Thermochemistry - work, heat, total energy

[CheMech]

Homework Statement

(a) A certain battery runs a toy truck and becomes partially discharged. In the process, it performs a total of 117J of work on its immediate surroundings. It also gives off 3.0J of heat, which the surroundings absorb No other work or heat is exchanged with the surroundings. Compute q, w, and ∆U of the battery, making sure each quantity has the proper sign.

(b) The same battery is now recharged exactly to its original condition. This requires 210J of electrical work from an outside generator. Determine q for the battery in this process. Explain why q has the sign that it does.

Homework Equations

∆U=q+w

The Attempt at a Solution

The main problem I'm having is determining what sign each quantity is. I know that q is negative if the system loses heat, and positive if heat flows into it. ∆U seems pretty self explanatory: if the system gains total energy, then ∆U is positive. If it loses total energy, then ∆U is negative.

I'm mostly confused with determining the sign of work. In part (a), the problem states that the battery performs 117J of work on its immediate surroundings. I can think of two arguments that explain why work should be positive or negative.

I could see work as positive because the system does positive work on the environment. An analogy could be that it applies a force to the environment, and the environment is affected.

I could see it as negative though because: (1) The system loses energy by performing work (exothermic), hence the negative sign, and (2) As the system performs work on the surroundings, the surroundings perform work on the system. If the system does 117J of work, then -117J of work is done on the system. Like a free body diagram in physics, are we supposed to focus on the work done to the system to determine the correct sign, or are we supposed to focus on the work the system does on the environment?

Any help is greatly appreciated :)

 

[nate808]

Hi there! I can help you with this problem.

First, let's define the quantities we are dealing with. q represents heat, w represents work, and ∆U represents the change in internal energy. In this problem, we are dealing with a closed system, meaning that no matter or energy can enter or leave the system. Therefore, the first law of thermodynamics tells us that the change in internal energy (∆U) is equal to the heat added to the system (q) plus the work done on the system (w).

Now, let's look at part (a) of the problem. The battery performs 117J of work on its surroundings, meaning that it loses 117J of energy. This work is considered positive because it is done by the system on the surroundings. In other words, the system is doing work on the surroundings, so the work is positive. Additionally, the battery gives off 3.0J of heat, which is absorbed by the surroundings. This heat is considered negative because it is lost by the system. Therefore, q = -3.0J and w = 117J. Plugging these values into the first law of thermodynamics, we get:

∆U = q + w
∆U = -3.0J + 117J
∆U = 114J

Since the battery is partially discharged, it loses a total of 114J of energy, hence the negative sign.

Moving on to part (b) of the problem, we are now recharging the battery back to its original state. This process requires 210J of electrical work from an outside generator. Since the battery is gaining energy, the work done on the system is considered negative. Therefore, w = -210J. We can then use the first law of thermodynamics to find q:

∆U = q + w
∆U = 0 (since the battery is back to its original state)
q = -w (we can rearrange the equation)
q = -(-210J) (substituting w = -210J)
q = 210J

This means that the battery absorbs 210J of heat during the recharging process. This makes sense because the battery is gaining energy, so heat must be flowing into the system.

I hope this helps clarify the signs for each quantity. Remember, in thermodynamics, we focus on the perspective of the