Chapter 20, Problem 20.73AP

Review. A 670-kg meteoroid happens to be composed of aluminum. When it is far from the Earth, its temperature is –15.0°C and it moves at 14.0 km/s relative to the planet. As it crashes into the Earth, assume the internal energy transformed from the mechanical energy of the meteoroid-Earth system is shared equally between the meteoroid and the Earth and all the material of the meteoroid rises momentarily to the same final temperature. Find this temperature. Assume the specific heat of liquid and of gaseous aluminum is 1 170 J/kg · °C.

To determine

The value of the final temperature.

Answer to Problem 20.73AP

The value of the final temperature is $5.87\times {10}^4°\text{C}$.

Explanation of Solution

Given info: The mass of the meteorite is $670\text{kg}$, the initial temperature of the meteorite is $-15.0°\text{C}$, the speed of the meteorite is $14.0\text{km}/\text{s}$ and the specific heat of liquid and gaseous aluminum is $1170\text{J}/\text{kg}\cdot °\text{C}$.

Write the expression for the gravitational potential enegy.

$U=\frac{G{M}_{\text{E}}m}{R_{\text{E}}}$

Here,

$m$ is the mass of the meteorite.

$G$ is universal constant.

$M_{\text{E}}$ is the mass of the earth.

$R_{\text{E}}$ is the radius of the earth.

Write the expression for the kinetic enegy.

$k=\frac{1}{2}m{v}^2$

Here,

$v$ is the speed of the meteorite.

Write the expression for the loss of mechanical energy of the meteorite.

$E=U+k$

Substitute $\frac{G{M}_{\text{E}}m}{R_{\text{E}}}$ for $U$ and $\frac{1}{2}m{v}^2$ for $k$ in the above equation.

$E=\frac{1}{2}m{v}^2+\frac{G{M}_{\text{E}}m}{R_{\text{E}}}$

Substitute $670\text{kg}$ for $m$, $14.0\text{km}/\text{s}$ for $v$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $5.98\times {10}^{24}\text{kg}$ for $M_{\text{E}}$ and $6.37\times {10}^6\text{m}$ for $R_{\text{E}}$ in the above equation.

$\begin{array}{c}E=\left[\begin{array}{l}\frac{1}{2}\left(670\text{kg}\right){\left(14.0\text{km}/\text{s}\times \frac{1000\text{m}}{1\text{km}}\right)}^2\\ +\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(670\text{kg}\right)}{6.37\times {10}^6\text{m}}\end{array}\right]\\ \approx 1.08\times {10}^{11}\text{J}\end{array}$

Thus, the loss of mechanical energy is $1.08\times {10}^{11}\text{J}$.

The value of the internal energy is half of the loss of mechanical energy.

Write the expression for the internal energy.

$\Delta E_{\text{int}}=\frac{1}{2}E$

Substitute $1.08\times {10}^{11}\text{J}$ for $E$ in above equation.

$\begin{array}{c}\Delta E_{\text{int}}=\frac{1}{2}\times 1.08\times {10}^{11}\text{J}\\ \approx \text{5}\text{.38}\times {\text{10}}^{10}\text{J}\end{array}$

Thus, the internal energy is $\text{5}\text{.38}\times {\text{10}}^{10}\text{J}$.

Write the expression for the energy required to increase the temperature.

$E=mc\left(T_{\text{f}}-T_{\text{i}}\right)$ (1)

Here,

$c$ is the specific heat capacity.

$T_{\text{f}}$ is the final temperature.

$T_{\text{i}}$ is the initial temperature.

Write the expression for the rate of energy required during a phase change.

$E=Lm$ (2)

Here,

$L$ is the latent heat of the substance.

For the energy required to raise the temperature to a melting point.

Substitute $E_1$ for $E$, $670\text{kg}$ for $m$, $900\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $660°\text{C}$ for $T_{\text{f}}$ and $-15°\text{C}$ for $T_{\text{i}}$ in equation (1).

