Question

An analytical laboratory balance typically measures mass to the nearest \(0.1 \mathrm{mg}\). What energy change would accompany the loss of \(0.1 \mathrm{mg}\) in mass?

Short answer

The energy change accompanying the loss of 0.1 mg in mass can be calculated using Einstein's mass-energy equivalence formula (E = mc^2). First, convert the mass to kilograms: \(0.1\ mg = 0.1 × 10^{-6}\ kg\). Then substitute the values of mass and the speed of light into the formula: \(E = (0.1 × 10^{-6}\ kg) \times (3 × 10^8\ m/s)^2\). Finally, evaluate the expression: \(E = 9 × 10^{10}\ Joules\). So, the energy change is \(9 × 10^{10}\ J\).

Step-by-step solution

Convert mass to kilograms

Given that the loss of mass is 0.1 mg, we first need to convert it to kilograms (the SI unit of mass) using the conversion factor: 1 mg = 1 x 10^{-6} kg. So, \[Mass (m) = 0.1\ mg = 0.1 × 10^{-6}\ kg.\]

Use Einstein's mass-energy equivalence formula

Now, we can apply Einstein's mass-energy equivalence formula (E = mc^2) to find the energy change associated with the loss of mass. The speed of light (c) is approximately 3 x 10^8 m/s. The formula is: \[E = m \times c^2\]

Calculate the energy change

Substitute the values of mass (m) and speed of light (c) into the formula: \[E = (0.1 × 10^{-6}\ kg) \times (3 × 10^8\ m/s)^2\]

Evaluate the expression

Now, let's calculate the value of E: \[E = (0.1 × 10^{-6}\ kg) \times (9 × 10^{16}\ m^2/s^2)\] \[E = 9 × 10^{10}\ kg\cdot m^2/s^2\]

Express the energy change in Joules

Since the energy (E) is in kg∙m^2/s^2, we can also express it in Joules (J), as 1 Joule = 1 kg∙m^2/s^2. \[E = 9 × 10^{10}\ J\] The energy change that would accompany the loss of 0.1 mg in mass is \(9 × 10^{10}\ Joules\).

Key concepts

Einstein's Mass-Energy Equivalence Equation

One of the most famous equations in physics, formulated by Albert Einstein, is the mass-energy equivalence equation: \( E = mc^2 \). This equation reveals that mass and energy are interchangeable. In simpler terms, any change in mass can be converted into an equivalent energy change and vice versa. Here, \( E \) stands for energy, \( m \) represents mass, and \( c \) is the speed of light in a vacuum, approximately \( 3 \, \times \, 10^8 \, \text{m/s} \).
This groundbreaking principle implies that even a small amount of mass can convert into enormous energy due to the large value of \( c^2 \). It's the foundational concept behind nuclear energy and explains phenomena in both astrophysics and particle physics.
Using this equation, we are able to calculate how much energy is released or absorbed when a certain amount of mass is "lost" or "gained." It highlights the direct connection between the mass of an object and its energy content.

Unit Conversion

In scientific calculations, it is crucial to use the correct units to ensure accuracy. In the problem, we have to convert the mass from milligrams (mg) to kilograms (kg) because the International System of Units (SI) standardizes mass in kilograms.
Unit conversion involves multiplying by factors that equal one, allowing you to change the units without changing the actual value. Knowing that \( 1 \ \text{mg} = 1 \times 10^{-6} \ \text{kg} \), a conversion from milligrams to kilograms simply requires multiplying the mg value by \( 10^{-6} \).
For this exercise, \( 0.1 \ \text{mg} \) is converted like this:

• First, remember the conversion factor: \( 1 \ \text{mg} = 1 \times 10^{-6} \ \text{kg} \).
• Multiply: \( 0.1 \ \text{mg} \times 10^{-6} = 0.1 \times 10^{-6} \ \text{kg} \).

This step ensures that the mass is in the correct unit to be used in Einstein’s equation.

Analytical Balance

An analytical balance is a highly precise instrument used in laboratories to measure mass. It can detect minute changes with high accuracy, often as small as \( 0.1 \ \text{mg} \) or even smaller. This high sensitivity makes it an indispensable tool in chemical, pharmaceutical, and many other scientific experiments.
The functionality of an analytical balance relies on balancing the gravitational force of an object with an opposing force, allowing precise readings. It is crucial for experiments that require a high degree of precision and accuracy, ensuring that measurements of very small masses are reliable.
In this particular exercise, the balance measured a small mass loss of \( 0.1 \ \text{mg} \). Despite its seeming insignificance in day-to-day measurements, such precision can be critical in scientific deductions, particularly those involving Einstein’s equation.

Energy Calculation

Energy calculation in this context refers to using the mass-energy equivalence principle to determine the energy associated with the loss of a certain mass.
We use \( E = mc^2 \) to determine how much energy corresponds to the lost mass. Here’s a step-by-step of the calculation:

• Substitute the mass in kilograms: \( m = 0.1 \times 10^{-6} \ \text{kg} \).
• Use the speed of light \( c = 3 \times 10^8 \ \text{m/s} \).
• Apply the formula: \( E = (0.1 \times 10^{-6}) \times (3 \times 10^8)^2 \).
• Calculate: \( E = 9 \times 10^{10} \ \text{J} \).

This calculation reveals how energy significantly increases due to the speed of light squared factor, which emphasizes why even a tiny mass can equate to a massive amount of energy.