Armageddon! Goal Link mechanical energy to thermal energy, phase changes, and the ideal gas law to create an estimate. Problem A comet half a kilometer in radius consisting of ice at 273 K hits Earth at a speed of 4.00 × 10^4 m/s. For simplicity, assume that all the kinetic energy converts to

Question

Armageddon!

Goal Link mechanical energy to thermal energy, phase changes, and the ideal gas law to create an estimate.

Problem A comet half a kilometer in radius consisting of ice at [latex]273 \mathrm{~K}[/latex] hits Earth at a speed of [latex]4.00 imes 10^{4} \mathrm{~m} / \mathrm{s}[/latex]. For simplicity, assume that all the kinetic energy converts to thermal energy on impact and that all the thermal energy goes into warming the comet. (a) Calculate the volume and mass of the ice. (b) Use conservation of energy to find the final temperature of the comet material. Assume, contrary to fact, that the result is superheated steam and that the usual specific heats are valid, though in fact they depend on both temperature and pressure. (c) Assuming the steam retains a spherical shape and has the same initial volume as the comet, calculate the pressure of the steam using the ideal gas law. This law actually doesn't apply to a system at such high pressure and temperature, but can be used to get an estimate.

Strategy Part (a) requires the volume formula for a sphere and the definition of density. In part (b), conservation of energy can be applied. There are four processes involved: (1) melting the ice, (2) warming the ice water to the boiling point, (3) converting the boiling water to steam, and (4) warming the steam. The energy needed for these processes will be designated [latex]Q_{ ext {melt }}, Q_{ ext {water }}, Q_{ ext {vapor }}[/latex], and [latex]Q_{ ext {steam }}[/latex], respectively. These quantities plus the change in kinetic energy [latex]\Delta K[/latex] sum to zero because they are assumed to be internal to the system. In this case, the first three [latex]Q[/latex] 's can be neglected compared to the (extremely large) kinetic energy term. Solve for the unknown temperature, and substitute it into the ideal gas law in part (c).

Accepted Answer

(a) Find the volume and mass of the ice.

Apply the volume formula for a sphere:

[latex] \begin{aligned} V & =\frac{4}{3} \pi r^{3}=\frac{4}{3}(3.14)\left(5.00 imes 10^{2} \mathrm{~m} ight)^{3} \ & =5.23 imes 10^{8} \mathrm{~m}^{3} \end{aligned} [/latex]

Apply the density formula to find the mass of the ice:

[latex] \begin{aligned} m & = ho V=\left(917 \mathrm{~kg} / \mathrm{m}^{3} ight)\left(5.23 imes 10^{8} \mathrm{~m}^{3} ight) \ & =4.80 imes 10^{11} \mathrm{~kg} \end{aligned} [/latex]

(b) Find the final temperature of the cometary material.

Use conservation of energy:

[latex] \begin{aligned} & Q_{ ext {melt }}+Q_{ ext {water }}+Q_{ ext {vapor }}+Q_{ ext {steam }}+\Delta K=0 & (1)\ & m L_{f}+m c_{ ext {water }} \Delta T_{ ext {water }}+m L_{v}+m c_{ ext {steam }} \Delta T_{ ext {steam }}+\left(0-\frac{1}{2} m v^{2} ight)=0 & (2) \end{aligned} [/latex]

The first three terms are negligible compared to the kinetic energy. The steam term involves the unknown final temperature, so retain only it and the kinetic energy, canceling the mass and solving for [latex]T[/latex] :

[latex] \begin{aligned} & m c_{ ext {steam }}(T-373 \mathrm{~K})-\frac{1}{2} m v^{2}=0 \ & T=\frac{\frac{1}{2} v^{2}}{c_{ ext {steam }}}+373 \mathrm{~K}=\frac{\frac{1}{2}\left(4.00 imes 10^{4} \mathrm{~m} / \mathrm{s} ight)^{2}}{2010 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}}+373 \mathrm{~K} \ & T=3.98 imes 10^{5} \mathrm{~K} \end{aligned} [/latex]

(c) Estimate the pressure of the gas, using the ideal gas law.

First, compute the number of moles of steam:

[latex]n=\left(4.80 imes 10^{11} \mathrm{~kg} ight)\left(\frac{1 \mathrm{~mol}}{0.018 \mathrm{~kg}} ight)=2.67 imes 10^{13} \mathrm{~mol}[/latex]

Solve for the pressure, using [latex]P V=n R T[/latex] :

[latex] \begin{aligned} P & =\frac{n R T}{V} \ & =\frac{\left(2.67 imes 10^{13} \mathrm{~mol} ight)(8.31 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})\left(3.98 imes 10^{5} \mathrm{~K} ight)}{5.23 imes 10^{8} \mathrm{~m}^{3}} \ P & =1.69 imes 10^{11} \mathrm{~Pa} \end{aligned} [/latex]

Remarks The estimated pressure is several hundred times greater than the ultimate shear stress of steel! This highpressure region would expand rapidly, destroying everything within a very large radius. Fires would ignite across a continent-sized region, and tidal waves would wrap around the world, wiping out coastal regions everywhere. The sun would be obscured for at least a decade, and numerous species, possibly including Homo sapiens, would become extinct. Such extinction events are rare, but in the long run represent a significant threat to life on Earth.

Question 11.8: Armageddon! Goal Link mechanical energy to thermal energy, p...

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