$\begin{array}{c}{E}_1=\left(670\text{kg}\right)\left(900\text{J}/\text{kg}\cdot °\text{C}\right)\left(660°\text{C}-\left(-15°\text{C}\right)\right)\\ \approx 4.07\times {10}^8\text{J}\end{array}$

Thus, the energy required to raise the temperature to a melting point is $4.07\times {10}^8\text{J}$.

For the energy required to melt the meteorite.

Substitute $E_2$ for $E$, $670\text{kg}$ for $m$ and $3.97\times {10}^5\text{J}/\text{kg}$ for $L$ in equation (2).

$\begin{array}{c}{E}_2=\left(670\text{kg}\right)\left(3.97\times {10}^5\text{J}/\text{kg}\right)\\ \approx 2.66\times {10}^8\text{J}\end{array}$

Thus, the energy required to melt the meteorite is $2.66\times {10}^8\text{J}$.

For the energy required to raise the temperature to a boiling point.

Substitute $E_3$ for $E$, $670\text{kg}$ for $m$, $1170\text{J}/\text{kg}\cdot °\text{C}$ for $c$, $2450°\text{C}$ for $T_{\text{f}}$ and $660°\text{C}$ for $T_{\text{i}}$ in equation (1).

$\begin{array}{c}{E}_3=\left(670\text{kg}\right)\left(1170\text{J}/\text{kg}\cdot °\text{C}\right)\left(2450°\text{C}-660°\text{C}\right)\\ \approx 1.40\times {10}^9\text{J}\end{array}$

Thus, the energy required to raise the temperature to a boiling point is $1.40\times {10}^9\text{J}$.

For the energy required to boil the meteorite.

Substitute $E_4$ for $E$, $670\text{kg}$ for $m$ and $1.14\times {10}^7\text{J}/\text{kg}$ for $L$ in equation (2).

$\begin{array}{c}{E}_4=\left(670\text{kg}\right)\left(1.14\times {10}^7\text{J}/\text{kg}\right)\\ \approx 7.64\times {10}^9\text{J}\end{array}$

Thus, the energy required to boil the meteorite is $7.64\times {10}^9\text{J}$.

Write the expression for the internal energy change for the meteorite.

$\Delta E_{\text{int}}=E_1+E_2+E_3+E_4+mc\left(T_{\text{f}}-T_{\text{i}}\right)$

Substitute $\text{5}\text{.38}\times {\text{10}}^{10}\text{J}$ for $\Delta E_{\text{int}}$, $4.07\times {10}^8\text{J}$ for $E_1$, $2.66\times {10}^8\text{J}$ for $E_2$, $1.40\times {10}^9\text{J}$ for $E_3$, $7.64\times {10}^9\text{J}$ for $E_4$, $670\text{kg}$ for $m$, $1170\text{J}/\text{kg}\cdot °\text{C}$ for $c$ and $2450°\text{C}$ for $T_{\text{i}}$ in above equation.

$\begin{array}{c}\text{5}\text{.38}\times {\text{10}}^{10}\text{J}=\left(\begin{array}{l}4.07\times {10}^8\text{J}+2.66\times {10}^8\text{J}+1.40\times {10}^9\text{J}+7.64\times {10}^9\text{J}\\ +\left(670\text{kg}\right)\left(1170\text{J}/\text{kg}\cdot °\text{C}\right)\left(T_{\text{f}}-2450°\text{C}\right)\end{array}\right)\\ 4.4087\times {10}^{10}\text{J}=\left(670\text{kg}\right)\left(1170\text{J}/\text{kg}\cdot °\text{C}\right)\left(T_{\text{f}}-2450°\text{C}\right)\\ T_{\text{f}}\approx 5.87\times {10}^4°\text{C}\end{array}$

Conclusion:

Therefore, the value of the final temperature is $5.87\times {10}^4°\text{C}$.