A cell phone or computer battery has three ratings marked on...

A cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts. (a) What are these three values for your cell phone? (b) Convert the charge capacity $Q$ into coulombs. (c) Convert the energy capacity $U$ into joules. (d) Multiply the charge rating $Q$ by the potential rating $V,$ and verify that this is equivalent to the energy capacity $U$. (e) If the charge $Q$ were stored on a parallel-plate capacitor with air as the dielectric, at the potential $V$, what would be the corresponding capacitance? (f) If the energy in the battery were used to heat $1 \mathrm{~L}$ of water, estimate the corresponding change in the water temperature? (The heat capacity of water is $4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .$ )

A cell phone or computer battery has three ratings marked on it: a charge capacity listed in mAh (milliamp-hours), an energy capacity in Wh (watt-hours), and a potential rating in volts.
(a) What are these three values for your cell phone? (b) Convert the charge capacity $Q$ into
coulombs. (c) Convert the energy capacity $U$ into joules. (d) Multiply the charge rating $Q$ by the potential rating $V,$ and verify that this is equivalent to the energy capacity $U$. (e) If the charge $Q$ were stored on a parallel-plate capacitor with air as the dielectric, at the potential $V$, what would be the corresponding capacitance? (f) If the energy in the battery were used to heat $1 \mathrm{~L}$ of water, estimate the corresponding change in the water temperature? (The heat capacity of water is $4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K} .$ )

Key Concepts

Energy-Charge-Voltage Relationship

This relationship is a fundamental concept stating that the energy stored or delivered by a system is the product of its charge and the potential difference (voltage). Understanding this connection is essential for verifying and calculating the energy based on the battery's charge and voltage ratings.

Specific Heat Capacity and Temperature Change

Specific heat capacity describes the amount of energy required to raise the temperature of a unit mass of a substance by one degree. This concept is instrumental when calculating temperature changes resulting from energy transfers, such as using battery energy to heat a given mass of water.

Unit Conversions in Electrical Systems

Unit conversions in electrical systems involve translating between various units, such as converting milliamp-hours to coulombs and watt-hours to joules. These conversions ensure that calculations are consistent with the SI units, facilitating accurate analysis in physics and engineering.

Capacitance

Capacitance is a measure of a component’s ability to store electrical charge per unit voltage, defined as the ratio of stored charge to the voltage across it. It is an important concept in both designing capacitors and relating energy storage in different electrical systems.

Energy Capacity

Energy capacity indicates the total amount of energy stored in a battery, often measured in watt-hours. This value reflects the battery’s ability to perform work or supply power and can be converted into joules, the SI unit of energy.

Electric Potential (Voltage)

Voltage is the electrical potential difference between two points and is a measure of the energy per unit charge. It is crucial for understanding how much work can be done by the battery when a charge moves through an electrical circuit.

Charge Capacity

Charge capacity represents the total amount of electric charge a battery can store and deliver over time, typically expressed in milliamp-hours. It is a key parameter that indicates how long a battery can power a device before needing a recharge.

Transcript

0:00 Hi there.

00:01 So for this problem, we are investigating a cell phone.

00:06 So we need to assume some values for the charge capacity, the energy, and the potential rating.

00:13 So i'm going to say first, and that's the first part of this problem, the information for a cell phone or for your cell phone.

00:23 So i'm going to say that the charge capacity is equal to two thousand.

00:32 600 million per hour that the energy is equal to 9 .88 watts per hour and the potential rating is equal to 3 .8 volumes.

01:03 So those are the values that we're going to set for this exercise.

01:10 So for part b of this problem, we are asked to convert the charge capacity into columns.

01:20 So we have in here 2 ,600 milanpers per hour.

01:32 We know that the definition of current is charge.

01:45 Charge over seconds.

01:49 So in here, as you can see, corresponds to the charge because we have units of umpers with our units of the current and ours, which are units of seconds.

02:11 So we already have the charge, but we need to convert that into columns.

02:17 So what i'm going to do is to write this in empers first.

02:22 We know that we have milly, so that it will be this times 10 to the minus 3 umpers per hour.

02:31 So if we multiply this in here, we're going to 10 .6 empers per hour.

02:38 Now we change that hour to seconds because colombs is umpers per cent umpours times.

02:47 Seconds so we need to multiply dot 2 .6 and in one hour there is 3 ,600 seconds.

03:00 That's what we obtain to multiply 60 by 60 and this will give us that the that charge capacity in columns is 9 ,360 columns.

03:17 So this is the solution for this first part of this problem.

03:21 Now, for part c, we are asked to multiply the charge rating q by the potential rating b and verify that this is equivalent to the energy capacity u.

03:40 So for that part, you're going to take the energy u.

03:45 So what are you going to do is multiplied the value that we are given for the energy 9 .88 whaps per hour this times 9 .8 joules per second and this times the time it takes it is one hour so it's going to be three three thousand and the in 600 seconds.

04:29 So when we multiply this, it will give us 35 .6 kilojoules.

04:36 Kilo is 10 to the 3.

04:39 So that's what we obtained for part c of this problem...

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