Chapter 3, Problem 3.14P

To determine

(a)

The number of grains in the metal cube.

Answer to Problem 3.14P

The number of grains in the metal cube is ${10}^{12}\text{grains}$.

Explanation of Solution

Given:

The side of the copper cube is $0.10\text{m}$.

The size of grain on each side is $10\times {10}^{-6}\text{m}$.

The average grain boundary energy of copper is $625\times {10}^{-3}\text{J}/{\text{m}}^{\text{2}}$.

Formula used:

The number of grain is given by,

$n=\frac{V_{\text{cube}}}{V_{\text{grains}}}$   ....... (I)

Here, $n$ is the number of grains, $V_{\text{cube}}$ is the volume of cube and $V_{\text{grains}}$ is the volume of grain.

The volume of a cube is given by,

$V_{\text{cube}}=a^3$   ....... (II)

Here, $a$ is the side of the cube.

The volume of grains is given by,

$V_{\text{grain}}={\left(a_{\text{g}}\right)}^3$   ....... (III)

Here, $a_{\text{g}}$ is the side of each grain.

Substitute ${\left(a_{\text{g}}\right)}^3$ for $V_{\text{grain}}$ and $a^3$ for $V_{\text{cube}}$ in equation (I).

$n=\frac{a^3}{{\left(a_g\right)}^3}$   ....... (IV)

Calculation:

The number of grains is calculated as,

Substitute $10\times {10}^{-6}\text{m}$ for $a_{\text{g}}$ and $0.10\text{m}$ for $a$ in equation (IV).

$\begin{array}{c}n=\frac{{\left(0.10\text{m}\right)}^3}{{\left(10\times {10}^{-6}\text{m}\right)}^3}\\ =\frac{1\times {10}^{-3}{\text{m}}^{\text{3}}}{1\times {10}^{-15}{\text{m}}^{\text{3}}}\\ ={10}^{12}\end{array}$

Conclusion:

Therefore, the number of grains present are ${10}^{12}\text{grains}$.

To determine

(b)

The total grain boundary area.

Answer to Problem 3.14P

The total grain boundary area is $300{\text{m}}^{\text{2}}$.

Explanation of Solution

Given:

The side of the copper cube is $0.10\text{m}$.

The size of grain on each side is $10\times {10}^{-6}\text{m}$.

The average grain boundary energy of copper is $625\times {10}^{-3}\text{J}/{\text{m}}^{\text{2}}$.

Formula used:

The total grain boundary area is given by,

$A_{\text{total}}=n\times \left(\frac{N}{\text{2}}\right){\left(a_{\text{g}}\right)}^2$   ....... (V)

Here, $A_{\text{total}}$ is the total grain boundary and $N$ is the number of sides.

Calculation:

The total grain boundary area is calculated as

Substitute ${10}^{12}$ for $n$, $6$ for $N$ and $10\times {10}^{-6}\text{m}$ for $a_{\text{g}}$ in equation (V).

$\begin{array}{c}{A}_{\text{total}}=\left({10}^{12}\right)\times \left(\frac{6}{\text{2}}\right){\left(10\times {10}^{-6}\text{m}\right)}^2\\ =\left({10}^{12}\right)\times \left(3\right)\left(1\times {10}^{-10}{\text{m}}^2\right)\\ =300{\text{m}}^2\end{array}$

Conclusion:

Therefore, the total ground boundary area is $300{\text{m}}^{\text{2}}$.

To determine

(c)

The total energy present in the copper cube due to grain boundaries.

Answer to Problem 3.14P

The total energy present in the copper cube due to grain boundaries is $188\text{J}$.

Explanation of Solution

Given:

The side of the copper cube is $0.10\text{m}$.

The size of grain on each side is $10\times {10}^{-6}\text{m}$.

The average grain boundary energy of copper is $625\times {10}^{-3}\text{J}/{\text{m}}^{\text{2}}$.

Formula used:

The total energy due to grain boundaries is given by

$E_{\text{GB}}=\gamma A_{\text{total}}$   ....... (VI)

Here, $E_{\text{GB}}$ is the total energy due to grain boundary and $\gamma$ is the grain boundary energy.

Calculation:

The total energy due to the grain boundary is calculated as

Substitute $625\times {10}^{-3}\text{J}/{\text{m}}^{\text{2}}$ for $\gamma$ and $300{\text{m}}^{\text{2}}$ for $A_{\text{total}}$ in equation (VI).

$\begin{array}{c}{E}_{\text{GB}}=\left(625\times {10}^{-3}\text{J}/{\text{m}}^{\text{2}}\right)\left(300{\text{m}}^{\text{2}}\right)\\ =188\text{J}\end{array}$

Conclusion:

Therefore, the total energy present in the copper cube due to grain boundaries is $188\text{J}$.

To determine

(d)

The height to which copper must be raised to increase the energy to an equal amount of energy in the grain boundaries.

Answer to Problem 3.14P

The height to which copper must be raised to increase the energy to an equal amount of energy in the grain boundaries is $2.12\text{m}$.

Explanation of Solution

Given:

The side of the copper cube is $0.10\text{m}$.

The size of grain on each side is $10\times {10}^{-6}\text{m}$.

The average grain boundary energy of copper is $625\times {10}^{-3}\text{J}/{\text{m}}^{\text{2}}$.

Formula used:

The expression for the equivalent gravitational energy is given by

$E_{\text{g}}=\rho V_{\text{cube}}gh$   ....... (VII)

Here, $E_{\text{g}}$ is the equivalent gravitational energy, $\rho$ is the density, $g$ is the acceleration due to gravity and $h$ is the height.

Calculation:

The height to which the copper cube must be raised is calculated by

Substitute $188\text{}\text{J}$ for $E_{\text{g}}$, $8.96\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$, ${10}^{-3}{\text{m}}^{\text{3}}$ for $V_{\text{cube}}$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation (VI).

$\begin{array}{c}188\text{}\text{J}=\left(8.96\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left({10}^{-3}{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)h\\ h=\frac{188\text{}\text{J}}{\left(8.96\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\right)\left({10}^{-3}{\text{m}}^{\text{3}}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)}\\ =\frac{188\text{J}}{87.9\text{J}/\text{m}}\\ =2.12\text{m}\end{array}$

Conclusion:

Therefore, the height to which copper must be raised to increased energy to an equal amount of energy in the grain boundaries is $2.12\text{m}$.