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, \(\Delta E\), is \(-21 \mathrm{~kJ}\).

Step-by-step solution

Write down the first law of thermodynamics

The formula for the first law of thermodynamics is: \[\Delta E = q + w\]

Plug in the given values for heat and work

We have \(q = -125 \mathrm{~kJ}\) and \(w = 104 \mathrm{~kJ}\). Plug these values into the formula: \[\Delta E = (-125 \mathrm{~kJ}) + (104 \mathrm{~kJ})\]

Calculate \(\Delta E\)

Add the heat and work values to find the change in internal energy: \[\Delta E = -21 \mathrm{~kJ}\] Therefore, the change in internal energy, \(\Delta E\), is \(-21 \mathrm{~kJ}\).

Key concepts

Change in Internal Energy

Internal energy is often symbolized as 'U' or 'E' and represents the total energy contained within a system, which in the context of thermodynamics typically refers to a gas or liquid. This energy is the sum of the potential and kinetic energy of the particles within the system. A change in internal energy, denoted as \( \Delta E \), occurs when there is an exchange of energy between the system and its surroundings.

The energy transfer can happen in two principal ways: heat transfer and work done on or by the system. These interactions are described by the first law of thermodynamics, which in a basic form can be written as \( \Delta E = q + w \) where \( q \) is the heat exchanged and \( w \) is the work done. It's crucial to note that the sign convention in thermodynamics is that heat absorbed by the system and work done on the system are positive, while heat released by the system and work done by the system are negative.

In the context of the given exercise, the system releases heat (shown by a negative sign) and has work done on it (positive because energy is added to the system), leading to a decrease in the system’s internal energy by \( -21 \text{kj} \). This example illustrates how the internal energy depends not only on heat but also on the work interactions.

Thermodynamics Heat Transfer

Heat transfer in thermodynamics is a fundamental concept representing the flow of energy due to a temperature difference. It's important to remember that heat always moves from a region of higher temperature to one of lower temperature. In thermodynamics, 'q' symbolizes heat transfer, and we assign it a positive value when the system gains heat and a negative value when the system loses heat.

There are three methods of heat transfer: conduction, convection, and radiation.

• Conduction occurs through direct contact between materials.
• Convection involves the movement of a fluid, which can be a gas or liquid.
• Radiation is energy transfer via electromagnetic waves and can occur through space.

However, in a thermodynamic process like the one in our exercise, we focus on the quantity of heat transferred rather than the method. The exercise presented a scenario where the system released heat, indicating that the surrounding environment absorbed that energy. This transfer was quantified as \( -125 \text{kj} \), with the negative sign indicating that energy is leaving the system.

Thermodynamics Work

Work in thermodynamics can be conceptualized as the energy transferred when a force acts over a distance. It is an integral part of energy changes within a system. Symbolized by 'w', work is positive when done on the system—meaning energy is added to the system—and negative when the system does work on the surroundings.

In the context of a thermodynamic system such as a gas in a piston, if the gas expands, it does work on the surroundings; conversely, if the gas is compressed, the surroundings do work on the gas. Work can be calculated using the equation \( w = P \Delta V \), where 'P' represents pressure and \( \Delta V \) represents the change in volume.

For the problem in question, the system had \( 104 \text{kj} \) of work done on it, meaning the surroundings used that amount of energy to compress or act upon the system. The positive value corroborates this transfer, hence increasing the system's internal energy by that amount. When we talk about work in a thermodynamic cycle, we refer to this kind of energy transfer, which, along with heat, affects the internal energy of the system.