Problem 47 - Fundamentals of Physics

Question

A \(5.00\) -grain aspirin tablet has a mass of \(320 \mathrm{mg}\). For how many kilometers would the energy equivalent of this mass power an automobile? Assume \(12.75 \mathrm{~km} / \mathrm{L}\) and a heat of combustion of \(3.65 \times 10^{7} \mathrm{~J} / \mathrm{L}\) for the gasoline used in the automobile.

Step-by-Step Solution

Answer

The aspirin's energy can power a car for about 10 million kilometers.

Step 1: Convert Mass from Milligrams to Kilograms

First, we need to convert the mass of the aspirin tablet from milligrams to kilograms because the energy calculation requires mass in kilograms. We know that 1 kilogram equals 1,000,000 milligrams.\[ \text{Mass in kg} = \frac{320 \text{ mg}}{1,000,000} = 0.00032 \text{ kg} \]

Step 2: Calculate Energy Equivalent using E=mc²

The energy equivalent of a mass can be calculated using the equation \(E=mc^2\), where \(E\) is energy, \(m\) is mass, and \(c\) is the speed of light in a vacuum \( (3 \times 10^8 \text{ m/s}) \).\[ E = 0.00032 \text{ kg} \times (3 \times 10^8 \text{ m/s})^2 \]\[ E = 0.00032 \times 9 \times 10^{16} \]\[ E = 2.88 \times 10^{13} \text{ J} \]

Step 3: Determine the Volume of Gasoline Equivalent to the Energy

Using the given heat of combustion (energy per liter of gasoline), we calculate the amount of gasoline that equates to the aspirin’s energy content.\[ V = \frac{E}{\text{Heat of Combustion}} = \frac{2.88 \times 10^{13} \text{ J}}{3.65 \times 10^{7} \text{ J/L}} \]\[ V \approx 7.89 \times 10^5 \text{ L} \]

Step 4: Calculate the Distance the Energy Can Power the Automobile

Finally, calculate how far this volume of gasoline would power the car, given the car's fuel efficiency.\[ \text{Distance} = V \times \text{Fuel Efficiency} = 7.89 \times 10^5 \text{ L} \times 12.75 \text{ km/L} \]\[ \text{Distance} \approx 1.0 \times 10^7 \text{ km} \]

Key Concepts

E=mc² Formula

The famous equation, \(E=mc^2\), teaches us that mass and energy are interchangeable. This concept, proposed by Albert Einstein, reveals how a small amount of mass can be converted into a huge amount of energy. Here:

• \(E\) stands for energy, measured in Joules (J).
• \(m\) is mass, measured in kilograms (kg).
• \(c\) is the speed of light in a vacuum, approximately \(3 \times 10^8\) meters per second (m/s).

By substituting mass and energy into this equation, we can see that a tiny bit of matter can transform into a large quantity of energy. This principle forms the basis for understanding energy conversion processes at molecular and atomic levels.

Mass-Energy Equivalence

Mass-energy equivalence explains that mass is a form of energy. This principle states that the mass of an object is a direct measure of its energy content. A particle's mass and the energy it contains are directly proportional.When applied to practical scenarios, mass-energy equivalence helps us understand phenomena like nuclear reactions, where substantial amounts of energy are released from a small mass change. This conversion is also vital for scientists and engineers when considering the energy potential stored within various materials for energy generation or storage.Mass-energy equivalence, in simple terms, asserts that energy and mass are two sides of the same coin. They are interchangeable via the speed of light squared \((c^2)\), connecting the two into a unified framework.

Fuel Efficiency

Fuel efficiency is a metric that describes how effectively a vehicle converts fuel into motion. It is usually measured in terms of kilometers per liter (km/L) or miles per gallon (mpg). The more kilometers a vehicle can travel on a liter of fuel, the more fuel-efficient it is considered to be. Factors affecting fuel efficiency include:

• Vehicle design and engine size.
• Driving habits and conditions.
• Fuel type and quality.

In our exercise, the car's fuel efficiency is given as 12.75 km/L, which indicates how far the car can travel per liter of gasoline. Thus, improving fuel efficiency helps reduce fuel consumption, lowers emissions, and saves on fuel costs.

Heat of Combustion

The heat of combustion refers to the energy released when a certain amount of fuel is burned. It is typically measured in Joules per liter (J/L) or Joules per kilogram (J/kg), depending on whether the fuel is liquid or solid.Understanding the heat of combustion is crucial for determining how much energy is available from burning a given amount of fuel.
In our exercise, we are given a heat of combustion of \(3.65 \times 10^7\) J/L for gasoline. This value helps calculate how much gasoline corresponds to the energy produced by the aspirin's mass. The higher the heat of combustion, the more energy a fuel contains per unit volume, allowing better performance and potentially reducing the quantity of fuel needed for the same energy output.