Chapter 18, Problem 18.40P

Interpretation Introduction

Interpretation:

The values of oxidation of copper for the given conditions should be determined. Also, the reason for the type of oxidation whether it is parabolic, linear, or logarithmic should be explained.

The time required for oxidizing the sheet of copper having the thickness $0.75\text{cm}$ should be calculated.

Concept introduction:

Oxidation: When metals react with oxygen to produce an oxide at the surface is known as oxidation. The factors on which oxidation depends are as follows:

1. The ease with which metal oxidized.
2. Nature of oxide
3. Rate at which oxidation occurs.

Parabolic Relation: When diffusion of ions or electrons occurs through a nonporous oxide layer and it acts as the controlling factor. Parabolic relation is given as,

  $y=\sqrt{kt}$

Where,

y is the thickness.

k is the value of oxidation occur with respect to time.

t is the time taken for oxidation at the given value of thickness.

Answer to Problem 18.40P

The required value of oxidation as per the given condition is shown in the table below:

Time (t in hours)	W(mass)	Thickness (y) in cm	Oxidation value (k)
$100\text{hr}$	$0.246\text{g}{\text{/cm}}^{\text{2}}$	$y_1=0.0275\text{cm}$	${\mathrm{k}}_1=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$
$250\text{hr}$	$\text{0}\text{.388}\text{g}\text{/}{\text{cm}}^{\text{2}}$	$y_2=0.04344\text{cm}$	${\mathrm{k}}_2=7.50\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$
$500\text{hr}$	$\text{0}\text{.550}\text{g}\text{/}{\text{cm}}^{\text{2}}$	${\text{y}}_{\text{3}}=0.615\text{cm}$	${\text{k}}_{\text{3}}=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$

The parabolic relation for oxidation of copper is considered.

The time required for oxidizing the sheet of copper having the thickness $0.75\text{cm}$ is $\text{t}\text{=}\text{18731}\text{hr}$

Explanation of Solution

Given information:

Copper is exposed to oxygen is ${1000}^{°}\text{C}$.

After 100 hours, the amount of copper lost per square cm of surface area is $0.246\text{g}{\text{/cm}}^{\text{2}}$.

After 250 hours, the amount of copper lost per square cm of surface area is $\text{0}\text{.388}\text{g}\text{/}{\text{cm}}^{\text{2}}$.

After 500 hours, the amount of copper lost per square cm of surface area is $\text{0}\text{.550}\text{g}\text{/}{\text{cm}}^{\text{2}}$.

Copper sheet is oxidized from both sides.

As per given information,

Calculate the rate of thickness after 100 hours having the amount of copper as $0.246\text{g}\text{/}{\text{cm}}^{\text{2}}$ ,

Thickness for this case is defined as the ratio of volume in cubic centimeter to area in square meter. The formula is stated below:

  $y_1={\left(\frac{\text{V}}{\text{A}}\right)}_{\text{Cu}}$

Where,

V is the volume in cubic cm.

A is the area in square cm.

  $y_1$ is the thickness copper.

Volume is defined as the ratio weight of component to density of component. The formula is given below:

  ${\text{V}}_{\text{Cu}}={\left(\frac{\text{W}}{\text{ρ}}\right)}_{\text{Cu}}$

Putting the values,

W = $0.246\text{g}\text{/}{\text{cm}}^{\text{2}}$

  ${\text{ρ}}_{\text{Cu}}=8.93\text{g}\text{/}{\text{cm}}^3$

  $\begin{array}{l}{\text{V}}_{\text{Cu}}={\left(\frac{0.246}{8.93}\right)}_{\text{Cu}}\\ {\text{V}}_{\text{Cu}}=0.0275{\text{cm}}^{\text{3}}\end{array}$

On putting the values in the formula of thickness taking the value of area as $\text{A}\text{=}\text{1}{\text{cm}}^{\text{2}}$.

