometerperhour$

Explanation of Solution

Given,

Speed of light $=70milesperhour$

By definition, $1milesperhour=1.60kilometerperhour$

Converting miles per hour to kilometer per hour,

$\begin{array}{c}70milesperhour=70\times 1.60kilometerperhour\\ =112kilometerperhour\end{array}$

(c)

Interpretation Introduction

Interpretation:

Distance in miles that light would travel in 1 hour has to be calculated.

Concept Introduction:

Answer to Problem 1.37QP

Distance in miles is ${1.72810}^{19}kilometer$

Explanation of Solution

Given,

Speed of light $=3\times {10}^{10}cm/s$

Speed of light in 1 hour

$\begin{array}{l}=3\times {10}^{10}cm/s\times 3600\mathrm{s}\\ =1.08\times {10}^{14}cm\end{array}$

Converting cm to km,

$1.08\times {10}^{14}cm\times \frac{{10}^{5}cm}{1km}=1.08\times {10}^{19}km$

By definition, $1miles=1.60kilometer$

Converting miles per hour to kilometer per hour,

$\begin{array}{c}1miles=1.08\times {10}^{19}\times 1.60kilometerperhour\\ ={1.72810}^{19}kilometer\end{array}$

(d)

Interpretation Introduction

Interpretation:

The amount of lead $\left(\text{Pb}\right)$ present in grams has to be calculated in the given amount of blood.

Concept Introduction:

$ppm=\frac{massofsolute}{massofsolution}\times {10}^6$

Answer to Problem 1.37QP

The amount of lead $\left(\text{Pb}\right)$ present in the given amount of blood is $\text{3}\text{.7}\text{×}{\text{10}}^{\text{-3}}\text{gram}\text{of}\text{lead}$

Explanation of Solution

The amount of blood is $\text{6}\text{.0}\text{×}{\text{10}}^{\text{3}}\text{gram}$

Content of lead is $\text{0}\text{.62}\text{ppm}$

Convert the unit of g blood into g lead

$\text{0}\text{.62}\text{ppm}\text{=}\frac{\text{0}\text{.62}\text{gram}\text{Pb}}{\text{1}\text{×}{\text{10}}^{\text{6}}\text{gram}\text{blood}}$

Calculation of lead content in gram unit

$\begin{array}{l}\text{6}\text{.0}\text{×}{\text{10}}^{\text{3}}\text{gram}\text{blood}\text{×}\frac{\text{0}\text{.62}\text{gram}\text{Pb}}{\text{1}\text{×}{\text{10}}^{\text{6}}\text{gram}\text{blood}}\\ \text{=}\text{3}\text{.72}\text{×}{\text{10}}^{\text{-3}}\text{gram}\text{Pb}\\ \text{=}\text{3}\text{.7}\text{×}{\text{10}}^{\text{-3}}\text{g}\text{Pb}\end{array}$

The amount of lead is $\text{3}\text{.7}\text{×}{\text{10}}^{\text{-3}}\text{g}\text{Pb}$

The amount of lead present in the given amount of blood was calculated by the use of given data and unit conversion.