Mastering the Closed System Energy Balance

The Great Sign Convention Debate

One of the first things that can trip up students is that the First Law equation sometimes appears differently in different textbooks. You might see $\Delta U=Q+W$ in a chemistry book, but $\Delta U=Q-W$ in an engineering or physics book. Why the difference? It all comes down to perspective.

To understand this, let's watch a short video that explains the two viewpoints.

As the video explained:

• Chemistry (ΔU = Q + W): Focuses entirely on the system. Any energy that crosses the boundary into the system, whether it's heat or work, is considered positive.
• Engineering (ΔU = Q - W): Takes a more practical, "device-oriented" view. We think about what we want to get out of a system (like work from an engine) versus what we have to put in (like heat from fuel).

From now on, we will exclusively use the engineering convention:
$\Delta U=Q-W$

Let's cement this with a simple analogy. Think of your thermodynamic system as a small business.

• Internal Energy ($U$) is your company's bank account balance. $\Delta U$ is the change in that balance.
• Heat ($Q$) is revenue you get from a client. When heat flows into the system, it's like money coming in. It's positive.
• Work ($W$) is a useful product or service you provide. When your business does work, you're producing something of value that goes out to the world. Work done by the system is considered a positive output.

So, the change in your bank account ($\Delta U$) is your revenue ($Q$) minus your valuable output ($W$).

The Official Sign Convention for Engineers

Let's formally define the signs for each term in $\Delta U=Q-W$. This is our rulebook for energy accounting.

Let's break it down:

1. Heat Transfer (Q)

• $Q>0$: Heat is transferred from the surroundings TO the system.
  • Analogy: Putting a pot of water on a hot stove. The water (system) gains heat.
  • This is an endothermic process.
• $Q<0$: Heat is transferred from the system TO the surroundings.
  • Analogy: A hot cup of coffee cooling on your desk. The coffee (system) loses heat.
  • This is an exothermic process.

2. Work (W)

• $W>0$: Work is done BY the system ON the surroundings.
  • Analogy: The hot gas in an engine cylinder expands and pushes a piston up. The gas (system) is doing work. This is usually the desired output of a heat engine.
• $W<0$: Work is done ON the system BY the surroundings.
  • Analogy: You use a pump to compress air into a tire. You (part of the surroundings) are doing work on the air (the system).

3. Internal Energy Change ($\Delta U$)
This one is straightforward. It's the result of the heat and work interactions.

• $\Delta U>0$: The internal energy of the system increased.
• $\Delta U<0$: The internal energy of the system decreased.

To see this all laid out, let's refer to a clear summary.

Recap Break

This is the core of today's lesson, so let's pause and lock it in.

• The Law: $\Delta U=Q-W$
• The Logic: Change in Stored Energy = Energy In (as Heat) - Energy Out (as Work)
• The Rules:
  • Heat IN $\to$ $Q$ is positive
  • Work OUT $\to$ $W$ is positive

Now, let's use this to solve some problems.

Applying the First Law: Worked Examples

The best way to get comfortable with the signs is to see them in action. We'll use the same resource as before, which has some excellent, clear examples.

Let's walk through our own examples step-by-step, just like you'd do on an exam.

Worked Problem 1: Basic Application

A closed, rigid tank contains a fluid. The fluid is stirred by a paddle wheel, which does 100 kJ of work ON the fluid. During this process, the fluid loses 25 kJ of heat to the surroundings. What is the change in the internal energy of the fluid?

1. Identify the System: The system is the fluid inside the tank.
2. Analyze Heat (Q): The fluid loses 25 kJ of heat. Heat is leaving the system.
   • $Q=-25\text{ kJ}$
3. Analyze Work (W): Work is done on the fluid. Work is entering the system.
   • $W=-100\text{ kJ}$
4. Apply the First Law:
   $\Delta U=Q-W$
5. Substitute Values:
   $\Delta U=(-25\text{ kJ})-(-100\text{ kJ})$
   $\Delta U=-25\text{ kJ}+100\text{ kJ}=+75\text{ kJ}$

Answer: The internal energy of the fluid increased by 75 kJ. Even though it lost some heat, the large amount of work done on it caused its overall energy to go up.

Worked Problem 2: Solving for Heat

A gas in a piston-cylinder assembly undergoes a process where its internal energy decreases by 200 kJ. During the process, the gas is compressed, and 50 kJ of work is done ON the system. How much heat was transferred, and in which direction?

1. Identify the System: The system is the gas in the piston-cylinder.
2. Analyze Internal Energy ($\Delta U$): The internal energy decreases by 200 kJ.
   • $\Delta U=-200\text{ kJ}$
3. Analyze Work (W): Work is done on the system. Work is entering.
   • $W=-50\text{ kJ}$
4. Apply the First Law:
   $\Delta U=Q-W$
5. Rearrange and Solve for Q:
   $Q=\Delta U+W$
   $Q=(-200\text{ kJ})+(-50\text{ kJ})=-250\text{ kJ}$

Answer: $Q$ is negative, which means 250 kJ of heat was transferred FROM the gas TO the surroundings. This makes sense: the gas was compressed (which tends to increase energy) but its internal energy dropped significantly, so it must have lost a lot of heat.

Time to Practice!

