Problem 25 - Chemistry: An Atoms-Focused Approach

Question

Calculate \(\Delta E\) when a. \(q=100.0 \mathrm{J} ; w=-50.0 \mathrm{J}\) b. \(q=6.2 \mathrm{kJ} ; w=0.70 \mathrm{L} \cdot\) atm c. \(q=-615 \mathrm{kJ} ; w=-3.25\) kilowatt-hours \((1 \mathrm{kWh}=\) \(3600 \mathrm{kJ})\)

Step-by-Step Solution

Answer

Question: Calculate the change in internal energy of the system for the following cases: a. Heat added to the system is 100.0 J, and the system does 50.0 J of work on the surroundings. b. Heat added to the system is 6.2 kJ, and the system does 0.70 L⋅atm of work on the surroundings. c. The system loses 615 kJ of heat to the surroundings, and the surroundings do 3.25 kWh of work on the system. Provide the change in internal energy for each case.

Step 1: Identify the given values

In each case, we have been given both the heat (\(q\)) and work (\(w\)) values. We will apply these in the internal energy equation to calculate \(\Delta E\). a. \(q = 100.0 \mathrm{J}\) and \(w = -50.0 \mathrm{J}\) b. \(q = 6.2 \mathrm{kJ}\) and \(w = 0.70 \mathrm{L} \cdot\) atm c. \(q = -615 \mathrm{kJ}\) and \(w = -3.25\) kilowatt-hours

Step 2: Convert units if required

We need to convert the units for part (b) and part (c) to ensure consistency in calculations. For part (b), \(1 \mathrm{L} \cdot\) atm \(= 101.33 \mathrm{J}\), so \(w = 0.70 \mathrm{L} \cdot\) atm \(\times 101.33 \frac{\mathrm{J}}{\mathrm{L} \cdot \mathrm{atm}} = 70.93 \mathrm{J}\). Convert \(q\) to Joules: \(q = 6.2 \mathrm{kJ} \times 1000 \frac{\mathrm{J}}{\mathrm{kJ}} = 6200 \mathrm{J}\) For part (c), \(1 \mathrm{kWh} = 3600 \mathrm{kJ}\), so \(w = -3.25 \mathrm{kWh} \times 3600 \frac{\mathrm{kJ}}{\mathrm{kWh}} = -11700 \mathrm{kJ}\). Keep \(q\) in kilojoules: \(q = -615 \mathrm{kJ}\)

Step 3: Calculate the change in internal energy

Now we have the \(q\) and \(w\) values in consistent units for all three cases. We will apply the formula \(\Delta E = q + w\) for each case. a. \(\Delta E = 100.0 \mathrm{J} + (-50.0 \mathrm{J}) = 50.0 \mathrm{J}\) b. \(\Delta E = 6200 \mathrm{J} + 70.93 \mathrm{J} = 6270.93 \mathrm{J}\) c. \(\Delta E = -615 \mathrm{kJ} + (-11700 \mathrm{kJ}) = -12315 \mathrm{kJ}\) So the answers are: a. \(\Delta E = 50.0 \mathrm{J}\) b. \(\Delta E = 6270.93 \mathrm{J}\) c. \(\Delta E = -12315 \mathrm{kJ}\)

Key Concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationship between heat and other forms of energy, like mechanical work. It is guided by laws that govern energy exchanges and conversions that occur in a system. The concept of internal energy is a central part of thermodynamics. Internal energy, denoted as \( E \), represents the energy contained within a system due to the motion and interactions of particles.

Within thermodynamics, we often deal with changes in internal energy. These changes occur due to heat exchange (\( q \)) and work done (\( w \)). Mathematically, this is expressed as the first law of thermodynamics:

• \( \Delta E = q + w \)

This formula highlights the idea that the change in a system's internal energy is equal to the sum of heat added to the system and the work done on it. Understanding this formula is crucial because it helps calculate energy changes that occur in various physical processes.

Work and Heat

In thermodynamics, work and heat are two primary ways energy is transferred in and out of a system.

Work (\( w \)) is the energy transferred when a force is applied to an object to move it. It can be positive or negative, depending on the direction of transfer:

• Positive work is done on the system when energy enters it.
• Negative work means energy leaves the system when it does work on the surroundings.

Heat (\( q \)) is the energy transferred between a system and its surroundings due to a temperature difference.

• Absorbing heat into the system increases internal energy, represented as positive \( q \).
• Releasing heat decreases internal energy, shown as negative \( q \).

Together, work and heat determine how a system's energy changes. The careful conversion between different units ensures accurate calculations, especially when different measurement units are involved, such as converting kJ to J for consistency.

Energy Conversion

Energy conversion is about transforming one form of energy into another. This is a fundamental concept in physics, especially when analyzing systems thermodynamically. Within the context of thermodynamics, we see that changes in internal energy involve conversions between

• Mechanical work
• Heat energy

For instance, when gas expands within a piston, chemical potential energy converts to mechanical energy, doing work on the piston and potentially producing heat.

It's critical to correctly convert energy units to ensure values are accurate when comparing different energy forms. For example,

• 1 kWh is equivalent to 3600 kJ,
• 1 L·atm equals 101.33 J.

Correct conversion enables precise calculations when applying the first law of thermodynamics, especially in exercises that handle different units of measurement. Understanding these conversions helps better analyze systems and predict the outcome of energy changes.