  $y_1=\left(\frac{{\text{V}}_{\text{Cu}}}{{\text{A}}_{\text{Cu}}}\right)$

  $\begin{array}{l}{y}_1=\left(\frac{0.0275{\text{cm}}^{\text{3}}}{1{\text{cm}}^2}\right)\\ y_1=0.0275\text{cm}\end{array}$

Thus, the value of thickness after 100 hours is $y_1=0.0275{\text{cm}}^{\text{2}}$.

Calculate the rate of thickness after 250 hours having the amount of copper as $\text{0}\text{.388}\text{g}\text{/}{\text{cm}}^{\text{2}}$.

Thickness for this case is defined as the ratio of volume in cubic centimeter to area in square meter. The formula is stated below:

  $y={\left(\frac{\text{V}}{\text{A}}\right)}_{\text{Cu}}$

Where,

V is the volume in cubic cm.

A is the area in square cm.

  $y_1$ is the thickness copper.

Volume is defined as the ration weight of component to density of component. The formula is given below:

  ${\text{V}}_{\text{Cu}}={\left(\frac{\text{W}}{\text{ρ}}\right)}_{\text{Cu}}$

Putting the values,

W = $\text{0}\text{.388}\text{g}\text{/}{\text{cm}}^{\text{2}}$

  ${\text{ρ}}_{\text{Cu}}=8.93\text{g}\text{/}{\text{cm}}^3$

  $\begin{array}{l}{\text{V}}_{\text{Cu}}={\left(\frac{0.388}{8.93}\right)}_{\text{Cu}}\\ {\text{V}}_{\text{Cu}}=0.04344{\text{cm}}^{\text{3}}\end{array}$

On putting the values in the formula of thickness taking the value of area as $\text{A}\text{=}\text{1}{\text{cm}}^{\text{2}}$.

  $y_2=\left(\frac{{\text{V}}_{\text{Cu}}}{{\text{A}}_{\text{Cu}}}\right)$

  $\begin{array}{l}{y}_2=\left(\frac{0.04344{\text{cm}}^{\text{3}}}{1{\text{cm}}^2}\right)\\ y_2=0.04344\text{cm}\end{array}$

Thus, the value of thickness after 250 hours is $y_2=0.04344\text{cm}$.

Calculate the rate of thickness after 500 hours having the amount of copper as $\text{0}\text{.550}\text{g}\text{/}{\text{cm}}^{\text{2}}$.

Thickness for this case is defined as the ratio of volume in cubic centimeter to area in square meter. The formula is stated below:

  $y={\left(\frac{\text{V}}{\text{A}}\right)}_{\text{Cu}}$

Where,

V is the volume in cubic cm.

A is the area in square cm.

  $y_1$ is the thickness copper.

Volume is defined as the ratio of the weight of component to density of component. The formula is given below:

  ${\text{V}}_{\text{Cu}}={\left(\frac{\text{W}}{\text{ρ}}\right)}_{\text{Cu}}$

Putting the values,

W = $\text{0}\text{.550}\text{g}\text{/}{\text{cm}}^{\text{2}}$

  ${\text{ρ}}_{\text{Cu}}=8.93\text{g}\text{/}{\text{cm}}^3$

  $\begin{array}{l}{\text{V}}_{\text{Cu}}={\left(\frac{0.550}{8.93}\right)}_{\text{Cu}}\\ {\text{V}}_{\text{Cu}}=0.615{\text{cm}}^{\text{3}}\end{array}$

On putting the values in the formula of thickness taking the value of area as $\text{A}\text{=}\text{1}{\text{cm}}^{\text{2}}$.

  $y_3=\left(\frac{{\text{V}}_{\text{Cu}}}{{\text{A}}_{\text{Cu}}}\right)$

  $\begin{array}{l}{y}_3=\left(\frac{0.615{\text{cm}}^{\text{3}}}{1{\text{cm}}^2}\right)\\ y_3=0.615\text{cm}\end{array}$

Thus, the value of thickness after 500 hours is $y_3=0.615\text{cm}$.

