Problem 13 - Pearson Physics

Question

Calculate A system's thermal energy decreases by \(20 \mathrm{~J}\) while the system performs \(10 \mathrm{~J}\) of work. How much heat was added to the system?

Step-by-Step Solution

Answer

Heat removed from the system is 10 J.

Step 1: Understanding the Problem

We need to calculate how much heat was added to a system when we know the change in thermal energy and the work done by the system. According to the first law of thermodynamics, the change in internal energy (or thermal energy) of the system \( \Delta U \) is equal to the heat added to the system \( Q \) minus the work done by the system \( W \).

Step 2: Applying the First Law of Thermodynamics

We use the formula: \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, \( Q \) is the heat added to the system, and \( W \) is the work done by the system. We have \( \Delta U = -20 \, \mathrm{J} \) (since thermal energy decreases) and \( W = 10 \, \mathrm{J} \).

Step 3: Rearranging the Formula

Solve the formula for \( Q \) to find the heat added to the system. This gives us \( Q = \Delta U + W \).

Step 4: Substituting Values

Substitute the values for \( \Delta U \) and \( W \) into the equation to get \( Q = -20 \, \mathrm{J} + 10 \, \mathrm{J} \).

Step 5: Calculating the Result

Calculate \( Q = -20 \, \mathrm{J} + 10 \, \mathrm{J} = -10 \, \mathrm{J} \). This means that instead of adding heat, \( 10 \, \mathrm{J} \) of heat was removed from the system.

Key Concepts

Internal Energy

Internal energy is a fascinating concept in thermodynamics. It represents the total energy contained within a system as a result of microscopic factors. This includes:

• The kinetic energy of particles moving randomly inside the system.
• The potential energy due to interactions among particles such as electromagnetic forces.

Internal energy is a state function, meaning it solely depends on the current state of the system, and not how the system reached that state. Therefore, it doesn’t matter what transformations the system underwent in the past.
The change in internal energy (\( \Delta U \)) is affected by heat added to or removed from the system and the work done by or on the system.

Thermal Energy

Thermal energy is a specific component of the internal energy that arises from the temperature of a system. It is all about how energetic the individual particles are due to their constant movement and collisions. Think of it as the energy you feel as warmth.
When a system loses thermal energy, it usually means its temperature drops, as there is less vibrational, translational, or rotational energy among its particles.
In our original problem, the system's thermal energy decreased by \(20 \mathrm{~J} \), indicating it released energy, likely cooling down in the process.

Heat Transfer

Heat transfer is the process of thermal energy moving from one place to another. In thermodynamics, heat (\( Q \)) is transferred between systems due to a temperature difference. This can happen in several ways:

• Conduction: Direct contact transfers heat.
• Convection: Heat moves with a fluid like air or water.
• Radiation: Transfer through electromagnetic waves without needing a medium.

In the exercise, it was asked how much heat was added to the system. Crucially, heat is treated as a transfer of energy, not a substance, affecting the system's internal energy upon entering or leaving.

Work Done

In physics, the concept of 'work' has a very specific meaning. When a system performs work, it transfers energy by moving something against an opposing force. The amount of work done depends on:

• Force applied.
• Distance over which it is applied.
• Direction of the force relative to movement.

In the first law of thermodynamics context, work has a direct impact on internal energy. Any work done by the system (\( W \)) decreases its internal energy as energy is leaving the system. Meanwhile, work done on the system would increase its internal energy. In the given exercise, the system did \(10 \mathrm{~J} \) of work, thus lowering its internal energy.