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Internal energy is a key concept when studying thermodynamics. It represents the total energy contained within a system due to the motion and interactions of its particles.
This includes both kinetic energy from the particles' movement and potential energy from their positions or interactions. In the context of our problem, internal energy change (\(\Delta U\)) helps us understand how energy is transferred within a system. When we say a system's internal energy increases by \(200 \mathrm{J}\), it means the total energy associated with the particles inside the system has gone up by this amount.
This change can happen through heat additions or by doing work, which are processes we'll explore further in the next sections.

Work done by a system is an important way it can transfer energy to its surroundings. In thermodynamics, work is often related to changes in volume, such as when a gas expands.
When a gas does work, like expanding and pushing against a piston, it transfers energy outward. In our problem, the system does \(100 \mathrm{J}\) of work. This means the energy is leaving the system compared to when the system gains energy from work done on it.
It's crucial to remember that in the equation \(\Delta U = Q - W\), \(W\) is the work done by the system. Hence, it's subtracted when calculating the internal energy change. This relationship highlights the balance between energy inputs and outputs.

Heat added to a system, denoted by \(Q\), is how energy is transferred based on a temperature difference.
Heat flows naturally from hotter to cooler regions, and when added, it increases the total energy within the system. In our problem setup, since the internal energy increases by \(200 \mathrm{J}\) while the system performs \(100 \mathrm{J}\) of work, we use the first law of thermodynamics to find out how much heat is necessary.
Rearranging the equation \(\Delta U = Q - W\), we solve for \(Q\):

• Given that \(\Delta U = 200 \mathrm{J}\)
• And \(W = 100 \mathrm{J}\),
• Substitute these into the equation: \(200 \mathrm{J} = Q - 100 \mathrm{J}\)
• Rearranging gives: \(Q = 200 \mathrm{J} + 100 \mathrm{J}\)
• Therefore, \(Q = 300 \mathrm{J}\).

This value represents the heat needed to be added to keep the balance in energy changes while accounting for both the increase in internal energy and the work done.