Chapter 1, Problem 1.74QE

(a)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  $\text{45}\text{s}\text{=}\underset{\_}{}\text{ms}\text{=}\underset{\_}{}\text{minutes}$

Explanation of Solution

Given,

  $\text{45}\text{s}\text{=}\underset{\_}{}\text{ms}\text{=}\underset{\_}{}\text{minutes}$

The conversion factor used to convert seconds to milliseconds is

  $\begin{array}{l}\text{1}\text{s}\text{=}{\text{10}}^{\text{3}}\text{ms}\\ \\ \text{then,}\\ \text{45}\text{s}\text{=}\text{45}\text{×}{\text{10}}^{\text{3}}\text{ms}\\ \text{=}\text{4}\text{.5}\text{×}{\text{10}}^{\text{4}}\text{ms}\\ \text{=}\text{45000}\text{ms}\end{array}$

The conversion factor used to convert seconds to minutes is

  $\begin{array}{l}\text{1}\text{minute}\text{=}\text{60}\text{s}\\ \\ \text{then,}\\ \text{45}\text{s}\text{=}\text{0}\text{.75}\text{minute}\end{array}$

The given blanks can be filled as follows:

  $\text{45}\text{s}\text{=}\underset{\_}{4.5\times {10}^4}\text{ms}\text{=}\underset{\_}{0.75}\text{minutes}$

(b)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  $\text{550}\text{nm}\text{=}\underset{\_}{}\text{cm}\text{=}\underset{\_}{}\text{m}$

Explanation of Solution

Given,

  $\text{550}\text{nm}\text{=}\underset{\_}{}\text{cm}\text{=}\underset{\_}{}\text{m}$

The conversion factor used to convert nanometers to centimeters is

  $\begin{array}{l}\text{1}\text{nm}\text{=}{\text{10}}^{\text{-9}}\text{m}\\ \text{1}\text{cm}\text{=}{\text{10}}^{\text{-2}}\text{m}\\ \\ \text{then,}\\ \text{550}\text{nm}\text{=}\text{5}\text{.5}\text{×}{\text{10}}^{\text{-5}}\text{cm}\\ \text{=}\text{0}\text{.000055}\text{cm}\end{array}$

The conversion factor used to convert nanometers to meters is

  $\begin{array}{l}\text{1}\text{nm}\text{=}{\text{10}}^{\text{-9}}\text{m}\\ \\ \text{then,}\\ \text{550}\text{nm}\text{=}\text{550}\text{×}{\text{10}}^{\text{-9}}\text{m}\\ \text{=}\text{5}\text{.5}\text{×}{\text{10}}^{\text{-7}}\text{m}\end{array}$

The given blanks can be filled as follows:

  $\text{550}\text{nm}\text{=}\underset{\_}{5.5\times {10}^{-5}}\text{cm}\text{=}\underset{\_}{5.5\times {10}^{-7}}\text{m}$

(c)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  ${\text{4}}^{\text{°}}\text{C}\text{=}\underset{\_}{}\text{K}\text{=}\underset{\_}{}{}^{\text{°}}\text{F}$

Explanation of Solution

Given,

  ${\text{4}}^{\text{°}}\text{C}\text{=}\underset{\_}{}\text{K}\text{=}\underset{\_}{}{}^{\text{°}}\text{F}$

The conversion factor used to convert degree Celsius to Kelvin is

  $\begin{array}{l}{\text{T}}_{\text{C}}\text{=}{\text{T}}_{\text{K}}\text{-}\text{273}\text{.15}\\ \\ \text{then,}\\ {\text{4}}^{\text{°}}\text{C}\text{=}\text{-269}\text{.15}\text{K}\end{array}$

The conversion factor used to convert degree Celsius to degree Fahrenheit is

  $\begin{array}{l}{\text{T}}_{\text{F}}\text{=}{\text{T}}_{\text{C}}\text{×}\left(\frac{\text{1}{\text{.8}}^{\text{°}}\text{F}}{\text{1}{\text{.0}}^{\text{°}}\text{C}}\right)\text{+}{\text{32}}^{\text{°}}\text{F}\\ \\ \text{then,}\\ {\text{4}}^{\text{°}}\text{C}\text{=}\text{39}{\text{.2}}^{\text{°}}\text{F}\end{array}$

The given blanks can be filled as follows:

  ${\text{4}}^{\text{°}}\text{C}\text{=}\underset{\_}{-269.15}\text{K}\text{=}\underset{\_}{39.2}{}^{\text{°}}\text{F}$

(d)

Interpretation Introduction

Interpretation:

The given conversions have to be performed to fill the blanks.  The operations have to be done without calculator or spreadsheet.

  $\text{2}\text{.00}\text{L}\text{=}\underset{\_}{}{\text{cm}}^{\text{3}}\text{=}\underset{\_}{}{\text{m}}^{\text{3}}\text{=}\underset{\_}{}\text{qt}$

Explanation of Solution

Given,

  $\text{2}\text{.00}\text{L}\text{=}\underset{\_}{}{\text{cm}}^{\text{3}}\text{=}\underset{\_}{}{\text{m}}^{\text{3}}\text{=}\underset{\_}{}\text{qt}$

The conversion factor used to convert liters to cubic centimeters is

  $\begin{array}{l}\text{1}\text{L}\text{=}{\text{10}}^{\text{3}}\text{mL}\\ \text{1}\text{mL}\text{=}\text{1}{\text{cm}}^{\text{3}}\\ \\ \text{then,}\\ \text{2}\text{.00}\text{L}\text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{3}}\text{mL}\\ \text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{3}}{\text{cm}}^{\text{3}}\end{array}$

The conversion factor used to convert liters to cubic meters is

  $\begin{array}{l}\text{1}\text{L}\text{=}{\text{10}}^{\text{3}}\text{mL}\\ \text{1}{\text{cm}}^{\text{3}}\text{=}\text{1}\text{mL}\\ \text{1}{\text{m}}^{\text{3}}\text{=}{\text{10}}^{\text{6}}{\text{cm}}^{\text{3}}\\ \\ \text{then,}\\ \text{2}\text{.00}\text{L}\text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{3}}{\text{cm}}^{\text{3}}\\ \text{=}\text{2}\text{.00}\text{×}{\text{10}}^{\text{-3}}{\text{m}}^{\text{3}}\end{array}$

The conversion factor used to convert liters to quart is

  $\begin{array}{l}\text{1}\text{L}\text{=1}\text{.057}\text{qt}\\ \\ \text{then,}\\ \text{2}\text{.00}\text{L}\text{=}\text{2}\text{.00}\text{×}\text{1}\text{.057}\text{qt}\\ \text{=}\text{2}\text{.114}\text{qt}\end{array}$

The given blanks can be filled as follows:

  $\text{2}\text{.00}\text{L}\text{=}\underset{\_}{\text{2}\text{×}{\text{10}}^{\text{3}}}{\text{cm}}^{\text{3}}\text{=}\underset{\_}{\text{2}\text{×}{\text{10}}^{\text{-3}}}{\text{m}}^{\text{3}}\text{=}\underset{\_}{\text{2}\text{.114}}\text{qt}$