Question

A battery has specific capacity \(252 \frac{\mathrm{C}}{\mathrm{g}}\) and mass of \(50 \mathrm{~g}\). Its overall density is \(2.245 \frac{\mathrm{g}}{\mathrm{m}^{3}}\). (a) Find the specific capacity in \(\frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{g}}\). (b) Find the capacity in mA.h. (c) Find the charge density in \(\frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{m}^{3}}\).

Short answer

(a) 70 mA.h/g; (b) 3500 mA.h; (c) 157.15 mA.h/m^3.

Step-by-step solution

Identify the Given Specific Capacity

The specific capacity of the battery is given as \(252 \frac{\mathrm{C}}{\mathrm{g}}\). The task is to convert this to \(\frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{g}}\).

Convert Coulombs to Milliampere-Hours

Use the conversion factor: 1 Coulomb (C) = 1 Ampere-Second (A.s) and 1 mA.h = 3.6 Coulombs. Therefore, \(\frac{252 \text{ C}}{\text{g}} \times \frac{1 \text{ A.s}}{1 \text{ C}} \times \frac{1 \text{ mA.h}}{3.6 \text{ C}} = \frac{70 \text{ mA.h}}{\text{g}}\).

Calculate the Battery Capacity

The mass of the battery is given as \(50 \text{ g}\). To find the overall capacity of the battery, multiply the specific capacity in \(\frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{g}}\) by the mass: \(70 \frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{g}} \times 50 \text{ g} = 3500 \text{ mA.h}\).

Calculate the Charge Density

The overall density of the battery is \(2.245 \frac{\mathrm{g}}{\mathrm{m}^3}\). To find the charge density in terms of \(\frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{m}^3}\), multiply the specific capacity converted in Step 2 by the density:\(70 \frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{g}} \times 2.245 \frac{\mathrm{g}}{\mathrm{m}^3} = 157.15 \frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{m}^3}\).

Key concepts

Understanding Specific Capacity

Specific capacity is an important metric in evaluating a battery's performance. It tells us how much electric charge a battery can store per unit mass.
This value is usually expressed in terms of Coulombs per gram (\(\frac{\mathrm{C}}{\mathrm{g}}\)) or in milliampere-hours per gram (\(\frac{\mathrm{mA}\cdot\mathrm{h}}{\mathrm{g}}\)).

To convert the specific capacity from \(\frac{\mathrm{C}}{\mathrm{g}}\) to \(\frac{\mathrm{mA}\cdot\mathrm{h}}{\mathrm{g}}\), we must use the relationship between Coulombs and milliampere-hours.
Remember, 1 Coulomb is equal to 1 Ampere-Second (A⋅s). Additionally, 1 milliampere-hour (mA⋅h) equals 3.6 Coulombs.

In our original problem, a specific capacity of \(252 \frac{\mathrm{C}}{\mathrm{g}}\) was given.
Using the relevant conversion factor, it was converted to \(70 \frac{\mathrm{mA}\cdot\mathrm{h}}{\mathrm{g}}\).
This conversion allows us to easily calculate and compare battery capacities in familiar units.

Exploring Charge Density

Charge density helps us understand how much charge is held in a specific volume of battery material.
It combines the specific capacity (charge per mass) with the overall material density.
For this reason, charge density is typically given in units of \(\frac{\mathrm{mA}\cdot \mathrm{h}}{\mathrm{m}^3}\).

To find the charge density, we multiply the specific capacity converted to \(\frac{\mathrm{mA}\cdot\mathrm{h}}{\mathrm{g}}\) by the battery's overall density.

• In our case, the specific capacity conversion yielded \(70 \frac{\mathrm{mA}\cdot\mathrm{h}}{\mathrm{g}}\).
• The battery's overall density was given as \(2.245 \frac{\mathrm{g}}{\mathrm{m}^3}\).

Multiply these two values to get \(157.15 \frac{\mathrm{mA}\cdot \mathrm{h}}{\mathrm{m}^3}\).
This reveals how effectively the battery material stores charge within a given volume.

Role of Conversion Factors

Conversion factors are essential for transforming quantities from one unit to another in scientific calculations. They make it easier to work with various units comfortably.
When dealing with electric charge and capacity in batteries, using the right conversion factors allows for a seamless transition between different measurements, such as from Coulombs to milliampere-hours.

In our example, we used the following key conversion factors:

• 1 Coulomb \(= 1 \text{ A ⋅ s}\)
• 1 milliampere-hour \(= 3.6 \text{ C}\)

Using these, we converted specific capacity from \(\frac{\mathrm{C}}{\mathrm{g}}\) to \(\frac{\mathrm{mA} \cdot \mathrm{h}}{\mathrm{g}}\), enabling further calculations for charge density.
Without these factors, accurately comparing and calculating dimensions would be challenging.
Understanding and applying conversion factors are vital skills in the study of batteries and electrical engineering.