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The capacity of batteries such as a lithium-ion battery is expressed in units of milliamp-hours (mAh). A typical battery of this type yields a nominal capacity of \(2000 \mathrm{mAh}\). (a) What quantity of interest to the consumer is being expressed by the units of \(\mathrm{mAh}\) ? (b) The starting voltage of a fresh lithium-ion battery is \(3.60 \mathrm{~V}\). The voltage decreases during discharge and is \(3.20 \mathrm{~V}\) when the battery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, estimate the total maximum electrical work the battery could perform during discharge.

Short Answer

Expert verified
(a) The unit of milliamp-hours (mAh) represents the capacity of a battery, expressing the amount of current (in milliamperes or mA) that a battery can deliver for one hour before it gets discharged. The higher the mAh value, the longer the battery life. (b) The total maximum electrical work the lithium-ion battery can perform during discharge is estimated to be 0.8 Joules.

Step by step solution

01

a) Meaning of milliamp-hours (mAh)

The unit of milliamp-hours (mAh) is used to express the capacity of a battery. It represents the amount of current (in milliamperes or mA) that a battery can deliver for one hour before it gets discharged. The higher the mAh value, the longer the batteriy life.
02

b) Estimate the total maximum electrical work.

To find the total maximum electrical work the battery can perform, we can use the formula: Electrical Work = Charge × Voltage Difference We are given the nominal capacity of the battery (2000 mAh), and the starting and ending voltages (3.60 V and 3.20 V, respectively). We first need to find the total charge and the voltage difference. Step 1: Convert the capacity (2000 mAh) to charge (in Coulombs). Since the current is given in milliamperes (mAh), we need to convert it to amperes (Ah) first, and then multiply it by the time (1 hour) to get the charge. 1 Ah = 1000 mAh Charge (in Ah) = Capacity (in Ah) × Time (in hours) Step 2: Calculate the voltage difference. Voltage Difference = Starting Voltage - Ending Voltage Step 3: Calculate the electrical work. Electrical work = Charge × Voltage Difference. Now, we'll plug in the values and solve the equations.
03

Convert the capacity to charge.

First, let's convert the capacity (2000 mAh) to Ah: Capacity = 2000 mAh ÷ 1000 = 2 Ah Now, we'll find the charge using the given time (1 hour): Charge = Capacity × Time = 2 Ah × 1 hour = 2 Coulombs
04

Calculate the voltage difference.

We'll calculate the voltage difference using the starting and ending voltages. Voltage Difference = Starting Voltage - Ending Voltage = 3.60 V - 3.20 V = 0.40 V
05

Calculate the electrical work.

Now, we have the charge (2 Coulombs) and the voltage difference (0.40 V). We'll use the formula to calculate the electrical work: Electrical work = Charge × Voltage Difference = 2 Coulombs × 0.40 V = 0.8 Joules The total maximum electrical work the lithium-ion battery can perform during discharge is estimated to be 0.8 Joules.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Milliamp-hours (mAh)
Milliamp-hours, often abbreviated as mAh, is a vital unit that describes the capacity of a battery. Simply put, it measures how long a battery can continuously provide a specific current before it is completely depleted. For instance, a battery rated at 2000 mAh can deliver a current of 2000 milliamperes for one hour, 1000 milliamperes for two hours, and so on.
It's a crucial indicator when choosing batteries for devices since it helps gauge how long your device can operate on a single charge.
  • Higher mAh ratings mean longer usage times.
  • Helps in comparing battery lifespan across different devices.
Understanding mAh is particularly important when selecting batteries for energy-demanding gadgets like smartphones, where battery life significantly impacts user experience.
Lithium-ion Battery
Lithium-ion batteries are a popular choice for many portable electronics due to their high energy density and lightweight design. These batteries can store a substantial amount of energy in a compact size, making them ideal for today's tech-savvy world.
Unlike other types of batteries, lithium-ion batteries have some distinct characteristics:
  • They have a slow loss of charge when not in use, ensuring longer standby times.
  • Minimal memory effect, which means they don’t require complete discharging before recharging.
  • Have a nominal cell voltage around 3.6-3.7 volts, providing consistent power.
The starting voltage of a lithium-ion battery may begin at 3.60 V and can decrease to around 3.20 V as the battery discharges. Monitoring this voltage drop is essential as it relates directly to the battery's health and ability to deliver its rated capacity.
Electrical Work Calculation
The concept of electrical work involves determining how much energy a battery can provide. Using the formula:\[\text{Electrical Work} = \text{Charge} \times \text{Voltage Difference}\]allows us to measure the total work done by a battery in terms of Joules.
To estimate this electrical work:
  • First, convert the battery capacity from mAh to Ampere-hours and then to Coulombs (since 1 Coulomb = 1 Ampere-hour).
  • Next, identify the voltage difference during discharge. It's the starting voltage minus the ending voltage.
  • Finally, multiply the total charge by the voltage difference to get the electrical work.
For example, using a capacity of 2000 mAh and a voltage difference from 3.60 V to 3.20 V, we calculated an electrical work of 0.8 Joules. This calculation is crucial for determining how efficiently a battery-powered device can perform its operations over time.