As for the case of copper, parabolic relationship is taken into consideration. The relation gives parabolic relationship,

  $y={\left(\frac{\text{V}}{\text{A}}\right)}_{\text{Cu}}$

Where,

V is the volume in cubic cm.

A is the area in square cm.

  $y_1$ is the thickness copper.

Calculation of oxidation value,

For copper having the value of thickness as $y_1=0.0275\text{cm}$ at time is equal to $100\text{hr}$.

Putting the value in the formula,

  $\begin{array}{l}{y}_1={\sqrt{k_{1}t}}_1\\ {\text{y}}_{\text{1}}=0.0275{\text{cm}}^{\text{2}}\\ {\text{t}}_{\text{1}}=100\mathrm{hr}\\ 0.0275=\sqrt{k_1\times 100}\\ {\mathrm{k}}_1=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}\end{array}$

The required value of oxidation after 100 hours is ${\mathrm{k}}_1=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$.

Calculation of oxidation value,

For copper having the value of thickness as $y_2=0.04344\text{cm}$ at time is equal to $250\text{hr}$.

Putting the value in the formula,

  $\begin{array}{l}{y}_2={\sqrt{k_{2}t}}_2\\ {\text{y}}_{\text{2}}=0.04344\text{cm}\\ {\text{t}}_{\text{2}}=250\text{hr}\\ \text{0}\text{.04344}=\sqrt{k_2\times 250}\\ {\mathrm{k}}_2=7.50\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}\end{array}$

The required value of oxidation after 250 hours is ${\mathrm{k}}_2=7.50\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$.

Calculation of oxidation value,

For copper having the value of thickness as ${\text{y}}_{\text{3}}=0.615\text{cm}$ at time is equal to $500\text{hr}$.

Putting the value in the formula,

  $\begin{array}{l}{y}_3={\sqrt{k_{3}t}}_3\\ {\text{y}}_{\text{3}}=0.615\text{cm}\\ {\text{t}}_{\text{3}}=500\text{hr}\\ 0.615=\sqrt{k_3\times 500}\\ k_3=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}\end{array}$

The required value of oxidation after 250 hours is ${\text{k}}_{\text{3}}=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$.

The required value of oxidation as per the given condition is shown in the table below:

Time (t in hours)	W(mass)	Thickness (y) in cm	Oxidation value (k)
$100\text{hr}$	$0.246\text{g}{\text{/cm}}^{\text{2}}$	$y_1=0.0275\text{cm}$	${\mathrm{k}}_1=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$
$250\text{hr}$	$\text{0}\text{.388}\text{g}\text{/}{\text{cm}}^{\text{2}}$	$y_2=0.04344\text{cm}$	${\mathrm{k}}_2=7.50\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$
$500\text{hr}$	$\text{0}\text{.550}\text{g}\text{/}{\text{cm}}^{\text{2}}$	${\text{y}}_{\text{3}}=0.615\text{cm}$	${\text{k}}_{\text{3}}=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$

Calculation of time required for oxidizing the sheet of copper having the thickness $0.75\text{cm}$. Using parabolic relation and copper is oxidized from both sides.

  $y=\sqrt{kt}$

As copper is oxidized from both sides, the calculation of thickness,

  $\begin{array}{l}y=\frac{0.75}{2}\\ \text{y}\text{=}\text{0}\text{.375}\text{cm}\end{array}$

Substituting the values in the formula,

  $\text{k}=7.56\times {10}^{-6}{\text{cm}}^{\text{2}}\text{/}\text{hr}$

  $\begin{array}{l}\text{y}\text{=}\text{0}\text{.375}\text{cm}\\ \text{0}\text{.375}\text{=}\sqrt{7.56\times {10}^{-6}\times \text{t}}\\ \text{t}=18731\text{hr}\end{array}$

The time required for oxidizing the sheet of copper having the thickness $0.75\text{cm}$ is $\text{t}\text{=}\text{18731}\text{hr}$.