Problem 35 - College Physics

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

Answer

The mass of water boiled away is calculated from the asteroid's kinetic energy and water's heat properties.

Step 1: Calculate the Asteroid's Kinetic Energy

The kinetic energy (KE) of an object can be calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. Substituting the given values, we get: \[ KE = \frac{1}{2} \times 2.60 \times 10^{15} \text{ kg} \times (32.0 \times 10^3 \text{ m/s})^2 = 1.33 \times 10^{23} \text{ J} \] This is the total kinetic energy of the asteroid upon impact.

Step 2: Calculate Energy Used to Boil Water

Only 1.00\( \% \) of the asteroid's kinetic energy is used to boil the water. Therefore, the energy used to boil the water is: \[ 1.33 \times 10^{23} \text{ J} \times 0.01 = 1.33 \times 10^{21} \text{ J} \] This is the energy used for boiling water.

Step 3: Calculate Energy Required to Heat and Boil Water

To find the mass of water boiled, we need to calculate the energy required to first heat the water from \(10.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C}\) and then to boil it away. The energy required to heat the water is: \( Q_{\text{heat}} = m c \Delta T \), where \( c = 4186 \text{ J/(kg}\cdot\text{C)} \) is the specific heat, and \( \Delta T = 90^{\circ} \mathrm{C} \). The energy required to vaporize the water is: \( Q_{\text{vaporize}} = m L \), where \( L = 2.26 \times 10^6 \text{ J/kg} \) is the latent heat of vaporization. The total energy is: \[ Q_{\text{total}} = m (c \Delta T + L) = m (4186 \times 90 + 2.26 \times 10^6) \text{ J/kg} \]

Key Concepts

Asteroid Impact

When we talk about an asteroid impact, it's all about understanding the intense energy involved. Imagine an asteroid with a massive size striking Earth at an immense speed. The kinetic energy released is breathtaking. The formula to grasp this is \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass of the asteroid, and \( v \) is its velocity. This gives us an idea of how much energy is generated upon impact.
For example, let's consider an asteroid that weighs \(2.60 \times 10^{15} \ \text{kg} \) and travels at \(32,000 \ \text{m/s} \). Such an impact would release \(1.33 \times 10^{23} \ \text{J} \) of energy. That's a phenomenal amount, enough to cause significant effects on Earth upon collision.
Rather than harming directly, some of this energy can do things like vaporizing water, as tiny as a thousandth of the total energy used for boiling in this scenario. Understanding the mechanism behind asteroids can help us assess their potential risks and benefits.

Boiling Water Calculation

Calculating the energy required to boil water might seem straightforward, but it encompasses several steps. First, you need to consider the initial temperature of the water. Then, estimate how much energy is needed to heat it to boiling point, and finally, convert it into steam. This requires two crucial calculations: heating the water and vaporizing it.
In our scenario, we're working with ocean water at \(10^\circ C \). To begin, we heat it up to \(100^\circ C \). For this, we use the equation \( Q_{\text{heat}} = mc\Delta T \), where \( m \) is the mass, \( c \) is the specific heat (\(4186 \ \text{J/(kg} \cdot \text{C)}\)), and \(\Delta T \) is the change in temperature.

• The change in temperature \(\Delta T\) = \(100^\circ - 10^\circ = 90^\circ C\).
• This means \( Q_{\text{heat}} = m \times 4186 \times 90 \ \text{J} \).

Next, to create steam, we use \( Q_{\text{vaporize}} = mL \), where \( L \) is the latent heat of vaporization \( (2.26 \times 10^6 \ \text{J/kg}) \). Putting it all together, the energy calculation becomes insightful for understanding the boiling process.

Specific Heat

Specific heat is a fascinating concept to master. It is the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius. Water has a high specific heat, which is \(4186 \ \text{J/(kg} \cdot \text{C)}\). This implies that water can absorb a great amount of heat without a significant temperature change.
Why is understanding specific heat important? Well, specific heat offers insights into the energy required for heating water, which is critical in thermal calculations. For example, heating 1 kg of water by \(1^\circ C\) requires \(4186 \ \text{J}\). If larger masses or larger temperature changes are involved, you simply multiply: \( Q = mc\Delta T \).
This property helps explain why oceans are such good heat reservoirs and play a crucial role in climate regulation. In our asteroid impact problem, accounting for water's specific heat helps calculate the exact energy needed to heat it. It connects the dots for understanding how substantial energy transfers impact our environment.

Latent Heat of Vaporization

Latent heat of vaporization is a key concept when transitioning water into gas form, such as steam. It's the energy required to transform a kilogram of water at its boiling point into steam without changing its temperature. For water, this is \(2.26 \times 10^6 \ \text{J/kg}\).
This concept is vital in processes like boiling, condensation, and understanding weather patterns. In the context of the asteroid impact, it helps determine how much energy is needed to vaporize water after it's been heated to the boiling point. After heating, the remaining energy transforms the liquid into gas

• Vaporization requires energy input described by \( Q_{\text{vaporize}} = mL \).
• The total energy to boil and vaporize combines heating and vaporization: \( Q_{\text{total}} = m(c\Delta T + L) \).

This gives us a comprehensive set of tools to understand thermal energy transformations and the physical state changes, particularly under intense situations like an asteroid strike.