Question

A battery harnesses a chemical reaction to extract energy in the form of useful electrical work.

(a) A certain battery runs a toy truck and becomes partially discharged. In the process, it performs a total of 117.0 J of work on its immediate surroundings. It also gives off 3.0 J of heat, which the surroundings absorb. No other work or heat is exchanged with the surroundings. Compute q, w, and $\mathbf{\Delta U}$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 210.0 J of electrical work from an outside generator. Determine qfor the battery in this process. Explain why qhas the sign that it does.

Short answer

a) The values of$∆U$ , w and q are -220.0 J, -117.0 J and -3.0 J respectively.

(b) The heat gained by the system has a positive sign. Hence, the value of q is 10.0 J.

Step-by-step solution

Heat, work and internal energy

Internal energy constitutes the energy connected with the molecules of a particular system. It comprises potential energy and kinetic energy. The system undergoes energy transfer and conversions when it proceeds via a change because of factors like heat, work and internal energy.

Computing the value of q, w and ΔU

(a) As per the first law of thermodynamics, the mathematical expression is:

$∆U=q+w$

Where,$∆U$is the internal energy, q is the heat gained or lost by the system and w is the work done on or by the system.

Here battery is the system. It does not work to function the toy truck. Hence, work (w) is done by the system. Therefore the sign must be negative.

Battery gives out heat to the surrounding. Hence, the sign of q is negative.

Substituting the value of -3 J for q and -117.0 J for w to calculate$∆U$of the battery:

$$\begin{gathered} \text{∆U =}\left(\text{- 117.0 J}\right)\text{+}\left(\text{- 3.0 J}\right) \\ \text{= - 220 J} \end{gathered}$$

In this process, the internal energy of the system decreases by 220 J.

Hence, the values of $∆U$, w and q are -220.0 J, -117.0 J and -3.0 J respectively.

(b)Now the battery is recharged into the original condition. As the system has lost 220 J of energy, it will have gain 220 J of energy to reach the original condition.

As the work is done on the battery by an external generator, it will have a positive sign.

The mathematical expression for the first law of thermodynamics can be used to solve for q,

$$\begin{gathered} \text{∆U = q + w} \\ \text{q =}\text{U∆- w} \end{gathered}$$

Substitute +220.0 J for$∆U$ and +210.0 J for w to evaluate the value of q.

$$\begin{gathered} \text{q =}\left(\text{220.0 J}\right)\text{-}\left(\text{210.0 J}\right) \\ \text{= 10.0 J} \end{gathered}$$

The heat gained by the system has a positive sign. Hence, the value of q is 10.0 J.