Problem 56
Question
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 \(2 \times 10^{15} \mathrm{kg}\) .)
Step-by-Step Solution
Verified Answer
Approximately \(1.11 \times 10^{12} \text{ kg}\) of water is boiled away.
1Step 1: 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.
2Step 2: 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.
3Step 3: 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 (\(4,186 \text{ J/kg}\cdot \text{°C}\)), \(\Delta T\) is the temperature change (\(90 \text{°C}\) as water boils at \(100 \text{°C}\)), \(m\) 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 \(E_{boil}\).
4Step 4: 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 \(E_{boil}\), \(c = 4,186 \text{ J/kg}\cdot \text{°C}\), \(\Delta T = 90 \text{°C}\), and \(L = 2,256,000 \text{ J/kg}\) to find the mass of water that is boiled off.
Key Concepts
Kinetic Energy CalculationBoiling Water ProcessHeat of Vaporization
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:
\[KE = \frac{1}{2}mv^2\]where:
- \( m \) is the mass of the object in kilograms (kg).
- \( v \) is the velocity of the object in meters per second (m/s).
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:
\[Q = mc\Delta T + mL\]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°C**.
- \( \Delta T \) is the change in temperature needed to reach boiling, which here is **90°C** (from **10°C** to **100°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.
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.
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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