Chapter 1, Problem 1.53QP

Interpretation Introduction

Interpretation: For the given data, it should be determined for the cost to supply enough vanillin &detection aroma in a large aircraft hangar with a volume of $\text{5}{\text{.0×10}}^{\text{7}}{\text{ft}}^{\text{3}}$.

Concept introduction:

• Volume is the ratio of mass to density and its SI unit is ${\text{m}}^{\text{3}}$.

        $Volume=\frac{Mass}{Density}$

• Conversion formula:

$\begin{array}{l}1\text{gallon}\text{=}\text{3}\text{.79}\text{L}\\ \text{1}\text{mL}\text{=}\text{0}\text{.001L}\\ \text{1}\text{mL}\text{=}\text{1}\text{g}\end{array}$

• $Thickness=\frac{Volume}{area}$
• Conversional formula from Celsius to Fahrenheit,

${\text{F}}^{\text{o}}\text{=}\left({}^{\text{o}}\text{C×}\frac{{\text{9}}^{\text{o}}\text{F}}{{\text{5}}^{\text{o}}\text{C}}\right)\text{+}{\text{32F}}^{\text{o}}$

• Conversional formula from Fahrenheit to Celsius

${}^{\text{o}}\text{C}=({\text{F}}^{\text{o}}-{\text{32F}}^{\text{o}})\times \frac{{\text{5}}^{\text{o}}\text{C}}{{\text{9}}^{\text{o}}\text{F}}$

• Conversional formula from Celsius to Kelvin,

$\text{K}\text{=}{}^{\circ }\text{C}\text{+}\text{273}\text{.15}$

Given:

The threshold limit is $\text{2}{\text{.0×10}}^{\text{-11}}\text{g vanilin}$ per litter of air.

Current price is 50g of vanillin =$\text{\$112}$