Question

\(\Delta U=100 \mathrm{J}\) for a system that gives off \(100 \mathrm{J}\) of heat and (a) does no work; (b) does 200 J of work; (c) has 100 J of work done on it; (d) has 200 J of work done on it.

Short answer

The system's internal energy changes as follows: In part (a), \(\Delta U = 100J\); in part (b), \(\Delta U = -100J\); in part (c), \(\Delta U = 200J\); and finally, in part (d), \(\Delta U = 300J\).

Step-by-step solution

Part (a) Calculations

For part (a), the system does no work. This means that \(W = 0\). Plugging this into the formula, we get \(\Delta U = Q - W = 100J - 0 = 100J\)

Part (b) Calculations

For part (b), the system does 200J of work. So, \(W = 200J\). Substitute this into the formula to get \(\Delta U = Q - W = 100J - 200J = -100J\)

Part (c) Calculations

For part (c), the system has 100J of work done on it. In this case, the work done on the system is negative, hence \(W = -100J\). Substituting this into the equation, we find \(\Delta U = Q - W = 100J - -100J = 200J\)

Part (d) Calculations

In part (d), the system has 200J of work done on it. Here, the work done on the system is also negative, so \(W = -200J\). This gives \(\Delta U = Q - W = 100J - -200J = 300J\)

Key concepts

Internal Energy Change

The concept of Internal Energy Change is fundamental in understanding thermodynamics. It represents the total change in a system's energy state. This change, denoted by \( \Delta U \), can result from the energy the system exchanges with its surroundings. It can occur in the form of heat \( Q \) or work \( W \) done on or by the system.

In our exercise, for instance, the system is said to have a change in internal energy of \( \Delta U = 100 \, \text{J} \). This change is resultant of the heat and work interactions with the environment. When no work is done by or on the system, it simply means all of the energy change is due to heat transfer, leaving the internal energy increased by the heat amount \( Q \).

Work Done by System

When discussing the topic of Work Done by System, it's essentially energy transferred by the system to its surroundings. If the system expands against an external pressure, for example, it does work on the surroundings, which leads to a decrease in the system's internal energy.

In the exercise, when the system does 200J of work, it loses that amount of energy, hence the negative sign in the mathematical expression \( W = 200J \). This action decreases its internal energy, which is why we calculate \( \Delta U \) as being \( -100J \) in part (b). The key is to recognize that work done by the system reduces its internal energy.

Heat Transfer

Heat Transfer is energy in transit due to a temperature difference. Whenever heat is added to a system, its internal energy increases unless the system does work to dissipate the energy. Conversely, if the system loses heat, its internal energy decreases.

In the context of our exercise, the system releases \( 100 \, \text{J} \) of heat. If no work is done, as in part (a), that energy is directly subtracted from the system's internal energy. However, if work is involved, as in parts (b), (c), and (d), we must consider both heat and work to determine the net change in internal energy.

Thermodynamic Processes

Exploring Thermodynamic Processes reveals how systems evolve between different states. These processes involve changes in internal energy (\( \Delta U \)), heat transfer (\( Q \)), and the work done (\( W \)) by or on the system. The first law of thermodynamics is the guiding principle, stating that energy can neither be created nor destroyed, only transformed.

The processes described in our exercise represent different thermodynamic scenarios. In parts (a) and (b), the system evolves without external work, while in (c) and (d), it experiences work done on it, illustrated by the positive change in internal energy. These variations show how energy balance is maintained according to the first law of thermodynamics.