Question

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{kJ}, w=+88 \mathrm{kJ}\) b. \(q=+82 \mathrm{kJ}, w=-47 \mathrm{kJ}\) c. \(q=+47 \mathrm{kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

Short answer

In summary, the change in internal energy for each part of the exercise is: a. \(\Delta E = 41\,\text{kJ}\) b. \(\Delta E = 35\,\text{kJ}\) c. \(\Delta E = 47\,\text{kJ}\) The surroundings do work on the system in part a only, as the work value is positive.

Step-by-step solution

a. Calculate ∆E for q = -47 kJ and w = +88 kJ.

First we plug the given values into the equation: \[\Delta E = (-47\,\text{kJ}) + (+88\,\text{kJ})\] Now, sum the values: \[\Delta E = 41\,\text{kJ}\] So, the change in internal energy for part a is 41 kJ.

b. Calculate ∆E for q = +82 kJ and w = -47 kJ.

Plug the given values into the equation: \[\Delta E = (+82\,\text{kJ}) + (-47\,\text{kJ})\] Now, sum the values: \[\Delta E = 35\,\text{kJ}\] So, the change in internal energy for part b is 35 kJ.

c. Calculate ∆E for q = +47 kJ and w = 0.

Plug the given values into the equation: \[\Delta E = (+47\,\text{kJ}) + (0)\] In this case, the value of work is zero, which means there is no work done. Therefore, the change in internal energy is equal to the amount of heat transferred: \[\Delta E = 47\,\text{kJ}\] So, the change in internal energy for part c is 47 kJ.

d. Determine in which case(s) the surroundings do work on the system.

The surroundings do work on the system when work (w) is positive. This is because a positive work value means that energy is being added to the system through work. Looking back at the given values, we find that: - In part a, w = +88 kJ (positive work) - In part b, w = -47 kJ (negative work) - In part c, w = 0 (no work performed) Therefore, the surroundings do work on the system in part a.

Key concepts

Thermodynamics

Thermodynamics is a branch of physics that deals with the relationships between heat, work, temperature, and energy. The fundamental principles of thermodynamics are encapsulated in four laws, which explain how energy is transferred between objects and transformed from one form to another. In the context of the exercise provided, the change in internal energy, denoted by \(\Delta E\), reflects a fundamental concept of the first law of thermodynamics, which states that energy cannot be created or destroyed, only transferred or converted.

The exercise exemplifies this by asking us to calculate \(\Delta E\) under different conditions of heat (\(q\)) and work (\(w\)). When the system gains or loses energy in the form of heat or work, its internal energy changes. If heat is added to the system (\(q > 0\)), the internal energy increases. Conversely, if the system loses heat (\(q < 0\)), the internal energy decreases. The same goes for work done on or by the system; when work is done on the system (\(w > 0\)), energy is added and internal energy increases. If the system does work (\(w < 0\)), it loses energy, and the internal energy decreases.

Heat Transfer

Heat transfer is the process by which heat energy moves from one place or material to another. This transfer can occur in three primary ways: conduction, convection, and radiation. In thermodynamics, when we consider systems at the molecular level, heat transfer is a vital concept as it affects the internal energy of a system.

In the exercise, \(q\) represents the heat transferred to or from the system. A positive \(q\) indicates that heat is absorbed by the system, whereas a negative \(q\) suggests that the system has released heat to its surroundings. The process of calculating \(\Delta E\) inherently involves considering how the heat transfer interacts with the work done, showing their interconnected nature. When no work is done (\(w = 0\)), such as in part c, the change in internal energy is entirely due to the heat transfer to the system.

Work Done by/on System

The work done by or on a system is an essential concept in thermodynamics, relating to energy transfer through force application over a distance. In practical terms, when a force acts upon a system to move its boundary, work is performed on the system, and its internal energy increases. Conversely, if the system applies a force to move its boundary outward, it's doing work on its surroundings, which results in a decrease in internal energy.

Within the exercise, \(w\) is used to denote the work done. As the solution indicates, positive work (\(w > 0\)) as in part a, means the surroundings do work on the system, increasing its internal energy. Negative work (\(w < 0\)), shown in part b, means the system does work on its surroundings and loses internal energy. Understanding this interplay is crucial for solving problems related to mechanical work in thermodynamic processes.