Problem 3 - Pearson Physics

Question

Give the change in thermal energy of a system if (a) \(W=50 \mathrm{~J}, Q=50 \mathrm{~J}\); (b) \(W=-50 \mathrm{~J}, Q=-50 \mathrm{~J}\); or (c) \(W=50 \mathrm{~J}, Q=-50 \mathrm{~J}\).

Step-by-Step Solution

Answer

(a) 0 J; (b) 0 J; (c) -100 J.

Step 1: Understanding the First Law of Thermodynamics

The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system. This is expressed mathematically as \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added, and \( W \) is the work done by the system.

Step 2: Substituting Values for Situation (a)

For part (a), we have \( W = 50 \mathrm{~J} \) and \( Q = 50 \mathrm{~J} \). Substituting these values into the formula \( \Delta U = Q - W \), we get: \( \Delta U = 50 \mathrm{~J} - 50 \mathrm{~J} \).

Step 3: Calculating \( \Delta U \) for Situation (a)

By solving the expression, \( \Delta U = 50 \mathrm{~J} - 50 \mathrm{~J} = 0 \mathrm{~J} \). So, the change in thermal energy is \( 0 \mathrm{~J} \).

Step 4: Substituting Values for Situation (b)

For part (b), \( W = -50 \mathrm{~J} \) and \( Q = -50 \mathrm{~J} \). Substituting these into \( \Delta U = Q - W \), we have: \( \Delta U = -50 \mathrm{~J} - (-50 \mathrm{~J}) \).

Step 5: Calculating \( \Delta U \) for Situation (b)

Simplifying the expression gives us \( \Delta U = -50 \mathrm{~J} + 50 \mathrm{~J} = 0 \mathrm{~J} \). Therefore, the change in thermal energy is \( 0 \mathrm{~J} \).

Step 6: Substituting Values for Situation (c)

In part (c), we have \( W = 50 \mathrm{~J} \) and \( Q = -50 \mathrm{~J} \). Substituting these values into the formula yields \( \Delta U = -50 \mathrm{~J} - 50 \mathrm{~J} \).

Step 7: Calculating \( \Delta U \) for Situation (c)

Performing the arithmetic, we find \( \Delta U = -50 \mathrm{~J} - 50 \mathrm{~J} = -100 \mathrm{~J} \). This indicates that the thermal energy decreases by \( 100 \mathrm{~J} \).

Key Concepts

Change in Internal Energy

The change in internal energy (\( \Delta U \)) of a system is a fundamental part of the First Law of Thermodynamics. Internal energy refers to the total energy contained within a system. This includes energy from all the microscopic components such as kinetic and potential energies of atoms and molecules.

According to the First Law of Thermodynamics, the change in internal energy is calculated by the difference between the heat (\( Q \)) added to the system and the work (\( W \)) done by the system:

• \( \Delta U = Q - W \)

When solving problems, it is important to clearly identify the signs of these quantities:

• Positive \( Q \): Heat is added to the system, increasing its internal energy.
• Negative \( Q \): Heat is removed from the system, decreasing its internal energy.
• Positive \( W \): Work is done by the system, reducing its internal energy.
• Negative \( W \): Work is done on the system, increasing its internal energy.

The example exercises show various scenarios of \( Q \) and \( W \), illustrating how \( \Delta U \) is determined, leading to changes in thermal energy.

Heat Transfer

Heat transfer is a critical concept in understanding the First Law of Thermodynamics. It involves the movement of thermal energy from one object or substance to another because of a temperature difference.

Heat can enter or leave a system in three main ways:

• Conduction: Direct transfer of heat through a substance, where thermal energy is passed from molecule to molecule.
• Convection: Heat transfer through a fluid (gas or liquid) where warmer fluid moves to cooler areas.
• Radiation: Transfer of heat in the form of electromagnetic waves, such as heat from the sun.

In thermodynamics problems like our exercise, heat transfer is often quantified and denoted by \( Q \). If heat enters the system, \( Q \) is positive, indicating an increase in internal energy. Conversely, if heat exits the system, \( Q \) is negative, indicating a decrease in internal energy. This changes the overall energy balance, dictating the change in the system's total internal energy.

Work Done by the System

Work done by the system is another key aspect of the First Law of Thermodynamics. In physics, work is defined as energy transfer that occurs when a force is applied over a distance.

In a thermodynamic system, work (\( W \)) can be positive or negative:

• Positive \( W \): Work is done by the system on its surroundings, like when a gas expands against external pressure. This results in a decrease in the internal energy of the system.
• Negative \( W \): Work is done on the system by the surroundings, such as compressing a gas. This means energy is being added to the system's internal energy.

In our example, we worked through scenarios where work performed various roles in the energy transfer process. The sign and magnitude of \( W \) significantly affect the outcome of the internal energy change, as seen in the exercises provided.