Work Done Calculation: Internal Energy Change Of 450J

Hey guys! Let's dive into a fascinating problem concerning thermodynamics, specifically dealing with internal energy, heat, and work done in a system. This is a core concept in physics and chemistry, and understanding it thoroughly will definitely help you ace your exams and grasp the real-world applications of these principles. We’re going to break down a scenario where the change in internal energy is 450 J, and the surroundings absorb 90 J of heat. Our main goal is to figure out the work done during this process and whether the work is done on the system or by the system. So, buckle up, and let's get started!

Understanding the First Law of Thermodynamics

To tackle this problem effectively, we need to recall the first law of thermodynamics. Think of this law as the energy conservation principle applied to thermodynamic systems. In simple terms, it states that energy cannot be created or destroyed, but it can be transferred or converted from one form to another. Mathematically, the first law of thermodynamics is expressed as:

ΔU = Q - W

Where:

• ΔU represents the change in internal energy of the system.
• Q is the heat added to the system.
• W is the work done by the system.

It’s super important to pay attention to the sign conventions here. A positive ΔU means the internal energy of the system has increased, while a negative ΔU means it has decreased. A positive Q means heat is added to the system, and a negative Q means heat is removed from the system (or in our case, absorbed by the surroundings). A positive W means work is done by the system on its surroundings, like a gas expanding and pushing a piston, and a negative W means work is done on the system by its surroundings, like compressing a gas.

Applying the First Law to Our Problem

Okay, so let’s apply this to our specific scenario. We’re given that the change in internal energy (ΔU) is 450 J. This means the system's internal energy has increased. We’re also told that the surroundings absorb 90 J of heat. This is where it gets a little tricky, so pay close attention! Since the surroundings are absorbing heat, the system is losing heat. Therefore, Q is -90 J. Remember, heat absorbed by the surroundings is equivalent to heat lost by the system.

Now we can plug these values into our equation:

450 J = -90 J - W

Our next step is to solve for W, which represents the work done. To do this, we’ll rearrange the equation:

W = -90 J - 450 J

W = -540 J

So, what does this -540 J tell us? The negative sign is crucial! It tells us that the work done is negative, which means work is done on the system, not by the system. The magnitude, 540 J, tells us the amount of work done. Therefore, 540 J of work is done on the system.

Interpreting the Results: Work Done on the System

Let’s break down what this result means in a more intuitive way. Imagine you have a container filled with gas. The internal energy of the gas (ΔU) has increased by 450 J. This could be due to the gas molecules moving faster or experiencing more vigorous interactions. Now, the surroundings absorb 90 J of heat. This means the gas lost 90 J of energy as heat transferred to the surroundings.

But how did the internal energy increase by 450 J if the system lost 90 J of heat? The answer is that work was done on the system. Think of it like this: someone or something is compressing the gas, pushing the walls of the container inward. This compression transfers energy to the gas, increasing its internal energy. The 540 J of work done on the system is more than enough to compensate for the 90 J of heat lost to the surroundings, resulting in a net increase of 450 J in internal energy.

Contrasting Work Done By the System

To further clarify, let's briefly contrast this with a situation where work is done by the system. If the gas were expanding against an external pressure (like pushing a piston outward), the gas would be doing work on its surroundings. In this case, W would be positive. The gas would be using some of its internal energy to perform this work, and if no heat were added, the internal energy would decrease.

Significance in Real-World Applications

Understanding these thermodynamic principles isn't just about solving textbook problems; it's about understanding how the world around us works. These concepts are vital in various fields, including:

• Engineering: Designing efficient engines, refrigerators, and other thermodynamic devices requires a solid grasp of these laws.
• Chemistry: Chemical reactions often involve changes in energy, and understanding thermodynamics helps predict whether a reaction will occur spontaneously.
• Meteorology: Weather patterns and climate change are governed by thermodynamic processes in the atmosphere.
• Biology: Living organisms rely on thermodynamic processes for energy transfer and metabolism.

Examples in Daily Life

Think about a car engine, for example. Fuel is burned, which releases heat (Q). This heat increases the internal energy (ΔU) of the gases in the cylinders. These gases then expand, pushing the pistons and doing work (W) to turn the wheels. Similarly, a refrigerator uses work to transfer heat from the inside (cooling it down) to the outside (warming it up).

Common Mistakes and How to Avoid Them

Thermodynamics can be tricky, and it’s easy to make mistakes if you're not careful. Here are some common pitfalls and how to avoid them:

1. Sign Conventions: As we highlighted earlier, getting the signs right for ΔU, Q, and W is absolutely crucial. Always remember: Positive ΔU means internal energy increases, positive Q means heat is added to the system, and positive W means work is done by the system.
2. Confusing Heat and Temperature: Heat (Q) is the transfer of energy, while temperature is a measure of the average kinetic energy of the molecules. They are related but not the same. You can add heat to a system without changing its temperature if the work is also done, or if the substance is undergoing a phase change (like melting or boiling).
3. Assuming Processes are Isothermal: An isothermal process is one where the temperature remains constant. Many problems don't specify this, so you can't automatically assume it. You need to consider all the variables—internal energy, heat, and work—to get the correct answer.
4. Forgetting Units: Always include units in your calculations and final answers. This helps you catch errors and ensures your answer is meaningful.

Tips for Mastering Thermodynamics

Here are some extra tips to help you truly master thermodynamics:

• Practice, practice, practice: Solve as many problems as you can. This will help you become comfortable with the concepts and develop your problem-solving skills.
• Draw diagrams: Visualizing the system can make it easier to understand what's happening.
• Explain concepts to others: Teaching someone else is a great way to solidify your own understanding.
• Use real-world examples: Connecting thermodynamic principles to everyday phenomena will make them more memorable and relevant.

Conclusion: Mastering Work Done Calculations

So, guys, in this article, we’ve dissected a problem involving the change in internal energy, heat transfer, and work done in a system. We found that when the internal energy increases by 450 J and the surroundings absorb 90 J of heat, 540 J of work is done on the system. This example highlights the crucial role of the first law of thermodynamics in understanding energy transformations.

Remember, mastering these concepts is not just about passing exams. It's about gaining a deeper understanding of how the world around us works. Keep practicing, stay curious, and you’ll be a thermodynamics pro in no time! If you have any questions, drop them in the comments below. Keep learning and keep exploring!