Question

A system releases \(125 \mathrm{~kJ}\) of heat while \(104 \mathrm{~kJ}\) of work is done on it. Calculate \(\Delta E\).

Short answer

The change in internal energy (ΔE) of the system is \(-229 \mathrm{~kJ}\).

Step-by-step solution

Identify the given values

The problem states that the system releases 125 kJ of heat (Q): that means, Q = -125 kJ (since heat is released, we consider it as negative). Furthermore, 104 kJ of work is done on the system (W): this implies, W = +104 kJ (since work is done on the system, we consider it as positive).

Apply the first law of thermodynamics

Now, we will use the first law of thermodynamics to find the change in the internal energy of the system (ΔE). The formula is: ΔE = Q - W.

Substitute the given values

We will plug in the given values of Q and W into the formula: ΔE = (-125 kJ) - (+104 kJ)

Calculate the change in internal energy (ΔE)

Now, perform the calculation: ΔE = -125 kJ - 104 kJ = -229 kJ The change in internal energy (ΔE) of the system is -229 kJ. Since the result is negative, it indicates that the internal energy of the system has decreased.

Key concepts

Internal Energy Change

The concept of internal energy is crucial in understanding how energy shifts within a system. Internal energy change, denoted as \( \Delta E \), refers to the difference in energy within the system between two states.
Internal energy is affected by heat exchange and work done on or by the system.

• When heat is added to the system, \( Q \) is positive, contributing positively to the internal energy.
• Conversely, when heat is released, \( Q \) becomes negative, reducing the internal energy.
• Similarly, if work is performed on the system, \( W \) is positive. If the system performs work, \( W \) is negative.

In our example, the system releases 125 kJ of heat, which means \( Q = -125 \text{ kJ} \). Simultaneously, 104 kJ of work is done on the system, meaning \( W = +104 \text{ kJ} \). Applying these to the first law of thermodynamics, we find that the internal energy change is \( \Delta E = Q - W = -125 \text{ kJ} - 104 \text{ kJ} \), leading to a result of \( -229 \text{ kJ} \). The negative value signifies a decrease in the system's internal energy.

Thermodynamic Calculation

Thermodynamic calculations often revolve around the principles set out by the laws of thermodynamics. The first law of thermodynamics, which is essentially a statement of energy conservation, provides a way to track energy changes.
The equation \( \Delta E = Q - W \) sums up the first law, where:

• \( \Delta E \) is the change in internal energy.
• \( Q \) represents the heat exchanged.
• \( W \) stands for the work done.

In practice, these calculations enable us to determine how much energy is conserved, transferred, or lost in a given process. In our specific exercise, substituting the given values directly into the equation provides a direct avenue to calculate the change in internal energy as \( -229 \text{ kJ} \). Such straightforward calculations illuminate the transformation processes within a system, making thermodynamic analysis accessible and comprehensible.

Work Done on a System

Understanding the nature of work done on a system is fundamental in thermodynamics. Work done on a system results in energy being transferred into the system, thereby impacting its internal energy.

• If \( W > 0 \), the system gains energy, as work is done on it.
• If \( W < 0 \), the system loses energy, as it performs work on its surroundings.

For our scenario, the work done on the system is \( 104 \text{ kJ} \), denoted positively since it adds to the internal energy. The balance of this work along with the heat exchange determines the total change in energy. This concept is essential as it helps in analyzing how systems interact with their environment and how energy inputs can modify a system's energetic state. By understanding these principles, students can predict changes in energy flows and understand the energetic demands of various processes.