Question

There are other forms of work besides \(\mathrm{P}-\mathrm{V}\) work. For example, electrical work is defined as the potential \(x\) change in charge, \(w=\phi d q\). If a charge in a system is changed from \(10 \mathrm{C}\) to \(5 \mathrm{C}\) in a potential of \(100 \mathrm{V}\) and \(45 \mathrm{J}\) of heat is liberated, what is the change in the internal energy? (Note: \(1 \mathrm{V}=1 \mathrm{J} / \mathrm{C})\).

Short answer

The change in internal energy is -455J.

Step-by-step solution

Calculate the work done by the system

The work is calculated using \(W=\phi dq\). Here, \(\phi\) is 100V and \(dq\) is the change in charge, which is \(10C - 5C = 5C\). So, the work \(W = 100V * 5C = 500J\). According to the convention, work done by the system is positive.

Calculate the change in internal energy

Applying the first law of thermodynamics, \(\Delta U=Q-W\). Here, \(Q = 45J\) (the heat liberated) and \(W = 500J\) (the work done by the system). So, \(\Delta U = 45J - 500J = -455J\). The change in internal energy is thus -455J.

Key concepts

Internal Energy

Internal energy is a fundamental concept in thermodynamics and refers to the total energy contained within a system. It includes all the kinetic and potential energies of the particles in the system.
This energy can change due to various forms of work or heat transfer.
In the exercise, the change in internal energy, \( \Delta U \), is calculated using the First Law of Thermodynamics, which states:

• First Law of Thermodynamics: \( \Delta U = Q - W \)

Here,

• \(Q\) represents the heat added to the system,
• \(W\) is the work done by the system.

These two factors can increase or decrease the internal energy depending on their values, leading to an understanding of how energy flows and transforms within a system.

Electrical Work

Electrical work is a type of work that involves the movement of electric charge within a potential field.
In the given exercise, this concept is crucial to understanding how work is done when charge moves in an electrical circuit.The formula for electrical work is given by:

• \( W = \phi \cdot dq \)

• \(\phi\) is the potential (voltage),
• \(dq\) is the change in charge.

In the exercise, the potential is 100V, and the change in charge is from 10C to 5C, resulting in 5C.
Thus, the work done \(W\) is \(100V \cdot 5C = 500J \).This illustrates how electrical work can change a system's internal energy by moving charges through a potential difference.

Heat Transfer

Heat transfer refers to the movement of thermal energy from one part of a system to another or between a system and its surroundings.
It occurs due to the temperature difference and is measured in Joules.In the problem, heat transfer is given as 45J, denoting energy liberated by the system.
When considering the First Law of Thermodynamics, heat transfer (

• \(Q\), in Joules) contributes to the calculation of the internal energy change.

In the formula \( \Delta U = Q - W \):

• \(Q\) is the heat added or liberated,
• leading to \( \Delta U = 45J - 500J = -455J \).

Understanding heat transfer allows students to recognize how thermal energy impacts the internal energy of a system, reflecting the intricate balance between heat and work.