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Most popular questions from this chapter

A voltaic cell similar to that shown in Figure 20.5 is constructed. One electrode half-cell consists of a magnesium strip placed in a solution of \(\mathrm{MgCl}_{2}\), and the other has a nickel strip placed in a solution of \(\mathrm{NiCl}_{2}\). The overall cell reaction is $$ \mathrm{Mg}(s)+\mathrm{Ni}^{2+}(a q) \longrightarrow \mathrm{Ni}(s)+\mathrm{Mg}^{2+}(a q) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half- reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode?(d) Indicate the signs of the electrodes. (e) Do electrons flow from the magnesium electrode to the nickel electrode or from the nickel to the magnesium? (f) In which directions do the cations and anions migrate through the solution?

A voltaic cell consists of a strip of cadmium metal in a solution of \(\mathrm{Cd}\left(\mathrm{NO}_{3}\right)_{2}\) in one beaker, and in the other beaker a platinum electrode is immersed in a NaCl solution, with \(\mathrm{Cl}_{2}\) gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions?

A voltaic cell similar to that shown in Figure 20.5 is constructed. One half- cell consists of an iron strip placed in a solution of \(\mathrm{FeSO}_{4}\), and the other has an aluminum strip placed in a solution of \(\mathrm{Al}_{2}\left(\mathrm{SO}_{4}\right)_{3} .\) The overall cell reaction is $$ 2 \mathrm{Al}(s)+3 \mathrm{Fe}^{2+}(a q) \longrightarrow 3 \mathrm{Fe}(s)+2 \mathrm{Al}^{3+}(a q) $$ (a) What is being oxidized, and what is being reduced? (b) Write the half- reactions that occur in the two half-cells. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. (e) Do electrons flow from the aluminum electrode to the iron electrode or from the iron to the aluminum? (f) In which directions do the cations and anions migrate through the solution? Assume the \(\mathrm{Al}\) is not coated with its oxide.

(a) What conditions must be met for a reduction potential to be a standard reduction potential? (b) What is the standard reduction potential of a standard hydrogen electrode? (c) Why is it impossible to measure the standard reduction potential of a single half-reaction?

Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at \(298 \mathrm{~K}\) : (a) \(\mathrm{Cu}(s)+2 \mathrm{Ag}^{+}(a q) \longrightarrow \mathrm{Cu}^{2+}(a q)+2 \mathrm{Ag}(s)\) (b) \(3 \mathrm{Ce}^{4+}(a q)+\mathrm{Bi}(s)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 3 \mathrm{Ce}^{3+}(a q)+ \mathrm{BiO}^{+}(a q)+2 \mathrm{H}^{+}(a q)\) (c) \(\mathrm{N}_{2} \mathrm{H}_{5}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{3-}(a q) \longrightarrow \mathrm{N}_{2}(g)+ 5 \mathrm{H}^{+}(a q)+4 \mathrm{Fe}(\mathrm{CN})_{6}^{4-}(a q)\)

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