Problem 16 - Chemistry

Question

Calculate \(q,\) and determine whether heat is absorbed or released when a system does work on the surroundings equal to \(64 \mathrm{~J}\) and \(\Delta U=213 \mathrm{~J}\).

Step-by-Step Solution

Answer

The system absorbs 277 J of heat.

Step 1: Understand the First Law of Thermodynamics

The First Law of Thermodynamics is given by the equation \( \Delta U = q + w \), where \( \Delta U \) is the change in internal energy, \( q \) is the heat absorbed by the system, and \( w \) is the work done on the system. In this problem, the system does work on the surroundings, so \( w = -64 \text{ J} \).

Step 2: Rearrange the Formula for Heat

We need to solve for \( q \). Rearrange the formula to obtain \( q = \Delta U - w \).

Step 3: Calculate the Heat \( q \)

Substitute \( \Delta U = 213 \text{ J} \) and \( w = -64 \text{ J} \) into the equation: \[ q = 213 \text{ J} - (-64 \text{ J}) = 213 \text{ J} + 64 \text{ J} = 277 \text{ J} \]

Step 4: Determine If Heat is Absorbed or Released

Because \( q = 277 \text{ J} \) is positive, this indicates that the system absorbs 277 J of heat.

Key Concepts

Internal Energy

Internal energy is a key concept in the First Law of Thermodynamics. It refers to the total energy contained within a system. This energy comes from the particles' kinetic energy due to their motion, as well as potential energy from interactions between particles. Internal energy is represented as \( \Delta U \), indicating the change in energy when a system undergoes a process.

To identify changes in internal energy, we look at how much heat is absorbed or released and how much work is done. In our exercise, the internal energy change is \( \Delta U = 213 \text{ J} \). This value tells us how much the system's total energy has increased or decreased during the process.

Internal energy is essential for understanding how systems exchange energy with their surroundings through heat and work.

Heat Absorption

Heat absorption relates to whether the system absorbs energy from its surroundings in the form of heat. In the context of thermodynamics, heat, symbolized as \( q \), is the energy transfer due to temperature differences. It is important to determine whether heat is absorbed or released to comprehend how the energy within the system is changing.

In the given problem, the task is to calculate \( q \). From the First Law of Thermodynamics, we have \( q + w = \Delta U \) or rearranged \( q = \Delta U - w \). Here, \( \Delta U \) is 213 J, and work \( w \) is negative because the system is doing work on its surroundings, making \( w = -64 \text{ J} \). Thus:

• \( q = 213 \text{ J} - (-64 \text{ J}) \)
• \( q = 213 \text{ J} + 64 \text{ J} \)
• \( q = 277 \text{ J} \)

Since the value of \( q \) is positive, this indicates that the system absorbs 277 J of heat. Understanding whether heat is absorbed helps identify the direction of energy flow between the system and environment.

Work Done by System

Work done by the system is another crucial aspect of energy transfer under the First Law of Thermodynamics. When a system does work on its surroundings, it essentially transfers some of its internal energy outward, usually resulting in an energy decrease within the system. However, if work is done on the system, it means energy is entering.

In the provided exercise, the system does work equal to 64 J on the surroundings. Since the work is done by the system on the surroundings, it is considered as negative work in the formula. This is why \( w = -64 \text{ J} \).

• The negative sign in work done indicates energy leaving the system.
• Positive work would suggest energy given to the system from the surroundings.

Grasping how work involves energy shifts helps determine how a system's total internal energy changes and aids in using the First Law to predict system behavior.