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Chapter 20: Problem 61

A 670-kg meteorite happens to be composed of aluminum. When it is far from the Earth, its temperature is \(-15^{\circ} \mathrm{C}\) and it moves with a speed of \(14.0 \mathrm{km} / \mathrm{s}\) relative to the Earth. As it crashes into the planet, assume that the resulting additional internal energy is shared equally between the meteor and the planet, and that all of the material of the meteor rises momentarily to the same final temperature. Find this temperature. Assume that the specific heat of liquid and of gaseous aluminum is \(1170 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}.\)

Short Answer

Expert verified
The final temperature of the meteorite is \(140000^{\circ} \mathrm{C}\).

Step by step solution

01

Write down what is given in the problem

1. Mass of the meteorite \(m = 670kg\), 2. Initial temperature of the meteorite \(T_{initial} = -15^{\circ} \mathrm{C}\), 3. Velocity of the meteorite \(v = 14.0 \mathrm{km} / \mathrm{s} = 14000 \mathrm{m}/\mathrm{s}\) (converted from km/s to m/s), 4. The specific heat capacity of aluminum \(c = 1170 \mathrm{J} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\)
02

Apply Conservation of Energy

The total energy of the meteorite before it crashes into the Earth equals the total energy after the crash. Before the crash, the meteorite has kinetic energy and some amount of thermal energy. After the crash, all energy transforms into thermal energy. Mathematically, it can be depicted as: \[\frac{1}{2} mv^2 + mc(T_{initial} - T_{0}) = mc(T_{final} - T_{0})\] where \(T_{0}\) is the reference temperature, typically considered to be \(0^{\circ} \mathrm{C}\).
03

Solve for the Unknown

Solving the above equation for the unknown \(T_{final}\), we get \[T_{final} = \frac{1}{2}\frac{v^2}{c} + (T_{initial} - T_{0})\] Plugging in the given values: \[T_{final} = \frac{1}{2}\frac{(14000)^2}{1170} - (15 - 0) = 140000^{\circ} \mathrm{C}\]
04

Check the Solution

The final temperature is positive and measured in degrees Celsius, aligning with our expectation in terms of the units and the system. Thus, the solution can be considered valid even though it seems to be quite high, suitable for a meteorite re-entry scenario.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In physics, it is calculated using the formula: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the object's mass in kilograms and \( v \) is its velocity in meters per second.
For the meteorite in our example, with a mass of 670 kg and a speed of 14,000 m/s, this kinetic energy becomes incredibly large.
  • Higher mass or speed results in greater kinetic energy.
  • In our scenario, the kinetic energy is transformed into thermal energy upon collision.

This transformation explains why the meteorite heats up significantly upon impact.
Specific Heat Capacity
The specific heat capacity of a material tells us how much energy is needed to change the temperature of a certain mass of the substance by one degree Celsius. For aluminum, the specific heat capacity is given as \( 1170 \ \mathrm{J} / \mathrm{kg}^{\circ}\mathrm{C} \).
In simple terms, it means:
  • To raise 1 kg of aluminum by 1 degree Celsius, 1170 Joules of energy is required.
  • Specific heat capacity varies across materials, influencing how they heat or cool.
Understanding specific heat capacity helps us assess how much a substance will heat up given an energy input, essential in solving this exercise. It aids in determining the final temperature of the meteorite post-collision.
Temperature Change
Temperature change is the difference between the initial and final temperatures of a substance. We often consider how different an object's end temperature is due to energy transfer.
In our problem, initially, the meteorite was at \(-15^{\circ} \mathrm{C}\) and absorbed heat until it hit a new high temperature calculated to be a startling \(140000^{\circ} \mathrm{C}\).
The calorie change is primarily due to the conversion of kinetic energy into thermal energy during the collision.
  • The formula for calculating temperature change is: \[ \Delta T = T_{\text{final}} - T_{\text{initial}} \]
  • In cases with massive energy influx like this meteorite collision, temperature changes can be extreme.
It illustrates the capacity of materials to undergo drastic thermal shifts when considerable energy is involved.
Meteorite Collision
Meteorite collisions are fascinating events where celestial objects enter Earth鈥檚 atmosphere, converting their kinetic energy into thermal energy. Upon impact, these bodies experience rapid heating, usually leading to vaporization or melting.
Several phenomena occur during such collisions:
  • The atmosphere generates friction, slowing the meteorite, which converts energy to heat.
  • Upon hitting Earth's surface, the remaining energy causes a sudden rise in temperature.
This drastic energy transformation during meteorite collisions often results in high temperatures, as seen in our exercise, where the meteorite's energy levels led to it reaching temperatures far beyond normal temperatures on Earth.

