Problem 49 - Pearson Physics

Question

Find the heat associated with each of the following processes: (a) \(W=50 \mathrm{~J}, \Delta E=50 \mathrm{~J}\); (b) \(W=-50 \mathrm{~J}\), \(\Delta E=-50 \mathrm{~J}\); (c) \(W=50 \mathrm{~J}, \Delta E=150 \mathrm{~J}\).

Step-by-Step Solution

Answer

(a) 100 J, (b) -100 J, (c) 200 J.

Step 1: Recall the First Law of Thermodynamics

The First Law of Thermodynamics is expressed by the equation \( \Delta E = Q - W \), where \( \Delta E \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system.

Step 2: Solve for Heat in Process (a)

For process (a), we have \( W = 50 \, \mathrm{J} \) and \( \Delta E = 50 \, \mathrm{J} \). Substitute these values into the formula: \( 50 = Q - 50 \). Rearranging gives us \( Q = 100 \, \mathrm{J} \).

Step 3: Solve for Heat in Process (b)

For process (b), \( W = -50 \, \mathrm{J} \) and \( \Delta E = -50 \, \mathrm{J} \). Substitute into the formula, \( -50 = Q + 50 \). Rearranging, \( Q = -100 \, \mathrm{J} \).

Step 4: Solve for Heat in Process (c)

For process (c), \( W = 50 \, \mathrm{J} \) and \( \Delta E = 150 \, \mathrm{J} \). Substituting, we have \( 150 = Q - 50 \), which gives \( Q = 200 \, \mathrm{J} \).

Key Concepts

Internal Energy Change

In thermodynamics, understanding the concept of internal energy change is crucial. Internal energy, typically represented by \( \Delta E \), represents the total energy contained within a system. It encompasses the kinetic and potential energy of all particles within the system. When a system undergoes a change, this energy either increases or decreases, which is referred to as the change in internal energy.

The change in internal energy \( \Delta E \) is influenced by two main factors: the heat added to the system and the work done by the system. According to the First Law of Thermodynamics, this relationship is expressed as:
\[ \Delta E = Q - W \]

• \( Q \) is the heat transferred into or out of the system.
• \( W \) is the work done by the system.

Understanding how these factors interact helps us calculate the internal energy change for any process.

Heat Transfer

Heat transfer is a fundamental concept in thermodynamics. It refers to the movement of thermal energy from one place to another. The First Law of Thermodynamics highlights the role of heat \( Q \) in changing a system's internal energy.

Heat transfer can occur in three ways: conduction, convection, and radiation, but in our equation, we're mainly concerned with the quantity of heat transferred. The sign of \( Q \) is crucial:

• When \( Q \) is positive, heat is added to the system.
• When \( Q \) is negative, the system loses heat.

This transfer is not just about temperature, but about energy. By determining \( Q \), we can understand how energy is exchanged in a thermodynamic process, whether the system absorbs energy from its surroundings or releases it.

Work Done by System

The concept of work done by a system is integral to the First Law of Thermodynamics. Work, represented by \( W \), is the energy exerted by a system when it moves or changes state. It's a form of energy transfer resulting from any force applied over a distance or through volume change in a system.

The sign and magnitude of \( W \) are important in calculations:

• When \( W \) is positive, the system does work on its surroundings, implying energy leaves the system.
• When \( W \) is negative, work is done on the system by its surroundings, meaning energy is added to the system.

This dual perspective is crucial since the work done affects the system's internal energy and how we interpret changes according to the equation: \[ \Delta E = Q - W \] When studying thermodynamics, comprehending work and its effect within the context of this equation helps us explain energy transformations accurately.