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Chapter 17: Problem 56

An asteroid with a diameter of 10 \(\mathrm{km}\) and a mass of \(2.60 \times 10^{15} \mathrm{kg}\) impacts the earth at a speed of 32.0 \(\mathrm{km} / \mathrm{s}\) , landing in the Pacific Ocean. If 1.00\(\%\) of the asteroid's kinetic energy goes to boiling the ocean water (assume an initial water temperature of \(10.0^{\circ} \mathrm{C} )\) , what mass of water will be boiled away by the collision? (For comparison, the mass of water contained in Lake Superior is about \(2 \times 10^{15} \mathrm{kg}\) .)

Short Answer

Expert verified
Approximately \(1.11 \times 10^{12} \text{ kg}\) of water is boiled away.

Step by step solution

01

Calculate the Kinetic Energy of the Asteroid

First, we need to find the asteroid's kinetic energy using the formula for kinetic energy \(KE = \frac{1}{2}mv^2\) where \(m\) is mass and \(v\) is velocity. In this case, \(m = 2.60 \times 10^{15} \text{ kg}\) and \(v = 32,000 \text{ m/s} \). Substitute these values into the equation:\[KE = \frac{1}{2} \times 2.60 \times 10^{15} \times (32 \times 10^3)^2\]Solving this gives the kinetic energy in joules.
02

Calculate the Energy Used to Boil Water

We know that only 1.00\% of the asteroid's kinetic energy will actually be used to boil the water. To calculate this energy, multiply the kinetic energy by 0.01:\[E_{boil} = 0.01 \times KE\]This formula calculates how much of the kinetic energy is dedicated to boiling the water.
03

Calculate the Heat Needed to Boil Water

The total energy needed to boil a certain mass of water consists of the energy required to raise the temperature of the water to its boiling point and then the heat of vaporization to convert it to steam. The equation for heating water is:\[Q = mc\Delta T + mL\]where \(c\) is the specific heat of water (\(4,186 \text{ J/kg}\cdot \text{掳C}\)), \(\Delta T\) is the temperature change (\(90 \text{掳C}\) as water boils at \(100 \text{掳C}\)), \(m\) is the mass of the water, and \(L\) is the latent heat of vaporization (\(2,256,000 \text{ J/kg}\)). We'll use this to find the mass \(m\) that can be vaporized by \(E_{boil}\).
04

Derive the Mass of Water Boiled Off

Re-arrange the equation from Step 3 to solve for \(m\):\[E_{boil} = mc\Delta T + mL \m = \frac{E_{boil}}{c\Delta T + L}\]Substitute \(E_{boil}\), \(c = 4,186 \text{ J/kg}\cdot \text{掳C}\), \(\Delta T = 90 \text{掳C}\), and \(L = 2,256,000 \text{ J/kg}\) to find the mass of water that is boiled off.