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Most popular questions from this chapter

The intensity of solar radiation reaching the top of the Earth's atmosphere is \(1340 \mathrm{W} / \mathrm{m}^{2} .\) The temperature of the Earth is affected by the so-called greenhouse effect of the atmosphere. That effect makes our planet's emissivity for visible light higher than its emissivity for infrared light. For comparison, consider a spherical object with no atmosphere, at the same distance from the Sun as the Earth. Assume that its emissivity is the same for all kinds of electromagnetic waves and that its temperature is uniform over its surface. Identify the projected area over which it absorbs sunlight and the surface area over which it radiates. Compute its equilibrium temperature. Chilly, isn't it? Your calculation applies to (a) the average temperature of the Moon, (b) astronauts in mortal danger aboard the crippled Apollo 13 spacecraft, and (c) global catastrophe on the Earth if widespread fires should cause a layer of soot to accumulate throughout the upper atmosphere, so that most of the radiation from the Sun were absorbed there rather than at the surface below the atmosphere.

Water in an electric teakettle is boiling. The power absorbed by the water is \(1.00 \mathrm{kW}\). Assuming that the pressure of vapor in the kettle equals atmospheric pressure, determine the speed of effusion of vapor from the kettle's spout, if the spout has a cross-sectional area of \(2.00 \mathrm{cm}^{2}.\)

A \(3.00-\mathrm{g}\) copper penny at \(25.0^{\circ} \mathrm{C}\) drops \(50.0 \mathrm{m}\) to the ground. (a) Assuming that \(60.0 \%\) of the change in potential energy of the penny-Earth system goes into increasing the internal energy of the penny, determine its final temperature. (b) What If? Does the result depend on the mass of the penny? Explain.

(a) In air at \(0^{\circ} \mathrm{C},\) a \(1.60-\mathrm{kg}\) copper block at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of ice at \(0^{\circ} \mathrm{C} .\) Friction brings the block to rest. Find the mass of the ice that melts. To describe the process of slowing down, identify the energy input \(Q,\) the work input \(W,\) the change in internal energy \(\Delta E_{\text {int }},\) and the change in mechanical energy \(\Delta K\) for the block and also for the ice. (b) A \(1.60-\mathrm{kg}\) block of ice at \(0^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over a sheet of copper at \(0^{\circ} \mathrm{C}\) Friction brings the block to rest. Find the mass of the ice that melts. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for the block and for the metal sheet during the process. (c) A thin 1.60-kg slab of copper at \(20^{\circ} \mathrm{C}\) is set sliding at \(2.50 \mathrm{m} / \mathrm{s}\) over an identical stationary slab at the same temperature. Friction quickly stops the motion. If no energy is lost to the environment by heat, find the change in temperature of both objects. Identify \(Q, W, \Delta E_{\text {int }},\) and \(\Delta K\) for each object during the process.

How much work is done on the steam when 1.00 mol of water at \(100^{\circ} \mathrm{C}\) boils and becomes \(1.00 \mathrm{mol}\) of steam at \(100^{\circ} \mathrm{C}\) at 1.00 atm pressure? Assuming the steam to behave as an ideal gas, determine the change in internal energy of the material as it vaporizes.
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