Now it's your turn. Use the same systematic approach: identify the system, determine the signs of the known quantities, and apply the First Law.

Conceptual Questions

1. Is it possible for the internal energy of a closed system to remain constant during a process ($\Delta U=0$) even if heat is added to it ($Q>0$)?

   Show answer

   Yes, absolutely. According to the First Law, $\Delta U = Q - W$. If $\Delta U = 0$, then $Q - W = 0$, which means $Q = W$. This describes a process where all the heat added to the system is immediately converted into work done by the system. For example, if you slowly add heat to a gas in a piston, and the gas expands to do work, it's possible for its temperature (and thus internal energy) to remain constant.

2. You have a perfectly insulated thermos containing hot coffee. You shake it vigorously. What happens to the internal energy of the coffee? What are the signs of $Q$, $W$, and $\Delta U$?

   Show answer

   * **Heat (Q):** The thermos is "perfectly insulated," which means it's adiabatic. No heat can be transferred. So, $Q = 0$. * **Work (W):** By shaking it, you are doing work *on* the coffee (this is a form of shaft work, stirring the fluid). Therefore, $W$ is negative. * **Internal Energy ($\Delta U$):** Using the First Law, $\Delta U = Q - W = 0 - W$. Since $W$ is negative, $\Delta U = -(-|W|) = +|W|$. Therefore, $\Delta U$ is positive. The internal energy (and temperature) of the coffee will increase slightly.

3. An electric heater is used to heat a room. Let's define the room (the air, furniture, etc.) as our system. Is this a heat or work interaction?

   Show answer

   This is a classic trick question! From an engineering thermodynamics perspective, it's a **work interaction**. The electricity crosses the system boundary and does work on the resistor inside the heater. The resistor then heats up and transfers heat to the air, but that heat transfer is *internal* to the system (room). The energy entered the system as electrical work. So, for the room system, $W$ is negative (work done on the system), and $Q$ is zero (assuming the room is well-insulated from the outside). The result is $\Delta U = Q - W = 0 - (-W_{elec}) = +W_{elec}$, and the room's internal energy increases.

Numerical Problems

1. Problem: During an expansion process, a closed system does 150 kJ of work. Its internal energy is found to have decreased by 50 kJ. Calculate the heat transfer and state its direction.

   Show step-by-step answer

   1. **System:** The closed system. 2. **Work (W):** The system *does* work, so it's an output. $W = +150 \text{ kJ}$. 3. **Internal Energy ($\Delta U$):** The energy *decreased*. $\Delta U = -50 \text{ kJ}$. 4. **First Law:** $\Delta U = Q - W \implies Q = \Delta U + W$. 5. **Solve:** $Q = (-50 \text{ kJ}) + (150 \text{ kJ}) = +100 \text{ kJ}$. **Answer:** The heat transfer is 100 kJ. Since the sign is positive, the heat was transferred **into** the system.

2. Problem: 2 kg of a substance is contained in a rigid, sealed vessel. It is heated by adding 500 kJ of heat. A paddle wheel inside the vessel is turned by an external motor, adding 100 kJ of work. What is the change in internal energy of the substance in kJ? What is the change in specific internal energy (energy per mass) in kJ/kg?

   Show step-by-step answer

   1. **System:** The 2 kg of substance in the vessel. 2. **Heat (Q):** Heat is *added*. $Q = +500 \text{ kJ}$. 3. **Work (W):** The vessel is rigid, so there is no expansion/compression work. However, the paddle wheel adds energy. This is work done *on* the system. $W = -100 \text{ kJ}$. 4. **First Law:** $\Delta U = Q - W$. 5. **Solve for $\Delta U$:** $\Delta U = (+500 \text{ kJ}) - (-100 \text{ kJ}) = 600 \text{ kJ}$. **Answer 1:** The change in internal energy is **600 kJ**.

   To find the specific internal energy change ($\Delta u$, lowercase), we divide by the mass.
   $\Delta u=\frac{\Delta U}{m}=\frac{600\text{ kJ}}{2\text{ kg}}=300\text{ kJ/kg}$
   Answer 2: The change in specific internal energy is 300 kJ/kg.

Conclusion

Excellent work today! We've established the universal rules for energy accounting in engineering. This lesson was all about building a solid, reliable method for applying the First Law.

Key Takeaways:

• We use the engineering convention for the First Law: $\Delta U=Q-W$.
• This equation is an energy balance: Change in Stored Energy = Heat In - Work Out.
• Sign Convention is King:
  • Heat INTO the system is positive Q.
  • Work done BY the system is positive W.
  • Energy flowing out of the system (heat loss or work input) is negative.
• Solving problems is a systematic process: identify the system, assign signs to the energy transfers, and then apply the equation.

Preview of the Next Lesson

We now have a robust way to balance the energy books for a closed system. But we're still missing a key piece of the puzzle. If $\Delta U=75\text{ kJ}$, what does that mean for the system? How much did its temperature change?

In our next lesson, we will build the bridge between the abstract concept of internal energy and the tangible, measurable property of temperature. We will introduce specific heats ($c_v$ and $c_p$), which are the crucial properties that tell us how much the energy of a substance changes when we heat it up. This will unlock our ability to solve a whole new class of powerful thermodynamic problems.