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Key Concepts

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

Kinetic Energy Calculation
Understanding how to calculate kinetic energy is crucial when analyzing the impact of an asteroid. Kinetic energy (\( KE \)) is the energy an object possesses due to its motion. The formula used to determine this energy is:
\[KE = \frac{1}{2}mv^2\]where:
  • \( m \) is the mass of the object in kilograms (kg).
  • \( v \) is the velocity of the object in meters per second (m/s).
For the asteroid in question, with a mass of **2.60 x 10^15** kg and a speed of **32,000** m/s, the kinetic energy can be calculated by substituting these values into the formula. This calculation reflects the staggering amount of energy the asteroid had upon impact. Comprehending this energy helps us quantify the potential effects of such an event, like the energy transferred to ocean water.
Boiling Water Process
To understand the boiling water process, we need to know how energy transforms water from a liquid to a gas. This transformation requires both heating the water to its boiling point and then converting it to steam. The formula that encompasses this process is:
\[Q = mc\Delta T + mL\]where:
  • \( Q \) is the total heat energy required.
  • \( m \) is the mass of the water.
  • \( c \) is the specific heat of water, which is **4,186 J/kg掳C**.
  • \( \Delta T \) is the change in temperature needed to reach boiling, which here is **90掳C** (from **10掳C** to **100掳C**).
  • \( L \) is the latent heat of vaporization, **2,256,000 J/kg**, representing the energy required to convert water at boiling point into steam without changing the temperature.
By applying these values, we can determine the energy needed to convert a specific mass of water into steam during any process, from cooking to going through an asteroid impact.
Heat of Vaporization
The heat of vaporization is the amount of heat energy required to convert a substance from a liquid into a vapor without changing its temperature. For water, this occurs at boiling point, and the latent heat of vaporization is a critical value when calculating energy needs in processes involving phase changes.
The latent heat of vaporization for water is **2,256,000 J/kg**. This means that to vaporize 1 kilogram of water, **2,256,000** joules of energy are needed, even if the temperature stays the same.
In the context of the asteroid impact, only **1%** of the asteroid's kinetic energy was used in vaporizing ocean water. Understanding how energy is divided and used can show us how much water could potentially be boiled away, helping to characterize the scale of transformation an event like this would cause. Calculations using the heat of vaporization underscore the immense energy applied, demonstrating nature's extraordinary power.

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Most popular questions from this chapter

A vessel whose walls are thermally insulated contains 2.40 \(\mathrm{kg}\) of water and 0.450 \(\mathrm{kg}\) of ice, all at a temperature of \(0.0^{\circ} \mathrm{C}\) The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to \(28.0^{\circ} \mathrm{C}\) ? You can ignore the heat transferred to the container.

The tallest building in the world, according to some architectural standards, is the Taipei 101 in Taiwan, at a height of 1671 feet. Assume that this height was measured on a cool spring day when the temperature was \(15.5^{\circ} \mathrm{C}\) . You could use the building as a sort of giant thermometer on a hot summer day by carefully measuring its height. Suppose you do this and discover that the Taipei 101 is 0.471 foot taller than its official height. What is the temperature, assuming that the building is in thermal equilibrium with the air and that its entire frame is made of steel?

What is the rate of energy radiation per unit area of a black-body at a temperature of (a) 273 \(\mathrm{K}\) and \((\mathrm{b}) 2730 \mathrm{K} ?\)

A refrigerator door is opened and room-temperature air \(\left(20.0^{\circ} \mathrm{C}\right)\) fills the \(1.50-\mathrm{m}^{3}\) compartment. A \(10.0-\mathrm{kg}\) turkey, also at room temperature, is placed in the refrigerator and the door is closed. The density of air is 1.20 \(\mathrm{kg} / \mathrm{m}^{3}\) and its specific heat is 1020 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . Assume the specific heat of a turkey, like that of a human, is 3480 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) . How much heat must the refrigerator remove from its compartment to bring the air and the turkey to thermal equilibrium at a temperature of \(5.00^{\circ} \mathrm{C}\) ? Assume no heat exchange with the surrounding environment.

Effect of a Window in a Door. A carpenter builds a solid wood door with dimensions \(2.00 \mathrm{m} \times 0.95 \mathrm{m} \times 5.0 \mathrm{cm} .\) Its thermal conductivity is \(k=0.120 \mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the inner and outer surfaces of the door have the same combined thermal resistance as an additional \(1.8-\mathrm{cm}\) thickness of solid wood. The inside air temperature is \(20.0^{\circ} \mathrm{C}\) , and the outside air temperature is \(-8.0^{\circ} \mathrm{C} .\) (a) What is the rate of heat flow through the door? (b) By what factor is the heat flow increased if a window 0.500 \(\mathrm{m}\) on a side is inserted in the door? The glass is 0.450 \(\mathrm{cm}\) thick, and the glass has a thermal conductivity of 0.80 \(\mathrm{W} / \mathrm{m} \cdot \mathrm{K}\) . The air films on the two sides of the glass have a total thermal resistance that is the same as an additional 12.0 \(\mathrm{cm}\) of glass.
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